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Book  No..!. 


APPLIED    MECHANICS 

A  Treatise  for  the  use  of  Students  who  have  time  to  work 

Experimental,    Numerical,    and  Graphical  Exercises 

illustrating  the  subject 


BY 


JOHN  PERRY,  M.E.,  D.Sc.,  LL.D.,  F.R.S. 

WHIT.  Sen.,  Assoc.  M.lNST.C.E. 

PROFESSOR     OF      MECHANICS      AND      MATHEMATICS     IN     THE     ROYAL    COLLEGE     OF 

SCIENCE,    SOUTH     KENSINGTON;     PAST     PRESIDENT    OF     THE    INSTITUTION 

OF   ELECTRICAL  ENGINEERS  ',    PRESIDENT  OF  THE   PHYSICAL   SOCIETY 


WITH  372  ILLUSTRATIONS 
NEW     EDITION      REVISED    AND     ENLARGED 


OF  THE 

UNIVERSITY 


NEW    YORK 
D.     VAN     NOSTRAND     COMPANY 

23  MURRAY,  AND   27  WARREN  STS 

1907 


T/ 
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n 


X»»^>V 

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\    UNIVERSITY   I 
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X£4L<FOR^£X 


PREFACE. 


Tins  book  describes  what  has  for  many  years  been  the  course 
of  instruction  in  Applied  Mechanics  at  the  Finsbury  Technical 
College.  All  mechanical  and  electrical  engineering  students 
in  their  first  year  had  two  lectures  a  week ;  the  substance  of 
these  lectures  is  here  printed  in  the  larger  type.  Mechanical 
engineers  had  three  lectures  a  week  in  their  second  year ;  the 
substance  of  these  lectures  is  here  printed  in  small  type.  As 
I  found  our  arrangement  of  hours  per  week  to  work  fairly 
well,  I  give  it  here  : — 


Mechanical  Engineers. 

Electrical  Engineers. 

1st  year. 

2nd  year. 

1st  year. 

2nd  year. 

Mathematics 

4 

4 

4 

4 

Graphics  and  Machine  Drawing 

7 

13 

4 

3 

Mechanics:  Lectures  ... 

2 

3 

2 

0 

Mechanical  Laboratory 

3 

2 

2 

0 

Mechanics:  Numerical  Exercises 

2 

1 

2 

0 

Mechanism 

1 

0 

0 

0 

Wood  and  Iron  Workshops    ... 

4 

7 

3 

6 

Chemists  and  building  trade  students  also  attended  in  the 
mechanical  department,  and  the  mechanical  engineering 
students  had  courses  of  study  in  the  physics  and  chemistry 
departments  of  the  College.  The  Mechanics  course  included 
work  on  the  steam  and  gas  engine  not  given  in  this  book. 
When,  after  much  experience  in  teaching  at  an  English  public 


173200 


IV  PREFACE. 

school,  in  the  Imperial  College  of  Engineering,  Japan,  and 
other  places,  I  ventured  sixteen  years  ago  to  publish  my 
method  of  teaching  mechanics,  it  was  met  with  some  ridicule. 
Even  without  encouragement  I  was  prepared  to  pursue  the 
course  which  I  had  tested  and  found  to  be  good,  but  I  met 
with  a  great  deal  of  encouragement  from  thoughtful  men  of 
about  my  own  age.  I  had  myself  to  scheme  out  and  make 
drawings  for  every  piece  of  apparatus  for  the  laboratory.  I 
knew  of  no  collection  of  numerical  and  graphical  exercises 
which  were  suitable  for  students,  but  gradually  a  collection 
was  made  from  those  given  out  in  the  lectures  which  seemed 
to  me  to  be  less  objectionable,  less  academic,  less  misleading, 
than  those  hitherto  available.  I  had  difficulty  in  getting 
clever  assistants  trained  in  academic  ways  to  sympathise 
with  me.  It  was  found  in  time  that  students  took  very  eagerly 
to  the  quantitative  experimental  work,  and  that  the  whole 
system,  faithfully  followed,  produced  men  whose  knowledge  was 
always  ready  for  use  in  practical  problems,  and  who  knew  the 
limits  of  usefulness  of  their  knowledge.  I  am  glad  to  say  that 
more  than  twenty  complete  sets  of  the  apparatus  have  been 
made  and  sent  to  various  institutions  by  my  workshop  assistant, 
Mr.  Shepherd. 

It  would  have  made  this  book  too  large  if  I  had  included 
in  it,  as  I  should  have  liked  to  do,  copies  of  the  instructions 
which  each  student  receives  when  he  begins  on  a  new  piece  of 
apparatus. 

Professor  (now  Sir  Robert)  Ball,  at  the  Royal  College  of 
Science,  Dublin,  started  quantitative  experimental  mechanical 
work.  He  used  the  well-known  frame  of  the  late  Professor 
Willis,  which  was  taken  to  pieces  and  built  up  in  new  forms 
for  fresh  experiments.  What  I  have  done  has  been  to  carry 
out  Professor  Ball's  idea,  using  a  distinct  piece  of  apparatus 
for  each  fresh  kind  of  experiment.  A  student  measures  things 
for  himself  ;  illustrates  mechanical  principles ;  finds  the  limits 
to  which  the  notions  of  the  books  as  to  friction  and  properties 
of  materials  are  correct ;  learns  the  use  of  squared  paper,  and 


PREFACE.  V 

the  accuracy  of  graphical  methods  of  calculation ;  and,  above 
all,  really  learns  to  think  for  himself.  Professor  Ewing,  at 
Cambridge,  has  developed  the  ideas  of  Professor  Ball  to  a  far 
greater  extent  than  what  I  have  had  opportunity  for,  and  I 
know  of  no  place  in  which  a  better  engineering  education  can 
be  obtained  at  the  present  time  than  at  Cambridge.  I  am 
glad  to  think  that  a  system  begun  under  the  Science  and  Art 
Department  by  Sir  Robert  Ball  is  now  likely  to  be  adopted 
generally  in  science  classes. 

I  am  under  great  obligations  to  my  assistant,  Mr.  G.  A. 
Baxandall,  who  has  been  to  great  trouble  in  adding  to  the 
exercises,  verifying  answers,  and  correcting  proofs.  Professor 
Willis,  D.Sc.,  has  been  kind  enough  to  read  through  the 
proofs,  and  consequently  I  feel  that  there  can  be  no  im- 
portant mistake  anywhere. 

I  should  like  to  think  that,  before  a  student  begins  the  part 
in  small  type,  he  has  worked  through  Thomson  and  Tait'a 
small  book  on  "  Natural  Philosophy,"  and  that  he  has  read  the 
early  part  of  my  book  on  "  The  Calculus  for  Engineers." 

JOHN  PERRY. 

IMh  July,  1897. 

Royal  College  of  Science, 

London  S.W. 


CONTENTS. 


CHAP.  PAGE 

I.  INTRODUCTORY 1 

II.  VECTORS  :  RELATIVE  MOTION 29 

III.  WORK  AND  ENERGY 38 

IV.  FRICTION 56 

V.  EFFICIENCY 88 

VI.  MACHINES:   SPECIAL  CASES 98 

VII.  ELEMENTARY  ANALYTICAL  AND  GRAPHICAL  METHODS  .        .  120 

VIII.  EXAMPLES  IN  GRAPHICAL  STATICS 151 

IX.  HYDRAULIC  MACHINES 170 

X.  MACHINERY  IN  GENERAL 220 

XL  KINETIC  ENERGY 242 

XII.  MATERIALS  USED  IN  CONSTRUCTION 275 

XIII.  TENSION  AND  COMPRESSION .290 

XIV.  SHEAR  AND  TWIST 323 

XV.  MORE  DIFFICULT  THEORY 361 

XVI.  BENDING 381 

XVII.  STRENGTH  AT  ANY  SECTION  OF  A  BEAM        .        .        .        .393 
XVIII.  SOME  WELL-KNOWN  RULES  ABOUT  BEAMS     ....  410 

XIX.  DIAGRAMS  OF  BENDING  MOMENT  AND  SHEARING  FORCE     .  427 

XX.  MORE  DIFFICULT  CASES  OF  BENDING  OF  BEAMS  .        .        .441 

XXI.  BENDING  AND  CRUSHING        .......  457 

XXII.  METAL  ARCHES 478 

XXIII.  MEASUREMENT  OF  A  BLOW      .....                       .  486 

XXIV.  FLUIDS  IN  MOTION 505 

XXV.  PERIODIC  MOTION 546 

XXVI.  MECHANISM 570 

XXVII.  CENTRIFUGAL  FORCE       .        . 597 

XXVIII.  SPRINGS 613 

XXIX.  RESILIENCE  OF  SPRINGS  .        . 641 

XXX.  CARRIAGE  SPRINGS                  ,               *       ....  646 

APPENDIX         .        .        .       .        .        .   *    .       ,       .       .  654 

.667 


LIST    OF    TABLES. 


PAGE 

I.  Normal  Pressure  of  Wind  against  Roofs 157 

II.  Moments  of  Inertia  and  the  M  and  k  of  Rotating  Bodies    .        .  251 

III.  Young's  Modulus  (Wertheim) 302 

IV.  Factors  of  Safety  for  Different  Materials  and  Loading        .         .  305 
V.  Ultimate  Stresses  for  Loads  Repeated  a  Great  Number  of  Times  310 

VI.  Moment  of  Inertia  and  Strength  Modulus  of  Various  Sections  . 

398-400 

VII.  Strength  and  Stiffness  of  Rectangular  Beams  Supported  at  the 

Ends  and  Loaded  in  the  Middle 411 

VIII.  Diagrams  of  Bending  Moment,  with  Strength  and  Stiffness,  of 
a  Uniform  Beam  when  Supported  or  Fixed  and  Loaded  in 

Various  Ways 414-16 

IX.  Breaking  Stress  of  Iron  and  Timber  Struts         .        .        .        .468 
X.  Values  of  the  Constant  B  for  Different  Lengths  of  Struts  and 

Different  Materials .468 

XI.  Values  of  the  Constant  n  for  Struts  of  Different  Sections         .     469 

XII.  Relative  Amplitudes  for  Different  Frequencies  in  a  Case  of 

Forced  Vibration 616 

XIII.  Spring  Materials  Subjected  to  Bending       .        ..        .        .        .     621 

XIV.  Materials  for  Cylindric  Spiral  Springs         .        .        ...        .    622 

XV.  Values  of  Torsional  and  Flexural  Rigidity  and  Axial  Load  for 

Various  Sections  of  Wire  used  in  Springs        ....     635 

XVI.  Relative  Elongations  of  Various  Springs  for  Same  Axial  Loads.     638 
XVII.  Proof  Load  and  Resilience  of  Various  Springs    ....     640 

XVIII.  Resilience  of  Spring  Materials  for  Different  Kinds  of  Loading   .     642 
XIX.  Useful  Constants         .        .        .        «        .        .        .        .         654-5 

XX.  Moduli  of  Rigidity       .  .       *,„    «    :,.        .        .        .     656 

XXI.  Moduli  of  Compressibility  ...        .         .        .        .        .     657 

XXII.  Melting  Points,  Specific  Gravity,  Strength,  etc.,  of  Materials  658-61 

XXIII.  Logarithms.        .        •,       V-     .•        •-     >      >•       -•        -          G62-3 

XXIV.  Antilogarithms    .        .        .        .        .        *;      .3      .        .         6G4-5 
XXV.  Angles  in  Degrees  and  Radians,  Sines,  Tangente,  etc.        .        .     606 


APPLIED  MECHANICS. 


CHAPTER  I. 

INTRODUCTORY. 

1.  THE  student  of  Applied  Mechanics  is  supposed  to  have  some 
acquaintance  already  with  the  principles  of  mechanics  ;  to  be 
able  to  multiply  and  divide  numbers  and  to  use  logarithms ; 
to  have  done  a  little  practical  geometry;  to  know  a  little 
algebra  and  the  definitions  of  sine,  cosine,  and  tangent  of  an 
angle ;  and  to  have  used  squared  paper.  He  is  supposed  to 
be  working  many  numerical  and  graphical  exercises ;  to  be 
spending  four  hours  a  week  at  least  in  a  mechanical  laboratory ; 
to  be  learning  about  materials  and  tools  in  an  iron  and  wood 
workshop ;  and  to  be  getting  acquainted  with  gearing  and 
engineering  appliances  in  a  drawing-office  and  elsewhere. 

Unfortunately,  many  students  are  deceived  as  to  their  fitness 
to  begin  the  study  of  A.pplied  Mechanics,  and  we  think  it 
necessary  in  this  introductory  chapter  to  suggest  some  pre- 
paratory exercise  work,  and  also  to  state  certain  definitions 
and  facts  which  will  be  afterwards  referred  to,  perhaps  as  if 
they  were  still  unknown. 

When  we  think  of  what  goes  on  under  the  name  of  teaching 
wo  can  almost  forgive  a  man  who  uses  a  method  of  his  own, 
however  unscientific  it  may  seem  to  he.  Nevertheless,  it  is  not 
easy  to  forgive  men  who,  because  they  have  found  a  study  interest, 
ing  themselves,  make  their  students  waste  a  term  upon  it,  when 
only  a  few  exercises  are  wanted — on  what  is  sometimes  called  the 
scientific  study  of  arithmetic,  for  example,  or  of  mensuration. 

In  our  own  subject  of  Applied  Mechanics  there  are  teachers 
who  spend  most  of  the  time  on  graphical  statics,  or  the  graph- 
ing of  functions  on  squared  paper,  or  the  cursory  examination  of 
thousands  of  models  of  mechanical  contrivances.  One  teacher  seems 
to  think  that  applied  mechanics  is  simply  the  study  of  kinematics 
and  mechanisms  ;  another,  that  it  is  simply  exercise  work  on  pure 
mechanics ;  another,  that  It  is  the  hreaking  of  j specimens  on  a  large 
testing  machine  ;  another,  that  it  is  the  trying  to  do  in  a  school  or 
college  what  can  only  he  done  in  real  engineering  works  ;  another, 
that  it  is  mere  graphics;  another,  that  it  is  all  calculus  and  no 


APPLIED   MECHANICS. 

graphics;  another,  that  it  is  all  shading  and  colouring  and  the 
production  of  pretty  pictures  without  centre  lines  or  dimensions. 
Probably  the  greatest  mistake  is  that  of  wasting  time  in  a  school 
in  giving  the  information  that  one  cannot  help  picking  up  in  one's 
ordinary  practical  work  after  leaving  school. 

We  belieAre  that  the  principles  which  an  engineer  really 
recollects  and  keeps  ready  for  mental  use  are  very  few.  By 
means  of  lectures,  models,  drawing-office  and  laboratory,  and 
numerical  exercise  work,  we  show  a  man  how  these  simple 
principles  enter,  in  curiously  different-looking  shapes,  into  his 
engineering  practice.  We  give  him  the  use  of  all  the  necessary 
.  methods  of  study,  and  we  send  him  out  into  practical  life  pre- 
pared to  study  things  for  himself.  We  ought  to  recognise  the 
fact  that  his  real  study  of  his  profession  is  not  at  school  or  college. 
We  ought  to  teach  him  how  to  learn  for  himself.  Any  child  can 
state  Newton's  second  law  of  motion,  and  the  other  half-dozen 
all-important  principles  of  mechanics,  so  as  to  get  full  marks  in  an 
examination  paper;  the  engineer  knows  that  the  phenomena  he 
deals  with  are  exceedingly  complex,  and  that  only  a  long  ex- 
perience will  enable  him  to  utilise  the  so  easily  stated  principles. 
Schools  and  colleges  are  the  places  in  which  men  ought  to  learn 
the  uses  of  all  mental  tools ;  they  are  sure  to  specialise  afterwards, 
but  in  the  meantime  we  ought  to  give  them  plenty  of  tools  to 
choose  from.  The  average  student  cannot  take  in  more  than  the 
elementary  principles  ,  the  best  students  need  not  take  in  more. 

2.  The  most  important  lesson  for  a  beginner,  however  he 
may  have  studied  mathematics  and  mechanics,  and  however  able 
he  may  be  as  a  mathematician,  is  this — that  he  must  not  go  on 
merely  assuming  that  he  knows  how  to  do  things  ;  he  must  know 
things  by  actual  trial  and  not  mere  hearsaj.  He  must  actually 
calculate  certain  numerical  results ;  he  must  actually  illustrate 
principles  with  laboratory  apparatus  :  and,  if  there  is  a  school 
workshop,  he  must  get  to  know  the  properties  of  materials  by 
chipping  and  filing  and  paring  and  planing  and  turning.  It 
is  just  the  same  as  in  one's  after-school  work.  There  is  no 
great  mechanical  engineer  who  has  not  himself  worked  like 
a  workman  with  other  workmen,  and  got  to  understand  men 
and  things  by  actual  contact  with  them.  The  man  who  shirks 
the  following  exercises  and  laboratory  work  will  lose  a  great 
deal  more  than  he  is  aware  of. 

Teachers  will  notice  that  things  requiring  even  a  little  pre- 
paration more  than  other  things  will  gradually  become  neglected. 
Therefore,  let  it  be  part  of  the  daily  work  for  every  student  to  use 
logarithms,  drawing  instruments  in  graphical  exercise  work,  and 
squared  paper.  In  the  drawing-office,  blue  prints  cught  to  be 
made  by  some  student  or  other  every  day  ;  the  planimeter  ought 


APPLIED    MECHANICS.  3 

to  be  used  every  day,  and  some  student  ought  to  be  resetting  his 
drawing-pens  and  other  instruments  every  day.  It  ought  to  be  a 
rule  that  all  apparatus  must  always  be  ready  for  use,  and  that  it  is 
always  in  use.  Teachers  can  arrange  their  own  work  in  such  a 
way  that  they  cannot  help  seeing  every  day  how  the  practical  work 
of  students  is  being  done.  When  we  find  our  system  to  be  going 
with  clockwork  regularity  and  we  feel  no  worry,  we  ought  to 
believe  that  some  change  is  necessary.  If  we  find  that  the  students 
are  not  absorbed  in  their  work,  we  must  understand  that  we 
teachers  are  in  fault.  (See  Appendix.) 

3.  Students  cannot  spend  too  much  time  in  multiplying, 
factoring,  and  simplifying  algebraical  and  trigonometrical 
expressions.     These  are  our  tools,  and  we  must  get  familiar 
with  them.     We  may  easily  spend  too  much  time  in  studying 
roots  of  equations,  permutations  and  combinations,  etc.,  and 
in  the  solution  of  triangles ;  and  therefore,  if  it  is  possible,  we 
try  to  learn   all   our   mathematics,   mechanics,   physics,   and 
chemistry  from  teachers  who  are  engineers.     What  acquaint- 
ance with  these  subjects  we  have,  ought  to  be  a  real  knowledge, 
not  the  glib  pretence  which  suffices  for  examinations  ;  it  must 
not  be  something  apart  from  our  life  and  work.     To  effect  this 
object  we  must  work  many  numerical  and  graphical  exercises, 
and  try  to  conquer  our  contempt  for  simple  laboratory  experi- 
ments, and  illustrate  the  forty-seventh  proposition  of  the  Fir^t 
Book  of  Euclid  by  actually  drawing  some  right-angled  triangles 
and  measuring  their  sides — to  illustrate  rules  about  triangles 
by  actual  measurement.      In  this  way  we  learn  much  more 
than  the  ordinary  geometrician  knows ;  among  other  things, 
we  obtain  a  valuable  knowledge  of  the  errors  we  are  likely  to 
make  in  graphical  calculation. 

4.  Every  student  is  supposed  to  be  able  to  calculate  the 
values  of  any  algebraical  or  trigonometrical  expression  when 
numerical  values  are  given.     He  ought  to  be  able,  when  given 
at  random  such  an  expression  as 

w  =  «2/3  fl-Va  j.   ,/m2  +  n2  a  loge  b  .  cos.  0, 

to  be  able  to  calculate  w  when  he  knows  the  values  of  a,  b,  m, 
n,  and  0.  He  ought  to  be  able  to  use  logarithms  in  multiplica- 
tion and  division  and  extraction  of  roots ;  to  know  all  the  usual 
mathematical  symbols,  and  how  it  is  sometimes  convenient  to 
use  ^/a  and  sometimes  ah.  It  is  pedantic  to  say  that  a 
man  must  not  use  a  formula  unless  he  is  able  to  prove  its 
truth.  It  is  usually  of  great  help  in  learning  to  prove  a 


*  APPLIED    MECHANICS. 

formula  to  have  previously  used  the  formula  and  know  the 
value  and  meaning  of  what  we  are  to  prove.  A  living 
Northern  professor  of  great  eminence  has  declared  that  a  boy 
ought  not  to  be  allowed  to  use  logarithms  until  he  is  able 
to  calculate  them ;  he  has  not  said  that  a  boy  ought  not  to 
use  a  watch  or  wear  a  coat  until  he  is  able  to  make  them. 

EXERCISES. 

1.  Find  4-326  x  0*003457  to  four  significant  figures,  leaving  out  all 
unnecessary  figures  in  the  work.     Find  p'01584  -j-  2-104  to  four  signifi- 
cant figures.     Also  do  these  using  logarithms.     Find  loge  7.     Calculate 

5  2'43,  3-°'246,  -042  °'476,    /\/246-3,    31-01*    x  0-02641** 

Ans.,     0-01495,  0'007529,  1-94591,  49-95,  0-7632,  0-2211,  3-008,  5'872. 

2.  If  m  =  (a3  +  2  a?b  +  *  -  '345)*  -j-  (a8  J2)5 

find  m  if  a  =  0'504,  6  =  0'309,  *  =  1'567. 

Ans.,  1-453  x  108. ' 

3.  What  errors  are  there  in  assuming 

(1  +  a)»  =  1  +  na 

to  be  true  in  (1-001)3  =  1-003,  (1-01)*  =  1-0033. 

(0-99)2  =  (1  -  -Ol)2  =  1  —  -02  =  -98. 

=  (1  —  -01)  ~  l  =  1  +  -01  =  1  01. 


=  (1  +  -01)  "~     =  (1  -  -0033)  =  -9967. 
'1-01 

-  -01)  =  10  (1  -  -01)*  =  10  (1  -  -005). 

=  9-95  ? 

The  above  answers  are  very  nearly  correct ;  the  student  is  expected  to 
find  the  correct  answers. 

4.  How  much,  error  is  there  in  the  assumptions 

1+Jj  =  1  +  a  _  IB,      (1  +  a)  (1  +  /B)  =  1  +  a  +  0, 

when         a  =  '01       j8  =  -01,  a  =  -  '003       0  =  -  '005  ? 

Ans.,  No  error  :  -01  per  cent.,  '004  per  cent.,  -0015  per  cent. 

5.  If  d  is  the  diameter  of  the  bore  or  the  "  calibre  "  of  a  gun,  it  is 
usually  assumed  that  the  weight  of  the  gun  is  proportional  to  ds,  and 
that  the  thickness  of  armour  which  its  projectile  will  pierce  is  propor- 
tional to  d.     If  an  8-inch  gun  weighs  14  tons  and  can  pierce  11  inches  of 
armour,  what  thickness  will  be  pierced  by  a  10-inch  gun,  and  what  is  the 
•weight  of  the  gun?  Am.,  13*75  inches,  27 '34  tons. 

5.  The  linear  expansion  of  bodies  by  heat  is  practically 
proportional  to  the  rise  of  temperature.  The  values  of  a,  the 
co-efficient  for  linear  expansion  (the  fractional  increase  in 
length  for  a  rise  in .  temperature  of  1°  Centigrade),  are  the 

*  See  Appendix. 


APPLIED    MECHANICS.  5 

following  numbers  divided  by  105  : — Aluminium,  2-34 ;  copper, 
1-79;  gold,  1-45;  iron,  1-2;  lead,  2-95  •  platinum,  0-9 ;  silver, 
1-94;  tin,  2-27  ;  zinc,  2-9  j  brass  (71  copper  to  29  zinc),  1-87; 
bronze  (86  copper  to  10  tin  to  4  zinc),  1-8;  German  silver, 
1-8;  steel,  1-11;  brick,  0-5;  glass,  0-9;  granite,  0-9;  sand- 
stone, 1*2;  slate,  1*04;  boxwood  (across  the  fibre),  6*1;  box- 
wood (along  the  fibre),  0'3 ;  oak  (across),  5 '4 ;  oak  (along)., 
0*5  ;  pine  (across),  3'4;  pine  (along),  0'5. 

The  co-efficient,  k,  of  cubical  expansion  is  three  times  the 
co-efficient  of  linear  expansion,  because  (1  +  a)3  =  1  +  3  a 
is  practically  correct  for  these  small  values  of  a.  The  average 
values  of  k  between  0°  and  100°  0.  are  the  following  numbers 
divided  by  103  :— Alcohol,  1'26  ;  mercury,  O18  ;  olive  oil,  0-8; 
petroleum,  1-04  ;  pure  water,  0*43  ;  sea- water,  0*5. 

The  student  is  supposed  to  have  worked  many  exercises 
like  the  following  ones : — 

1.  Steel  rails  of  0°  0.  have  an  aggregate  length  of  1  mile.     What  is 
the  length  at  33°  0.  ?  Ans.,  1  mile  24-2  inches. 

2.  A  ring  of  wrought  iron  has  an  inside  diameter  of  5  feet  when  at  a 
temperature  of  970°  C.     What  is  the  diameter  at  0°  C.  ?      Ans.,  4-9  feet. 

3.  A  cylindric  plug  of  copper  just  fits  into  a  hole  4  inches  diameter  in 
a  piece  of  cast  iron.     After  heating  the  mass  to  1,240°  C.,  by  how  much 
is  the  diameter  of  the  hole  too  small  for  the  plug?        Ans.^  -0293  inch. 

4.  A  har  of  iron  is  70  centimetres  long  at  0°  C.     What  is  its  length  in 
boiling  water  (100°  C.)  ?    What  is  its  length  at  50°  C.  ? 

Ans.,  70'079  centimetres,  70*039  centimetres. 

5.  Two  rods — one  of  copper,  the  other  of  iron — measure  98  centimetres 
each  at  0°  C.     What  is  the  difference  in  their  lengths  at  57°  0.  ? 

An,*.,  '027  centimetre. 

6.  Bars  of  wrought  iron,  each  3 '4  metres  long,  are  laid  down  at  a 
temperature  of  10°  C.     What  space  is  left  between  every  two  if  they  are 
intended  to  close  up  completely  at  40°  0.  ?  Ans.,  1*26  millimetres. 

7.  A  wrought-iron  connecting-rod  is  12  feet  long  at  10°  C.     What  is 
the  increase  of  length  at  80°  C.  ?  Ans.,  0-121  inch. 

8.  A  wrought-iron  Cornish  toiler  is  33  feet  long ;  the  shell  is  at  0°  0., 
the  flue  at  100°  C.     What  would  the  difference  of  the  lengths  be  if  the  flue 
were  not  prevented  from  expansion  ?  Ans.,  0*475  inch. 

9.  A  steel  pump  rod  is  1,000  feet  long.     What  is  its  change  of  length 
for  a  change  of  10°  C.  Ans.,  1'44  inch. 

10.  In  a  thermometer  "01  cubic  inch  of  mercury  at  10°  C.  is  raised  to 
15°  C.,  and  rises  1  inch  in  the  tube.  What  is  the  cross- section  of  the  tube  ? 

Ans.,  9  x  10~6  S(luare  inch- 

11.  The  volume  of  a  lump  of  iron  being  5  cubic  feet  at  10°  0.,  find  its 
volume  at  80°  C.  Ans.,  5-0126  cubic  feet. 

6.  A  student's  knowledge  of  mathematics  ought  to  be  such 
that  he  can  work  out  for  himself  all  the  rules  given  in  such  an 


6  APPLIED    MECHANICS. 

excellent  book  on  mensuration  as  that  of  Professor  A.  Lodge. 
The  thorough  study  of  such  a  book  is  one  of  several  ways 
which  may  be  recommended  of  getting  familiar  with  mathe- 
matical principles.  But  nobody's  life  is  long  enough  to  use 
all  these  ways,  and,  besides,  unnecessary  study  leads  to  dul- 
ness.  Hence,  if  a  student  has  taken  some  other  way,  he  need 
not  be  alarmed  at  his  ignorance  of  the  more  complex  rules  in 
mensuration  ;  he  may  feel  absolutely  certain  that  he  can  work 
out  such  rules  for  himself,  given  time  and  necessity.  He  will 
study  the  more  complex  rules,  such  as  prismoidal  formulae,  if 
he  needs  to  use  them  practically,  not  otherwise.  The  following 
rules  are  in  constant  use  and  must  be  familiar  to  the  student, 
whether  or  not  he  knows  the  reasons  for  them.  If  he  is 
familiar  with  the  rules  and  does  not  anxiously  search  for  the 
reasons  for  them,  he  lacks  the  necessary  spirit  of  the  practical 
engineer. 

RULES    IN    MENSURATION. 

An  area  is  found  in  square  inches  if  all  the  dimensions  are 
given  in  inches.  It  is  found  in  square  feet  if  all  the  dimen- 
sions are  given  in  feet. 

Area  of  a  parallelogram. — Multiply  the  length  of  one  side 
by  the  perpendicular  distance  from  the  opposite  side. 

The  centre  of  gravity  of  a  parallelogram  is  at  the  point  of 
intersection  of  its  diagonals. 

Draw  a  right-angled  triangle  ;  measure  very  accurately  the 
lengths  of  the  sides.  You  will  find  that,  no  matter  what  scale 
of  measurement  you  use,  the  square  of  the  length  of  the 
hypothenuse  is  equal  to  the  sum  of  the  squares  of  the  lengths 
of  the  other  two  sides. 

B  Area  of  a  triangle. — Any  side  multiplied 
by  its  perpendicular  distance  from  the  oppo- 
site corner  and  divided  by  two. 

The  centre  of  gravity,  or,  rather,  the  centre 
Jig.  i.  of   area,  of   a  triangle  is  found   by  joining 

(Fig.  1)  any  corner,  A,  with  the  middle 
point,  D,  of  the  opposite  side,  B  c,  and  making  D  G  one-third 
of  D  A.  G  is  the  centre  of  gravity. 

Area  of  an  irregular  figure. — Divide  into  triangles,  and 
add  the  areas  of  the  triangles  together. 

Circumference  of  a  circle. — Multiply  the  diameter  by 
3-1416. 


APPLIED    MECHANICS.  7 

Arc  of  a  circle. — From  eight  times  the  chord  of  half  the 
arc  subtract  the  chord  of  the  whole  arc ;  one-third  of  the 
remainder  will  give  the  length  of  the  arc,  nearly. 

Area  of  a  trapezium. — Half  the  sum  of  the  parallel  sides 
multiplied  by  the  perpendicular  distance  between  them. 

Area  of  a  circle. — Square  the  radius,  and  multiply  by 
3 '141 6  ;  or  square  the  diameter,  and  multiply  by  O7854. 

Area  of  a  sector  of  a  circle. — Multiply  half  the  length  of 
the  arc  by  the  radius  of  the  circle. 

Area  of  a  segment  of  a  circle. — Find  the  area  of  the  sector 
having  the  same  arc,  and  the  area  of  the  triangle  formed  by 
the  chord  of  the  segment  and  the  two  radii  of  the  sector. 
Take  the  sum  or  difference  of  these  areas  as  the  segment  is 
greater  or  less  than  a  semicircle. 

Otherwise,  for  an  approximate  answer : — Divide  the  cube 
of  the  height  of  the  segment  by  twice  the  chord,  and  add  the 
quotient  to  two-thirds  of  the  product  of  the  chord  and  height 
of  the  segment.  When  the  segment  is  greater  than  a  semi- 
circle, "subtract  the  area  of  the  remaining  segment  from  the 
area  of  the  circle. 

The  areas  of  curves  may  be  found  by  Simpson's  rule. — 
Divide  the  area  into  any  even  number  of  parts  by  an  odd 
number  of  equidistant  parallel  lines  or  ordinates,  the  first  and 
last  touching  the  bounding  curve.  Take  the  sum  of  the 
extreme  ordinates  (in  many  cases  each  of  the  extreme 
ordinates  is  of  no  length),  four  times  the  sum  of  the  even 
ordinates,  and  twice  the  sum  of  the  odd  ordinates  (omitting 
the  first  and  last) ;  multiply  the  total  sum  by  one-third  of  the 
distance  between  any  two  successive  ordinates. 

The  ordinary  rule  for  an  indicator  diagram  is : — Draw 
lines  at  right  angles  to  the  atmospheric  line,  touching  the 
extreme  ends  of  the  diagram.  Divide  the  distance  between 
them  into  ten  equal  parts  (a  parallel  ruler  with  ten  pieces  is 
sometimes  supplied),  and  at  the  middle  of  each  part  draw  a 
line  at  right  angles  to  the  atmospheric  line.  Measure  the 
breadth  of  the  diagram  on  each  of  these  ten  lines,  and  take 
one-tenth  of  their  sum.  This  gives  the  average  breadth,  and 
represents  the  average  pressure  to  scale.  The  better  plan  is 
to  find  the  area  by  a  planimeter.  The  earnest  student  will 
practise  the  use  of  the  planimeter,  finding  its  error  by  tests  on 
rectangles  and  circles.  The  average  breadth  of  a  diagram  is 
used  in  many  ways. 


8 


APPLIED    MECHANICS. 


Exercise  for   Advanced  Students. — Prove   that  Simpson's 
rule  gives  the  area  correctly  if  the  arcs  of  curve  between  the 
odd  ordinates  follow,  each  of  them,  any  such  law  as 
y  ^=.  a  +  bx  +  ex2 . 

Surface  of  a  sphere. — Multiply  the  diameter  by  the 
circumference.  (See  Appendix.) 

Surface  of  a  cylinder. — Multiply  the  circumference  by  the 
length,  and  add  the  areas  of  the  two  ends. 

Surface  of  a  right  circular  cone. — Multiply  half  the 
circumference  of  the  base  by  the  slant  side,  and  add  the  area 
of  the  base. 

Lateral  surface  of  the  frustum  of  a  right  cone. — Multiply 
the  slant  side  by  the  circumference  of  the  section  equidistant 
from  its  parallel  faces/ 

Area  of  an  ellipse. — Multiply  the  product  of  the  major  and 
minor  axes  by  -7854. 

The  areas  of  two  similar  figures  are  as  the  squares  of  their 
like  dimensions.  The  volumes  are  as  the  cubes  of  their  like 
dimensions. 

The  cubic  content  of  a  body  is  calculated  in  cubic  inches  if 
all  the  dimensions  are  given  in  inches ;  in  cubic  feet  if  all  the 
dimensions  are  given  in  feet. 

Cubic  content  of  a  plate. — Multiply  area  of  plate  by  its 
thickness. 

Cubic  content  of  a  sphere. — Cube  the  diameter,  and  multiply 
by  -5236. 

Cubic  content  of  the  segment  of  a  sphere. — Subtract  twice 


Fig.  2. 


APPLIED    MECHANICS. 


the  height  of  the  segment  from  three  times  the  diameter  of  the 
sphere ;  multiply  the  remainder  by  the  square  of  the  height, 
and  this  product  by  -5236. 

The  cubic  content  and  surface  of  a  sphere  are  each  two- 
thirds  of  that  of  the  cylindric  vessel  which  just  encloses  it. 

Cubic  content  of  any  prismatic  body  (Fig.  2). — Multiply 
the  area  of  the  base  by  the  perpendicular  height.  This  will 
give  the  same  product  as,  Multiply  the  area  of  cross  section  by 
length  along  the  axis  of  the  prism.  (The  axis  of  a  prismatic 
body  goes  from  centre  of  gravity  of  base  to  centre  of  gravity 
of  top.)  The  centre  of  gravity  of  a  prismatic  body  is  half-way 
along  the  axis. 

Cubic  content  of  any  pyramidal  or  conical  body  (Fig.  3). 


Fig.  3. 

— Multiply  the  area  of  the  base  by  one-third  of  the  perpen- 
dicular  height. 

Centre  of  gravity  is  one-quarter  of  the  way  along  the  axis 
from  the  base.  (The  axis  of  any  such  body  joins  the  centre  of 
area  of  base  with  the  vertex.) 

The  cubic  content  of  the  rim  of  a  wheel  is  found  by  multi- 
plying the  area  of  a  cross  section  by  the  circumference  of  the 
circle  which  passes  through  the  centres  of  gravity  of  the  cross 
sections. 

The  weight  of  a  cubic  inch  of  each  of  the  following  mate- 
rials is  given  in  Ibs. : — Cast  iron,  -26;  wrought  iron,  '28;  steel, 
28  :  brass,  -3 ;  copper,  '32  ;  bronze,  '3 ;  lead,  '4;  tin,  -27 ;  zina, 
•26.  Hence  to  find  the  weight  of  a  body  of  cast  iron  or  any 
other  of  these  substances,  find  the  volume  in  cubic  inches  and 
multiply  by  one  of  the  above  numbers. 

Very  often  it  is  only  the  approximate  weight  that  is 
wanted,  so  that  a  moulder  may  know  how  much  metal  to  melt, 
or  for  other  purposes.  Now,  suppose  we  want  the  approxi- 
mate weight  of  a  cast-iron  beam.  Find  roughly  the  average 
section,  and  get  its  area  in  square  inches,  multiply  by  the  length 

B* 


10  APPLIED    MECHANICS. 

in  inches,  add  to  this  the  cubic  content  of  any  little  gusset 
plates  or  other  excrescences,  multiply  by  *26  and  we  have  the 
weight  in  pounds. 

The  specific  gravity  of  a  substance  means  its  weight  as 
compared  with  the  weight  of  the  same  bulk  of  water.  Now, 
it  is  known  that  a  cubic  foot  of  water  weighs  very  nearly  1,000 
ounces,  or  rather  62-3  Ibs.  The  specific  gravity  of  a  brick 
varies  from  2  to  2 '16  7,  and  therefore  the  weight  of  a  cubic  foot 
of  brick  varies  from  2  x  1,000  or  2,000  to  2-167  x  1,000,  or 
2,167  ounces. 

We  see,  then,  that  from  a  table  of  specific  gravities  we  can 
get  the  weight  of  a  cubic  foot  of  a  substance,  and  therefore  if 
we  know  the  cubic  content  of  a  body  formed  of  this  substance 
we  can  calculate  its  weight. 

Various  plans  for  saving  labour  in  calculation  suggest 
themselves  to  people  working  at  any  particular  trade.  For 
instance,  if  a  pattern  has  no  prints  for  cores,  the  weight  of  the 
pattern  bears  nearly  the  same  proportion  to  the  weight  of  the 
casting  as  the  weight  of  a  cubic  inch  of  the  wood  bears  to  the 
weight  of  a  cubic  inch  of  cast  iron.  This  is  not  always  a  con- 
venient rule,  because  the  pattern  is  a  little  larger  than  the 
casting,  and  the  density  of  wood  alters  as  it  dries. 

The  area  of  an  irregular  figure  may  be  obtained  approxi- 
mately by  cutting  it  out  of  a  uniform  sheet  of  cardboard  and 
weighing  it.  Now  cut  out  a  rectangle  or  square  whose  area  it 
is  easy  to  calculate.  Weigh  this  also.  The  areas  are  in  the 
same  proportion  as  the  weights. 

The  area  of  cross  section  of  a  fine  wire  in  square  inches  can 
be  determined  with  some  accuracy  by  weighing  a  considerable 
length  of  the  wire,  dividing  by  the  weight  of  the  material  per 
cubic  inch,  and  dividing  by  the  length  of  the  wire  in  inches. 
EXERCISES. 

1.  Find  the  number  of  revolutions  per  mile  made  by  a  rolling  wheel 
4|  feet  diameter.  Am.,  373. 

2.  Find  the  area  and  circumference  of  a  circle  of  4  inches  radius. 
Find  the  circumference  and  diameter  of  a  circle  whose  area  is  20  square 
inches.     Ans.,  50-27  sqxiare  inches,  25-13  inches;  15-85  inches,  5-046  inches. 

3.  Find  the  area  of  a  parallelogram  whose  adjacent  sides  are  50  and 
30  feet  and  the  angle  between  them  65°.  Ans.,  1359-45  square  feet. 

4.  Find  the  area  of  a  sector  of  a  circle,  radius  4  in-ches,  angle  50°. 

Ans.,  6-982  square  inches. 

5.  Find  the  area  of  the  segment  of  a  circle,  chord  20  inches,  height 
3  inches.     If  this  were  a  parabolic  segment,  its  area  would  be  -frds  of 
chord  multiplied  by  height.  Ans.,  40'6  square  inches. 


APPLIED    MECHANICS.  11 

6.  Ordinates  of  a  curve,  1-5  inches  apart,  are  2  -30,  2-35,  2-46,  2-57, 
2-42,  2-21,  2-10.     Find  the  area  between  the  first  and  last  by  Simpson's 
rule.     Test  your  answer  by  drawing  the  curve  and  using  a  planimeter. 

Ans.,  21-34. 

7.  Find  the  area  of  the  surface  of  a  sphere,  its  radius  being  8  inches. 

Ans.,  804-2  square  inches. 

8.  A  boiler  has  300  tubes  8  feet  long,  3  inches  diameter.     What  is  the 
total  cross-sectional  area  ?    What  is  the  area  of  tube-heating  surface  ? 

Ans.,  2,121  square  inches,  1,885  square  feet. 

9.  Right  circular  cone,  base  3  inches  radius,  height  8  inches  ;  find  its 
curved  surface  and  volume.     Ans.,  80-52  square  inches,  75-41  cubic  inches. 

10.  Segment  of  sphere,  height  4  inches,  diameter  of  base  15  inches : 
find  volume.  Ans.,  309-9  cubic  inches. 

11.  Find  by  Simpson's  rule  the  number  of  cubic  feet  in  a  log  of 
timber  36  feet  long,  the  cross- sections  at  intervals  of  6  feet  being  8-20, 
5-68,  4-04,  2-92,  2-16,  1-54,  1-02  square  feet.  Ans.,  124-36. 

12.  A  railway  cutting  is  ^  mile  long;    the  thirteen  cross-sectional 
areas,  measured  at  equal  intervals  of  220  feet,  are  respectively  280,  462, 
594,  685,  757,  742,  500,  346,  320,  418,  512,  626,  560  square  feet.     Find 
the  volume  by  Simpson's  rule.  Ansn  52,480   cubic  yards. 

13.  Two  models  of  terrestrial  globes ;  the  areas  of  Africa  are  in  the 
ratio  3  to  2.     What  is  the  ratio  of  the  diameters  of  the  globes  ?    What  is 
the  ratio  of  their  volumes  ?  Ans.,  1-225,  1*837. 

14.  Find  the  surface  and  volume  of  a  sphere  of  2-361  inches  radius. 
If  it  is  of  cast  iron,  what  is  its  weight  ?    Find  the  weight  of  a  segment 
of  this  sphere  1  inch  in  height. 

Ans.,  70*1  square  inches,  55-12  cubic  inches;  14-44  Ibs.,  T67  Ib. 

15.  Find  the  volume  and  weight  of  the  rim  of  a  cast-iron  wheel ; 
section  circular,  outside  and  inside  radii  20  feet  and  18  feet  6  inches. 

Ans.,  213-7  cubic  feet,  42-87  tons. 

16.  Find  the  circumference  and  area  of  a  circle  whose  radius  is 
6  inches.     This  is  the  base  of  a  cylinder  of  11  inches  height.     What  is 
its  volume  ?    Find  its  curved  surface.     If  made  of  cast  iron,  what  is  its 
weight?    What  is  the  area  of  a  section  making  70°  with  the  axis? 
(A  x  cos.  20°  is  the  area  of  the  circular  base  which  is  its  projection.) 

Ans.,  31-42  inches,  78 -54  square  inches,  863'9  cubic  inches,  2-4  square 
feet,  226-4  Ibs.,  83'58  square  inches. 

17.  A  cone  on  an  elliptic  base,  whose  principal  diameters  are  12  inches 
and  8  inches,  is  20  inches  in  vertical  height.    What  is  its  volume  ?   What 
is  its  weight  if  made  of  cast  iron  ?       Ans.,  5027  cubic  inches,  13T7  Ibs. 

18.  A  plate  of  wrought  iron  fa  inch  thick  weighs  0*6  Ib.     What  is  its 
area  ?  Ans.,  34-3  square  inches. 

19.  A  disc  of  zinc  10  inches  diameter,  with  a  hole  of  2  inches  diameter 
in  it,  weighs  0-2  Ib.     Find  its  thickness.  Ans.,  0-0106  inch. 

20.  4  Ibs.  of  copper  is  drawn  into  wire  014  inch  diameter.     Find  its 
length.  Ans.,  6 7 '66  feet. 

21.  A  piece  of  round  copper  wire  100  feet  long  weighs  5  Ibs.   What  is 
its  diameter  ?  Ans.,  -128  inch. 

22.  A  spherical  shell  10  inches  outside  and  8  inches  inside  diameter 
of  cast  iron  :  what  is  its  weight  ?  AM.,  66-45  Ibs. 


12  APPLIED    MECHANICS. 

23.  A  rolled  girder  of  wrought  iron  is  12  inches  outside  depth,  flanges 
are  6  inches  by  ^  inch,  web  is  ^  inch  thick :  what  is  its  weight  if  18  feet 
long?  Ans.,  695-6  Ibs. 

24.  Cylindric  boiler  30  feet  long,  7  feet  diameter.     Two  cylindric  dues 
each  2  feet  6  inches  diameter:  what  is  the  area  of  the  plates?     Plate 
everywhere  £  inch  thick :  what  is  its  weight  ? 

When  in  this  boiler  water  covers  the  flues  and  its  level  is  a  quarter 
of  the  diameter  of  the  shell  from  the  top,  what  are  the  volumes  of  water 
and  steam?  Ans ^  1>189  gquare  feet?  10.7  ton8j  2-8  to  1. 

25.  A  circular  plate  of  lead  2  inches  thick,  8  inches  diameter,  is  con- 
verted into  spherical  shot  of  the  same  density,  each  of  '075  inch  radius. 
How  many  shot  does  it  make  ?  Ans.,  56,889. 

26.  Walking  from  the  centre  towards  the  end  of  one  span  of  a  lattice 
girder  railway  bridge,  I  count  10  wrought  iron  bars  of  rectangular  section, 
each  14  feet  long ;  the  cross  section  of  the  first  is  5  inches  x  §  inch.     If 
the  widths  increase  by  J  inch  and  the  thicknesses  by  ^  inch,  find  the 
total  weight  of  bars.    (See  Appendix.)  Ans.,  1'08  ton. 

7.  The  idea  of  velocity  involves  two  things — the  direction 
and  the  speed.  When  the  direction  does  not  alter,  we  speak  of 
the  speed  as  if  it  were  the  whole  idea.  Find  the  time  in  seconds 
taken  by  a  body  to  traverse  a  certain  distance  measured  in 
feet.  This  distance  divided  by  the  time  is  called  the  average 
velocity.  Thus,  if  a  railway  train  moves  through  200  feet  in 
4  seconds,  its  average  velocity  during  this  time  is  200  -^  4,  or 
50  feet  per  second.  If  we  find,  with  careful  measuring 
instruments,  that  it  moves  through  20  feet  in  '4  second,  or 
through  2  feet  in  *04  second,  the  velocity  is  20  4-  '4,  or 
2  -r  '04,  or  50  feet  per  second.  It  is  important  to  remember 
that  the  velocity  may  be  always  changing  during  an  interval 
of  time,  however  short.  To  get  the  velocity  at  any  instant, 
we  must  make  very  exact  measurements  of  the  time  taken  to 
pass  over  a  very  short  distance,  and  even  this  will  only  give  us 
che  average  velocity  during  this  short  time.  But  if  we  make 
a  number  of  measurements,  using  shorter  and  shorter  periods  of 
time,  the  average  velocity  becomes  more  and  more  nearly  the 
velocity  which  we  want.  Thus,  at  10  o'clock,  a  man  in  a 
railway  train  making  a  careful  measurement  finds  that  the 
train  passes  over  200  feet  in  the  next  4  seconds.  He  finds 
the  average  speed  for  4  seconds  after  10  o'clock  to  be  200  -r-  4, 
or  50  feet  per  second.  Another  man  finds  that  it  passes  over 
100-4  feet  in  the  two  seconds  after  10  o'clock,  and  finds  during 
these  two  seconds  an  average  velocity  of  100 '4  -f-  2,  or  50-2 
feet  per  second.  Another  man  finds  50'25  feet  passed  over  in 


APPLIED    MECHANICS.  13 

one  second  after  10  o'clock,  which  shows  a  velocity  of  50 '2 5 
feet  per  second.  Another  man  finds  25-132  feet  passed  over 
in  half  a  second  after  10  o'clock,  and  finds  25-132  -4-  0-5,  or 
50-264  feet  per  second.  Another  man  finds  12-567  feet  in  a 
quarter  second  after  10  o'clock,  and  his  observation  gives 
50 '268  feet  per  second,  and  so  on.  It  is  evident  that  the 
values  given  by  these  various  observations  are  approaching  the 
real  value  of  the  velocity  at  10  o'clock. 
Tabulating  these  results,  we  have  : — 


Intervals  of  Time  in  Seconds 
after  10  o'clock. 

Average  Velocity  in  Feet  per 
Second. 

4 

50-00 

2 

50-20 

1 

50-25 

i 

50-264 

* 

50-268 

Plot  the  two  sets  of  numbers  on  squared  paper,  and  draw 
a  curve  through  the  points  so  found.  Produce  the  curve,  and 
we  have  the  means  of  finding  the  average  velocity  for  an 
infinitely  small  interval  of  time  after  10  o'clock.  This  is  the 
required  velocity. 

8.  Acceleration. — This  is  the  time  rate  of  change  of  the 
velocity  of  a  body.  Thus  it  is  known  that  the  velocity  of  a 
body  falling  freely  in  London — 

At  the  end  of  one  second  is  32 -2  feet  per  second. 
„         „      two  seconds  is  64*4         „         „ 
»      t^ee      „          96-6 
„      four        „        128-8 

and  we  see  that  there  is  an  increase  to  the  velocity  of  32-2 
every  second.  The  acceleration  in  this  case  is  always  of  the 
same  amount — hence  we  call  it  uniform  acceleration,  and  say 
it  is  32-2  feet  per  second  per  second. 

EXERCISES. 

1.  One  mile  per  hour;    also  one  knot;    convert  each  of  these  into 
feet  per  minute  and  feet  per  second.  Ans.y  88,  1-467  ;  101-3,  1-689 

2.  A  torpedo-catcher  travels  at  32  knots ;  convert  this  into  Eneiish' 
miles  per  hour.  -4iw.,  36-85. 

3.  Prove  that  60  miles  per  hour  means  -0268  kilometres  *  per  second. 

*  See  tables  on  page  654. 


14  APPLIED    MECHANICS. 

4.  An  eccentric  disc  is  10  inches  diameter;    the   shaft  makes   300 
revolutions  per  minute.     "What  is  the  rubbing  velocity  on  the  straps  in 
feet  per  second  ?  Ans.,  13*09. 

5.  In  running  a  race  of  1  mile  long,  A  beats  B  by  100  yards,  and  B 
beats  0  by  90  yards.     By  how  many  does  A  beat  C  ? 

Ans.,  185  yards,  nearly. 

6.  Ten  miles  per  hour;  state  this  in  feet  per  second  and  in  centimetres 
per  second.  Ans.,  14§  ;  447. 

7.  An  acceleration  of  32-2  feet  per  second  per  second ;  state  this  in 
miles  per  hour  per  second  ;  state  it  in  centimetres  per  second  per  second. 

Ans.,  21-96;  981-4. 

8.  200  gallons  of  water  per  minute ;  how  many  pounds  per  second  ? 
How  many  cubic  feet  per  second  ?  Ans.,  33-3  ;  -535. 

9.  A  round  pipe  6  inches  diameter  has  30  gallons  per  second  flowing 
through  it.     What  is  the  velocity  ?    If  the  diameter  becomes  10  inches, 
what  is  the  velocity  ? 

Calculate  in  the  two  cases  the  kinetic  energy  of  one  pound  of  water, 
this  being  the  square  of  the  velocity  divided  by  64 '4. 

Ans.,  24-5  feet  per  second ;  8*8  feet  per  second ;  9'3  foot-pounds  ;  1-2 
foot-pound. 

9.  Example. — Two  fine  wires  are  10  feet  apart;  a  bullet 
breaks  them  both.  The  breaking  of  each  wire  causes  an 
electric  spark  to  make  a  mark  underneath  a  fixed  platinum 
pointer  on  a  revolving  drum.  If  the  drum  is  4  feet  in 
diameter,  and  revolves  at  300  revolutions  per  minute,  and 
when  the  drum  is  at  rest  the  spark-marks  are  found  to  be  1-32 
feet  asunder  on  the  curved  surface,  assuming  that  the  intervals 
of  time  between  the  breaking  of  the  wires  and  making  the 
marks  were  the  same,  find  the  time  between  the  breaking  of 
the  wires,  and  find  the  velocity  of  the  bullet.  The  surface 

velocity  is ^ ,  or  62 '83  feet  per  second;  1-32  divided 

by  this  gives  '02101  seconds ;  dividing  this  into  10  feet  gives 
476  feet  per  second  as  the  velocity  of  the  bullet. 

Exercise. — In  some  gun  experiments  screens  150  feet  apart 
were  cut  by  a  bullet  at  the  following  times  (in  seconds), 
counting  from  the  time  of  cutting  the  first  screen  : — 0,  0'0666, 
0-1343,  Q-2031,  0-2729,  0-3439,  0-4159.  Find  the  average 
velocity  between  every  two  successive  screens. 

Ans.,  2,252,  2,216,  2,180,  2,149,  2,113,  and  2,083  feet  per 
second. 

10.  Speed  or  velocity  is  a  rate — the  rate  of  increase  of  space  with 
regard  to  time.  If  a  body  has  passed  through  the  space  s  at  the  time  t, 
and  if  it  goes  over  the  additional  space  Ss  in  the  additional  time  of 

J<,  then  —  is  the  average  velocity.    Observe  that  Ss  is  one  symbol ; 


APPLIED    MECHANICS. 


15 


it  does  not  mean  a  quantity  called  5  multiplied  by  a  quantity  called  *. 
When  we  imagine  52  to  get  smaller  and  smaller  without  limit,  the 
average  velocity  becomes  what  we  call  the  real  velocity,  and  we 

indicate  it  by  the  symbol  -3- .  When,  instead  of  space  and  time, 
we  have  other  quantities,  we  generally  use  y  for  the  dependent 
and  x  for  the  independent  variable,  and  the  symbol  •—-  means 

" the  rate  of  increase  of  y  with  regard  to  x"  Thus,  if  y  is  the 
ordinate  of  a  curve  and  x  is  the  abscissa  at  the  point  p  (Fig. 
4),  and  Q  R  =  8y,  P  R  =  8x,  then  5y/Sx  is' the  tangent  of  the  angle 
Q  v  x.  As  5x  and  Sy  become  smaller  and  smaller,  so  that  Sy/Sx  is 
to  be  written  dy/dx,  then  dyjdx  is  evidently  tangent  of  the  angle 
which  the  tangent  at  P  makes  with  the  axis  of  x.  We  usually 
call  this  the  slope  of  the  curve  at  the  point  P. 

If  y  is  any  quantity  that  depends  upon  another,  xt  and  if  we 
plot  the  values  of  x  and  y  to 
scale  on  a  sheet  of  squared 
paper,  the  curve  shows  by 
its  slope  everywhere  the  rate 
of  increase  of  y  with  regard 
to  x.  If  we  know  the  alge- 
braic law  connecting  y  and 
x,  we  can  find  this  rate  by 
certain  easy  rules.  Thus,  if 


(1), 

-!  (2), 


y  =  Axn  4-  B 


dx 


whatever  kind  of  number  n  ^ig-  *• 

may  be. 

11.  In  teaching  beginners  it  is  well  to  start  on  the  assumption  that 
students  already  possess  the  notions  of  the  differential  and  integral 
calculus,  and  it  is  a  teacher's  duty  to  put  before  them  the  symbols 
used  in  the  calculus  at  once.  It  is  surely  much  better  to  do  this 
than  to  evade  the  calculus  in  the  fifty  usual  methods  which  we 
sometimes  see  adopted.  Unfortunately  many  readers  of  this  book 
are  likely  to  be  preparing  for  examinations  in  which  only  academic 
methods  of  elementary  study  are  recognised,  so  we  must  keep  our 
calculus  symbols  for  the  smaller  print  paragraphs.  We  may  say, 
however,  that  we  think  even  beginners  in  this  subject  will  be  able 
to  understand  the  author's  book  on  the  calculus  ;  and  if  so,  they 
will  find  the  study  of  applied  mechanics  very  greatly  simplified. 
The  language  of  the  calculus  is  the  natural,  easy,  simple  language 
of  the  engineer  ;  but  it  is  in  writing,  whereas  most  engineers  only 
tpeak  of  rates  and  integrals. 

If  y  is  known  as  a  function  of  x,  we  can  find  -^,  which  we 
may  call  u.  Conversely,  if  u  is  known  as  a  function  of  x,  we  can 


16 


APPLIED    MECHANICS. 


6nd  y.     The  two  symbols  are  — ,  called  a  rate,  or  "the  differential 
ax 

co- efficient  of  y  with  regard  to  x  "  ;  and  I  u  .  dx,   called  "  the  in- 
tegral of  u  with  regard  to  #."  J 

12.  The  following  formula)  will  suffice  for  nearly  all  engineering 
calculations  : — 


y 

dx 

N 

\n  .  Sx 

c 

A  xn 

0 

n  Ax  n  ~  l 

C 

B*m 

C  x 

*n  +  1 

A  log.  x 

A 
x 

A*-' 

A  log.  * 

A  eax 

a  A  en* 

A*- 

1     A  ^az 
a 

A  log.  (x  +  a) 

A 

A 

A  log.  (z  -f  a) 

a?  -f-  a 

z  +  a 

A  sin.  (ax  +  b} 

a  A  cos.  (a*  +  b) 

A  sin.  (##  +  i) 

—  —  cos.  (i*^;  +  fr> 
a 

A  cos.  (ax  +  b) 

—  a  A  sin.  (ax  +  ^) 

A  cos.  (a#  +  b} 

—  sin.  (as  +  &) 
a 

and  all  the  important  part  of  the  author's  book  on  the  calculus  is 

devoted  to  illustrating  the  use  of  them. 

13.  For  example,  if  x  is  the 
abscissa  o  B,  and  y  is  the  ordinate 
p  B  of  the  curve,  shown  in  Fig. 
5 ;  if  the  area  between  the 
curve  and  o  x  and  p  B,  and  any 
other  ordinate  nearer  o,  say  D  s,  be 
called  A  ;  and  if  the  area  to  Q  T  be 
called  A  +  S  A,  o  T  being  called 
x  +  Sx,  the  area  SA  =  p  Q  T  B,  and 
as  8x  is  made  smaller  and  smaller 
this  area  gets  more  and  more  nearly 

(y    +  -  Sy)  5a;,  as  Q  T  is  y  -f  5y. 
Thus 


V  3 

Fig.  6. 


B    T 


APPLIED    MECHANICS. 


17 


and  as  5#  gets  smaller  and  smaller  this  in  the  limit  gets  to  be 

dA 

dx 

Hence,  if  we  know  y  in  terms  of  x,  we  can  find  A  by  integration. 
If  we  know  A,  we  can  find  y  by  differentiation.  Thus,  if  we  have 
the  parabolic  curve, 

y  =  ax*  (3) 

A=.l«*»  +  o....(4), 

a  constant  is  added,  because  the  rate  of  change  of  a  constant 
is  0,  and  we  wish  our  answer  to  be  as  general  as  possible.  The 
value  of  c  is  fixed  as  soon  as  we  settle  from  which  ordinate  A  is  to 
be  calculated.  Thus,  if  A  is  0  when  x  is  0,  c  is  0.  If  F  (#)  is  the 
general  integral  of  u, 


f 

Ja 


u  .  dx 


tells  us  to  use  x2  for  x  in  the  general  integral,  then  to  use  xl  for  x 
and  subtract,  or 


i: 


u  .  dx  =  F 


-  F 


(5). 


If  u  is  the  ordinate  of  a  curve,  (5)  means  the  area  between  the 
curve,  the  axis  of  x,  and  the  two  ordinates  at  x\  and  x2. 

Any  summation  which  may  be  indicated  by  an  area  can  be 
effected  in  this  way  by  integration.     It  is  very  eaey  to  recollect 

the  rule  that  the  rate  of  change  of   xn    is    nxn  "  *,   and  that 

1 
the    integral  of    xm    is  — jr-7    #m  +   .      This  rule  suffices  for 

most  engineering  problems.  There  is  one  case  in  which  the  rule 
for  integrating  x  m  is  useless  to  us — namely,  when  m  is  —  1 : 
in  this  particular  case 


i 


dx 


14.  Integrations. — In  a 

gr  Dat  number  of  problems  we 
have  the  ordinates  like  AP, 
called  the  y,  of  every  point 
of  a  curve,  given  us",  the 
abscissa  o  A  of  the  point 
being  called  x. 

If  we  are  given  a  list  of  B 
corresponding  values  of  y  R 
and  x,  we  can  draw  the  curve. 

Now,  if  A  Q  represents  to 
scale  the  area  of  A  P  B  o,  and      C 
if  we  find  many  points  like  Q 


Fig.  6. 


18 


APPLIED    MECHANICS. 


and  join  them,  we  get  a  new  curve,  which  is  said  to  represent  the 
integral  or  area  of  the  first.  The  areas  may  be  found  by  means 
of  a  planimeter.  I  always  use  the  following  method  myself.  It  is 
inaccurate  to  this  extent,  that  the  curve  BP  is  taken  to  be  a  many- 
sided  polygon  instead  of  a  curve.  But  it  leads  to  a  very  quick 
solution  of  complicated-looking  problems.  Thus  the  following 
values  of  y  are  given  for  the  corresponding  values  of  x. 


» 

y 

A* 

0 

0-5 

1-869 

1 

1-869 

1-5 

1-756 

2 

3-625 

2-5 

1-657 

3 

5-282 

3-5 

1-571 

4 

6-853 

4-5 

1-497 

5 

8-350 

55 

1-432 

6 

9-782 

6-5 

1-375 

7 

11-157 

7-5 

1-326 

8 

12-483 

8-5 

1-247 

9 

13-730 

It  will  be  seen  that  we  assume  the  ordinate  at  any  point  like 
#=3  -5  to  be  the  average  height  of  the  curve  from  x  =  3  to  #=4. 
Notice  that  to  get,  say  6-853,  we  add  1-571  times  8x,  which  is  1, 
to  5-282.  We  always  take  the  values  of  y  at  equal  intervals  in 
the  values  of  x.  If  the  above  values  of  x  had  been  0,  0'05, 
0-10,  0-15,  etc.,  beginning  with  the  first,  we  should  have  had  to 
divide  all  the  numbers  in  the  last  column  by  10. 

Exercise.  —  For  values  of  #=0'5,  1'5,  2-5,  ...,  calculate  various 
values  of  y  irorny  =  2  +  3  x  +0  -5  x2,  and  find  the  integrals.  The 
true  value  of  the  integrals  is  A  =  2  x  -f  1  *5  #2+£  xs.  Calculate  this 
for  x=Q,  1,  2,  ...,  and  note  the  errors  in  our  numerical  method. 


means  the  rate  of  change  of  ~  with  regard 


15.  —  The  symbol 

to  x  ;  -y^  means  the  rate  of  change  of  -      with  regard  to  x. 

It  will  be  found  that  much  of    the  difficulty   which  some 


APPLIED    MECHANICS.  19 

students  find  in  using  the  calculus  is  due  to  their  not  being  able  to 
differentiate  f1  with  regard  to  t,  although  they  know  how  to 
differentiate  xn  with  regard  to  x  —  the  mere  use  of  another  letter 
than  the  one  to  which  they  have  been  accustomed  causing  the 
difficulty.  It  is  well,  therefore,  from  the  beginning  to  get  used  to 
many  other  letters,  such  as  t,  v,  s,  p,  w,  etc. 

If  *  is  space  and  t  time,  ^  is  velocity  v,.and  -^  =  -jr   is 

acceleration.  Newton's  symbols  were:  a  for  space  (length), 
s  for  velocity,  s  for  acceleration. 


.—  -If  acceleration  —  ^  =  a,  a  constant  ;  integrate  with 

regard  to  t,  and  we  have  ^-=  velocity  =  at  +  b  .  .  .  .  (6),  where  I 

is  the  constant  which  we  always  add  to  make  the  answer  general. 
We  note  the  meaning  of  b  to  be  the  velocity  when  t  =  0  ;  and 

perhaps  we  had  better  use  v0  instead  of  b,  so  that  -^  =  at  +  v0 

at 

Integrate  again,  and  we  have  s  =  \at'i  +  v0t  -f-  e  .  .  .  .  (7).  Evi- 
dently c  means  the  value  of  *  when  t  ==  0,  (6)  and  (7)  are  well- 
known  laws  of  uniformly  accelerated  motion. 

16.  Example,  —  A  chain  of  length  x,  hanging  vertically  down- 
wards, is  being  lowered  from  a  capstan.     Its  weight  per  foot  of  its 
length  is  w.    When  the  weight  wx  is  lowered  through  the  distance 
fcc,  the  work  done  is  wx  5#,  and  the  whole  work  done  by  it  from 
the  time  it  is  of  no  length,  till  it  is  of  length  I,  is  — 

ri  ri      -. 

I    wx  dx,  or       |  wx*    ,  or, 

or  its  total  weight  multiplied  by  half  its  vertical  length. 

Suppose  the  chain  to  get  thicker  as  more  of  it  is  let  down  ;  say 
that  w  =  a  +  b  x,  then  the  above  summation  becomes  — 

I    (a  +  bx)  x  dx  =  I    (ax  +  bx*)  dx  =  j    \ay?  +  \  bx*  1  = 

|  aP  +  J  bP.  These  two  answers  represent  also  the  work  done  in 
lifting  the  chain. 

17.  Example.  —  Fluid  expanding  through  the  volume  3v  at  the 
pressure  p  does  work  p.  5v,  and 


is  the  work  done  in  expanding  from  volume  v\  to  volume  v^,   Thus, 
suppose  that  expanding  fluid  follows  the  law 

pvk  =  e          or        p  =  cv  ~  fc, 


20  APPLIED    MECHANICS. 

where  Jc  and  c  are  constants,  we  have  to  integrate  cv  '  *  with 
regard  to  v,  and  according  to  our  rule  the  answer  is 


Putting  in  the  limits,  our  work  is 

£ (Va  -  fc  +  l  __  Vi  -  Tc  +  l^ 

and  this  may  be  put  in  other  shapes.  When  the  law  is  pv  =  ct 
the  work  is  c  loge  V—  .  » 

18.  The  Compound  Interest  Law. — If  we  are  told  that  the  rate 
of  increase  or  diminution  of  y  with  regard  to  x  is  proportional  to  y, 

say  that  J^  =  ay,  then  we  know  that  y  =  Aeax,  where  A  is 
any  constant. 

19.  The  Harmonic  Law.— If  y  =  A  sin.  («#  +  b),  we  find  that 

-^  =  A  a  cos.  («c  +  i).  It  will  be  found  that  this  includes  the 
ax 

case: — 

If          y  =  A  cos.  (<&»  -f  4), 

—  :=  —  A  a  sin.  (<z#  -j-  b). 
Thus  if          y  =  A  sin.  («z  -f  i), 

—  =  A  a  cos.  («#  -f  b),     . 

— -^  =  —  A  a2  sin.  (ax  +  b)  =  —  a2y. 

With  this  knowledge  a  student  can  put  in  a  few  words  all  the 
theory  of  Chapter  XXV.  He  will  see  that  simple  harmonic  motion 
is  stated  by  x  =  a  gin.  (qt  +  e) (1),  where  a  is  the  amplitude, 

q  is  2  vf  or  — ,  where  /  is  the  frequency  or  r  is  the  periodic 

time,  and  e  is  the  lead,  a  quantity  introduced  because  x  is  not  0 
when  t  =  0 ;  that  is,  we  give  generality  to  (1)  by  assuming  that 
we  may  begin  to  count  time  from  any  position  of  the  body. 
Observe  that  the  motion  may  be  angular,  x  and  a  being  angles. 

—  will  then  be  angular  velocity,  and  — ?  will  be  angular 
acceleration.  Differentiating  twice  we  get 

velocity         =  -^L  =  aq  cos.  (qt  +  i)  .  .  .  .  (2), 
at 

acceleration  =  ~  =  —  aq*  sin.  (qt  +  e) .  .  .  .  (3). 
Notice  the  sign  of  — T,  and  see  how  it  agrees  with  the  statements 
of  Chapter  XXV.     If  we  only  think  of  the  numerical  value  of  -^ 


APPLIED    MECHANICS. 


21 


we  notice  at  once  that  x  -r-  -TO  =  •-,> 
at*          £ 


.  (4),  and  the   square 


root  of  this  is  T/2  w,  which  is  the  rule  found  in  another  way  in 
Chapter  XXV. 

20.  Students  will  do  well  to  graph  on  squared  paper  some 
curves  like  the  following  : — 

1.  If  y  =  0-1  x"  +  3,  take  x  =  0,  x  —  1,  x  =  2,  etc.,  and  in  each 
case  calculate  y.     Plot  the  values  of  x  and  y  as  co-ordinates  of  points  on 
squared  paper,  and  draw  the  curve  passing  through  these  points. 

2.  Graph  y  =  a  +  bx. 

1st,  when  a  =  0,  b  —  1. 
2nd,  when  a  =  1,  b  —  1. 
3rd,  when  a  •=.  1,  b  =  1-5. 
4th,  when  a  =  -  1,  b  =  1-5. 
5th,  when  a  =  1,  b  =  —  1. 

3.  Graph  y  =  -1  ec. 

4.  Graph  y  =  10  sin.  (  -=-  x  -f 


5.  Graph  y  =  120  x  ~  *. 

6.  Graph  y  ==  120  x  "  12. 

7.  Graph  y  =  120  x  ~  °*7. 

8.  Graph  y  =  10  x  i 

9.  Graph  y  =  +  -v/25  —  #2, 

and  also  y  =  +  1-3  \/  25  —  z8. 

21.  The  following  observed  numbers  are  known  to  follow  a  law  like 
y  -  a  -{•  bx;  but  there  are  errors  of  observation.  Find  by  the  use  oi 
squared  paper  the  most  probable  values  of  a  and  b. 


X 

2 

3 

4 

6 

7 

9 

12 

13 

y 

5-6 

6-85 

9-27 

11-65 

12-75 

16-32 

20-25 

22-33 

Ans.,  y  =  2-5  -f-  1*5  #. 

22.  The  following  observed  numbers  are  known  to  follow  a  law  like 
y  :=«*/(!  +  **):— 

Find  by  plotting  the  values  of  y/#  and  y  on  squared  paper  that  these 
follow  in  a  law  y\x  +sy  =  a,  and  so  find  the  most  probable  values  of  a  and  s. 


x 

•5 

1 

2 

•3 

1-4 

2-5 

y 

•78 

•97 

1-22 

•55 

•1-1 

1-24 

2*). 


22  APPLIED    MECHANICS. 

23.  Vertical  Line.  —  A  line  showing  the  direction  in  which 
that  force  which  we  call  the  resultant  force  of  gravity  acts.     It 
is  a  line  at  right  angles  to  the  surface  of  still  water  or  mercury. 

24.  Level  Surface.  —  A  surface  like  that  of  a  s  till  lake, 
everywhere  at  the  same  level,  and  everywhere  at  right  angles 
to  the  force  of  gravity  or  other  volumetric  force  which  is  acting 
upon  matter.     It  is  not  a  plane  surface. 

25.  Curvature.  —  For  any  curve  we  can  find  at  any  place 
what  circle  will  best  coincide  with  the  curve  just  there.     The 
radius  of  this  circle  is  called  the  radius  of  curvature  at  the 
place.     But  since  we  say,  for  instance,  that  a  railway  line 
curves  much,  when  we  mean  that   the  radius  is  small,   the 
name  curvature  is  always  given  to  the  reciprocal  of  the  radius. 
Thus,  if  the  radius  is  8  feet,  we  say  that  the  curvature  is  ^. 
If  at  another  pla,ce  the  curvature  is  -|-.  the  change  of  curva- 
ture in  going  from  the  one  place  to  the  other  is  the  difference 
between  these  two  fractions. 

Curvature  may  also  be  denned  as  the  angle  turned  through  by 
a  tangent  to  the  curve  per  unit  of  its  length.  A  student  ought  to 
see  for  himself  that  the  two  definitions  are  the  same.  If  in  Fig.  5 
the  distance  along  the  curve  between  p  and  o,  is  called  5s,  and  if 
50  is  the  angle  which  the  tangent  at  o,  makes  with  the  tangent  at 
p,  then  the  average  curvature  between  p  and  o,  is  SQjSs.  As  p  and 

j/» 

Q  become  closer  and  closer,  without  limit,  the  curvature  is  -=- 

as 

If  a  curve  is  defined  by  its  x  and  y  co-ordinates,  the  curvature 


Example.  —  Find  the  curvature,  where  x  =  0,  of  the  parabola 
y  =  ax*.     Here  -jt  =  2lax   and  ^J[  =r  2  a,  so  that  the  curvature 

anywhere  is  2  a  •£•  j  1  +  4  a?x2  ]  i  and  at  the  vertex  where  x  =  0 
the  curvature  is  2  a. 

When,  as  uTordinary  beams,  -j-  is  small,  it  is  evident  that  the 

curvature  may  be  taken  to  be  -r-f  .     This  is  what  we  take  to  be 
true  in  the  discussion  of  beams  and  struts. 

Example.  —  In  making  100  steps  round  a  curve,  my  compass, 
showing  the  direction  of  motion,  changes  from  N.  to  N.E. 
What  is  the  average  curvature?  Answer  —  from  N.to  N.E.  is 
45  degrees,  or  0'7S54  radians,  and  this  divided  by  100  steps, 
or  '007854  radians  per  step,  is  the  average  curvature.  The 


a        a 

e    +e 


at 


APPLIED    MECHANICS.  23 

reciprocal  of  this,  or  127 '3  steps,  is  the  radius  of  curvature, 
if  the  curvature  is  constant — that  is,  if  the  curve  is  an  arc 
of  a  circle. 

Many  unpractical  rules  will  be  found  in  books,  requiring 
us  to  draw  a  tangent  to  a  given  curve  at  a  given  point,  or  to 
find  its  curvature  there  by  trial.  These  are  only  academic 
suggestions.  If  half  a  dozen  students  get  tracings  of  the  same 
curve,  and  two  points  to  measure  the  angle  between  the 
tangents  there,  they  will  obtain  six  very  different  answers. 

EXERCISES. 

1.  Through  what  angle  must  a  rail  10  feet  long  be  bent  to  fit  a  curve 
of  half  a  mile  radius?  Ans.,  0'22  degrees. 

2.  Arc  *  is  10  feet  long,  radius  r  half  a  mile  ;  find  the  versed  sine  x — 
that  is,  the  greatest  deflection  from  straightness.     Prove  that,  practically, 
«3  =  8  rx  ;  so  that  in  this  case  x  =  *0047  feet. 

x     —x 

3.  Find  the  radius  of  curvature  of  the  catenary  y=-^ 
the  vertex. 

Ans.,  a. 

4.  Find  the  radius  of  curvature  of  the  ellipse  -g  -f-  ^  =  1.     (1)  At  the 
end  of  the  major  axis ;  (2)  at  the  end  of  the  minor  axis. 

Ans.,  (1)  ~;  (2)  £ 

26.  Angle. — An  angle  can  be  drawn  :  First,  if  we  know  its 
magnitude  in  degrees ;  a  right  angle  has  90  degrees.  Second, 
if  we  know  its  magnitude  in  radians ;  a  right  angle  contains 
1-5708  radians.  Two  right  angles  contain  3'1416  radians. 
One  radian  is  equal  to  57*2958  degrees.  One  radian  has  an 
arc,  B  c,  equal  in  length  to  the  radius  A  B  or  A  c.  It  some- 
times gets  the  clumsy  name  "  a  unit 
of  circular  measure."  Third,  we  can 
draw  an  angle  if  we  know  either  its  sine 
cosine,  or  tangent,  etc.  Draw  any  angle, 
BAG  (Fig.  7).  Take  any  point,  p  in 
AB,  and  draw  PQ  at  right  angles  to 
A  c.  Then  measure  P  Q,  A  p  and  A  Q  Fig.  7? 

in   inches    and   decimals   of    an   inch. 

p  Q  -j-  A  P  is  called  the  sine  of  the  angle.  A  Q  -f-  A  p  is  called 
the  cosine  of  the  angle,  p  Q  -j-  A  Q  is  called  the  tangent  of  the 
angle.  Calculate  each  of  these  for  any  angle  we  may  draw, 
and  measure  with  a  protractor  the  number  of  degrees  in 


24  APPLIED    MECHANICS. 

the  angle.  We  shall  find  from  a  book  of  mathematical  tables 
whether  our  three  answers  are  exactly  the  sine,  cosine,  and 
tangent  of  the  angle.  This  exercise  will  impress  on  our 
memory  the  meaning  of  the  three  terms.  It  will  also 
impress  upon  us  the  fact  that  if  we  know  the  angle  in 
degrees,  we  can  find,  by  means  of  a  book  of  tables,  its  sine,  or 
cosine,  or  tangent ;  and  if  we  know  any  one  of  the  sides  A  p, 
or  P  Q,  or  A  Q,  of  the  right-angled  triangle  A  P  Q  and  the  angle 
A,  we  can  find  the  other  sides. 

Divide  the  number  of  degrees  in  an  angle  by  57 '2958,  and 
we  find  the  number  of  radians.  Suppose  we  know  the  number 
of  radians  in  the  angle  BAG,  and  we  know  the  radius  A B  or 
A  c,  then  the  arc  B  c  is 

A  B  x  number  of  radians  in  the  angle. 

Given,  then,  a  radius  to  find  the  arc,  or  given  an  arc  to 
find  the  radius,  are  very  easy  problems. 

A  student  becomes  accustomed,  on  seeing  an  angle  drawn 
on  paper,  to  judge  from  a  mere  glance  how  many  degrees  the 
angle  contains.  It  would  be  an  advantage  to  acquire  the 
habit  of  judging  how  many  radians  there  are  in  the  angle. 
What  we  mean  is,  that  he  ought  to  be  as  ready  to  think  in 
radians  as  in  degrees,  and  to  do  this  he  requires  to  be  familiar 
with  the  size  of  a  radian. 

EXERCISES. 

1.  Draw  an  angle  of  35°.     Find  "by  measurement  the  sine,  cosine, 
tangent,  cotangent,  secant,  and  cosecant  of  the  angle,  and  compare  with 
the  numbers  in  a  book  of  tables.     Calculate  the  number  of  radians. 
Try  if  sin.2  35°  +  cos.2  35°  ==  1 ;  if  sin.  35°  -=-  cos.  35°  =  tan.  35° ;    if 
tan.2  35°  +  1  =  sec.2  35° ;  and  if  cot.2  35°  +  1  =  cosec.2  35°. 

2.  If  a  =  55°,  ft  =  20°,  illustrate  the  following  important  formula)  by 
numerical  calculation : — 

Sin.  (a  -+-  j8)  —  sin.  a  cos.  ]8  +  cos.  a  sin.  ft. 
Sin.  (a  —  /3)  =  sin.  a  cos.  £  —  cos.  a  sin.  ft. 
Cos.  (a  -f-  ft)  =  cos.  a  cos.  ft  —  sin.  a  sin.  ft. 
Cos.  (a  —  j8)  =  cos.  a  cos.  £  +  sin.  a  sin.  /3. 

Sin.  a  cos.  ft  =  ^  {  sin.  («  +  £)+  sin.  (a  -  ft}  }  . 
Cos.  a  cos.  ft  =  g  j  cos.  (a  +  ft)  -f-  cos.  (a  —  ft) 

Sin.  a  sin.  ft  =  |  {  cos.  (a  -  j8)  -  cos.  (a  +  ft)  J  . 
Cos.  2a  =  2  cos.2a  —1  =  1—2  sin.2a. 


APPLIED    MECHANICS.  25 

3.  What  are  sin.  150°,  cos.  130°,  tan.  170°,  cos.  240°,  sin.  220°,  tan. 
218°,   sin.    290°,   cos.    310°,  tan.  320°?     Express  all  these  angles  in 
radians. 

Ans.,  -5,  -  -6428,  -  -1763,  -  -5,  -  -6428,  -7813,  -  -9397,  '6428, 
-  -8391;  2-6180,  2-2689,  2-9671,  4-1888,  3-8397,  3-8048,  5-0614,  5-4105, 
5-5851. 

4.  The  sine  of  an  angle  is  0'25;  find  its  cosine,  tangont,  cotangent, 
secant,  and  cosecant.     Find  the  angle  hy  actual  drawing.     How  many 
radians?  Am.,  '9683,  -2582,  3-875,  1-033,  4;  14°-5 ;  -2528. 

5.  What  are  the  sine,  tangent,  and  radians  of  1  \  degrees  ? 

Ans.,  Each  '0262. 

6.  If  in  Fig.  7  A  is  47°,  and'AP  5-23  feet,  find  AQ  and  p  Q. 

Ans.,  A  Q  =  3-567,  P  Q  =  3'824. 

27.  Angular  Velocity. — If  a  wheel  makes  90  turns  per 
minute,  this  means  that  it  makes  1-5  turns  per  second.     But 
in  making  one  turn  any  radial  line  moves  through  the  angle  ot 
360  degrees,  which  is  6-2832  radians;    so  that  1-5  turns  per 
second  means    6-2832  x  1*5,  or    9*4248  radians  per  second. 
This   is   the    common    scientific  way  in  which   the  angular 
velocity  of  a  wheel  is  measured — so  many  radians  per  second. 
If  a  wheel  makes  30  turns  per  minute,  its  angular  velocity  is 
3-1416  radians  per  second;    n  turns  per   minute   mean  2irn 
radians  per  minute,  or  2-n-n  +  60  radians  per  second.       One 
turn  is  the  angular  space  traversed  in  one  revolution. 

Exercise. — Show  that  the  linear  speed  in  feet  per  second  of  a  point  in 
a  wheel  is  equal  to  the  angular  velocity  of  the  wheel  multiplied  hy  the 
distance  in  feet  of  the  point  from  the  axis. 

28.  Angular   Acceleration.  —  The    increase    of    angular 
velocity  per  second.     If  a  wheel  starts  from  rest,  and  has  an 
angular  velocity  of  1  radian  per  second  at  the  end  of  the  first 
second,  its  average  angular  acceleration  during  this  time  is 
1  radian  per  second  per  second. 

EXERCISES. 

1.  A  shaft  revolves  at  800  revolutions  per  minute.      What  is  it* 
angular  velocity  in  radians  per  second  ?  Ans.,  83*79. 

2.  A  point  is  3,000  miles  from  the  earth's  axis,  and  revolves  once  in 
23  hours  56  minutes  4  seconds.     What  is  its  velocity  in  miles  per  hour  ? 

'Ans.,  787-5. 

3.  The  average  radius  of  the  rim  of  a  fly-wheel  is  10  feet.     When  the 
wheel  makes  150  revolutions  per  minute  what  is  the  average  velocity  of 
the  iim  ?  Ans.,  157*1  feet  per  second. 

4.  An  acceleration  of  1  turn  per  minute  every  second ;  how  much  is 
this  in  radians  per  second  per  second  ?  Ans.,  -1047. 

5.  A  wheel  is  revolving  at  the  rate  of  90  turns  a  minute.     What  is 
its  angular  velocity  ? 


26  APPLIED    MECHANICS. 

A  point  on  the  wheel  is  6  feet  from  the  axis  ;  what  is  its  linear  speed  ? 
If  its  distance  from  the  centre  be  increased  by  50  per  cent.,  what  does  its 
speed  become  ?  If  at  the  same  time  the  speed  of  the  wheel  increases 
50  per  cent.,  what  is  now  the  linear  speed  of  the  point  ? 

Ans.,  9*425  radians  per  second  ;  56*55  feet  per  second  ;  84*82  feet  per 
second  ;  127*2  feet  per  second. 

6.  There  is  a  lever,  A  o,  30  inches  long,  works  about  an  axis  at  o.  The 
lever  is  made  to  turn  by  applying  a  force  at  a  point  B  in  A  o,  15  inches 
from  o,  so  that  B  receives  a  velocity  of  2  feet  per  second.  What  is  the 
angular  velocity  of  the  lever  ? 

If  the  same  velocity  had  been  given  to  the  point  A  instead  of  B,  what 
would  the  angular  velocity  have  been  ? 

Ans.,  1*6  radians  per  second  ;  0*8  radian  per  second. 

29.  Length  of  Belt.  —  Let  D  and  d  be  the  diameters  of  pulleys,  c 
the  distance  between  their  centres,  L  the  length  of  belt,  let  D  +  d 
be  called  s.  Prove  that  for  a  crossed  belt 


where  sin.  6  r=  s  -7-  2<?. 

Example.  —  Find  the  length  of  a  crossed  belt  for  pulleys  of 
20  and  15  inches  diameter,  the  distance  of  their  centres  apart  being 
120  inches.  First  find  6,  if  sin.  6  =  35  -r  240  =  -1458.  So  that 
6  =  8°  23',  or  -1463  radians.  Also  cos.  0  =  *9893.  Hence  we 
have 

L  =  (1*5708  +  -1463)  35  +  240  x  '9893  —  297*5  inches. 
Notice  from  the  formula  (1)  that  as  L  depends,  only  on  c  and  s, 
if  c  and  s  are  the  same,  the  same  length  of  belt  will  do.  Thus  the 
same  crossed  belt  does  for  any  two  corresponding  steps  in  stepped 
cones  if  D  +  d  is  the  same.  It  is  exceedingly  easy  to  calculate  the 
sizes  of  these  steps  when  we  know  the  speeds.  Thus  the  above- 
mentioned  pulleys  were  20  and  15  inches,  and  their  speeds  were  as 
3  to  4.  If  there  were  two  steps,  and  we  want  another  pair  to  give 
a  speed  ratio,  say,  of  1  to  2  using  the  same  belt,  we  know  that  if 
their  diameters  are  D  and  d 

D  +  d  =:  35,          B  =  2d. 
Hence,     3  d  =  35  and  d  =  llf  ,   D  =  23  J. 

Exercise.  —  The  centres  of  two  pulleys,  3|  and  If  feet  in 
diameter  respectively,  are  10  feet  apart.  Find  the  length  of 
crossed  belt  required.  Ans.,  28-48  feet. 

If  the  belt  is  an  open  belt,  let  the  student  prove  that  the  length 
L  is 

L  =|  (D  +  d)  +  6  (D  -  d]  +  1c  cos.  e, 

where  sin.  6  =  (D  —  d]  -*•  2c.  It  is  easy  to  show  that  the 
following  answer,  which  is  much  easier  to  deal  with  later,  is 
practically  correct  : 

L  =  £(D  +  <*)  +  *  +  1 

I 
The  length  of  belt  depends  upon  D 


APPLIED    MECHANICS.  27 

Exercise.  —  Pulleys  of  20  and  15  inches  diameter,  whose   axes 
are  120  inches  apart,  are  connected  by  an  open  belt.  Find  its  length. 

Am.,  L  =  I  (fc,  '+  240  (l  +  I  (1|L  )  =  295.03". 


The  student  ought  to  find  the  correct  answer,  and  see  how  small 
is  the  error  in  the  use  of  our  approximate  formula. 

Suppose  we  have  a  pair  of  pulleys  DJ  and  dv  and  we  want 
another  pair  D2  and  d%  on  the  same  shafts  to  work  with  the  same 
length  of  belt,  the  ratio  of  D2  to  d%  being  known.  Putting  the  two 
expressions  for  L  equal,  we  have 

D2  +  d,  +  2-^  (D2  -  dtf  =  D,  +  ^  +  JL  („,  -  dtf  ....  (3). 

The  right-hand  side  is  known,  and  we  know  D2  in  terms  of  d2,  so 
that  it  is  easy  to  calculate. 

Example.  —  In  the  above  example  of  DJ  =  20,  d1  =  15,  c  =  120, 
let  us  calculate  D2  and  dzy  if  D2  =  2  dz.  Our  equation  (3)  becomes 

38  +  <6>2  =  3  *  + 


2120          ;  0       - 

The  solution  of  this  quadratic  gives  us  d%  —  33  ;  and,  therefore, 
D2  a=  66.  In  practice  we  usually  calculate  D2  and  dz  as  if  the  belt 
were  crossed.  These  are  nearly  the  right  answers.  Then,  taking 
the  D2  —  «?2  so  found,  we  find  from  (3)  a  corrected  value  of 
D2  -4-  ^2»  an(l  we  use  this,  knowing  the  ratio  of  D2  to  dz,  to  find 
them  more  correctly.  We  may,  if  we  please,  approximate  more 
nearly  by  repeating  this  process,  but  this  is  seldom  necessary. 

Example.  —  On  the  driving  shaft,  going  at  100  revolutions  per 
minute,  the  diameter  of  the  first  step  is  20  inches.  From  this  step 
the  driven  shaft,  which  is  120  inches  away,  is  to  go  at  200  revolu- 
tions. From  other  steps  the  speed  of  the  driven  shaft  is  to  be  150, 
100,  75,  and  50  revolutions.  Find  the  sizes  of  the  steps  if  the  belt 
is  crossed. 

Here  Dx  —  20,  d1  =  10,  D2  -f  d2  =  30,  i>2  =  j  dv  Hence  \  d2  -f  d9 
=  30,  2£  d2  =  30,  d2  =  12  inches,  D2  =  18  inches. 

In  the  same  way  we  find  d3  =  15,  D3  =  15  ;  d^  =  I7j,  D4  =  12f  ; 
^  =  20,  D5^10. 

30.  If  a  line,  AB,  makes  an  angle  d  with  the  horizontal, 
the  projection  of  its  length  on  the  horizontal  is  A  B  cos.  0. 
Its  projection  on  a  vertical  line  is  A  B  sin.  6. 

Exercise.—  Draw  two  lines  ox,  OY  at  right  angles  to  each  other. 
Now  draw  lines  o  p,  o  Q,  o  K,  o  s,  of  lengths  3,  5,  2J,  and  4  inches,  and 
making  angles  of  35°,  72°,  130°,  and  220°  with  o  x.  Find  the  projection 
of  each  line  on  o  x  and  on  o  Y,  and  the  sum  of  the  projections  on  o  x 
and  on  o  Y. 

Am.,  2-457;  1-545;  -1-607;  -3-064  ;  1-721  ;  4-755  ;  1-915;  -2-571; 
-0-669;  5-S20. 

If  a  plane  area  of  A  square  inches  is  inclined  at  an  angle  9 
with  the  horizontal,  its  area,  as  projected  on  the  horizontal, 
is  A  cos.  9  square  inches. 


28  APPLIED    MECHANICS. 

Try  to  prove  that  this  must  be  so  by  dividing  the  area  into 
strips  by  horizontal  lines. 

EXERCISES 

1.  A  plane  area  of   35  square^feet  is  inclined  at  20°   to  the  hori- 
zontal :  find  its  horizontal  and  vertical  projections. 

Ans.,  32-89  square  feet;  11-97  square  feet. 

2.  The  cross-section  (a  cross-section  always  means  a  section  by  a 
plane  at  right  angles  to  the  axis  or  line  of  centres  of  sections)  of  a 
cylinder  is  a  circle  of  0*7  inch  radius.     Find  the  areas  of  sections  which 
make  angles  of  25°  and  45°  with  the  cross-section.     Note  that  the  cross- 
section  is  a  projection  of  any  other  section. 

Ans.y  1-699,  2-177  square  inches. 

3.  The  above  cylinder  is  a  tie  bar  of  wrought  iron.     The  total  tensile 
load  is  12,000  Ib. ;  how  much  is  this  per  square  inch  of  the  cross-section? 
How  much  is  it  per  square  inch  of  either  of  the  other  sections  ? 

Am.,  7794  Ibs.,  7063  Ibs.,  5512  Ibs. 

4.  The  cross-section  of  a  pipe  is  a  circle  of  15  inches  diameter;  what 
is  the  area  in  square  feet?    If  13  gallons  flow  per  second,  what  is  the 
velocity  VQ  ?    What  is  the  area  of  a  section  at  28°  to  the  cross-section  ? 
What  is  the  velocity  v,  normal  to  this  section,  if  normal  velocity  x  area  = 
cubic  feet  per  second  ?     Show  that  v  is  the  resolved  part  of  vo  normal  to 
the  section.       Ans.,  1-228,  1'7  feet  per  second;  1-39,  1*5  feet  per  second. 

5.  Part  of  a  roof,  shown  in  plan  as  4,000  square  feet,  is  inclined  at 
24°  to  the  horizontal ;  what  is  its  area  ?  Ans.,  4378-7  square  feet. 

6.  A  tie  bar  or  short  strut  of  2  square  inches  cross- section ;  what  is 
the  area  of  a  section  making  45°  with  the  cross-section?     If  the  total 
tensile  or  compressive  load  is  20,000  Ibs.,  how  much  is  this  per  square 
inch  on  each  of  the  sections?    Resolve  the  total  load  normal  to  and 
tangential  to  the  oblique  section,  and  find  how  much  it  is  per  square  inch 
each  way.      An».t  2-828  square  inches ;  10,000  Ibs.,  7,070  Ibs.,  5,000  Ibs. 


Fig.  8. 


CHAPTER    II. 

VECTORS.       RELATIVE    MOTION. 

31.  ANY  quantity  which  is  directive  is  called  a  vector  quan- 
tity —  for  example,  a  velocity  or  a  force.    It  can  be  represented 
by  a  line.     Its  amount  can  be  represented  to  some  scale  by  the 
length  of  the  line.     The  clinure  of  the  line  and  an  arrow-head 
represent  the  clinure  and  sense  of  the  vector.    Vector  quanti- 
ties are  distinguished  from  scalar  quantities,  such  as  a  sum  of 
money,  the  mass  of  a  body,  energy,  temperature,  etc. 

The  resolved  part  of  a  vector  in  any  new  direction  is 
represented  by  the  projection  of 
its  representative  length  in  the  new 
direction.  Thus  in  Fig.  8,  if  OP 
represents  to  scale  a  velocity  or  a 
force,  its  resolved  part  in  the  direc- 
tion o  x  is  o  A,  the  amount  of  which 
is  o  P  cos.  A  o  P,  and  its  resolved  part  „ 
in  the  direction  o  Y  is  o  B,  the 
amount  of  which  is  o  P  cos.  BOP. 
Thus,  if  a  ship  is  going  at  9  knots  north-eastward,  the  northerly 
component  of  its  velocity  is  9  cos.  45°,  or  6-363  knots,  and  its 
easterly  component  is  6-363  knots. 

If  a  body  has  an  acceleration  of  20  feet  per  second  per 
second  in  the  direction  25°  east  of  north,  the  northerly  com- 
ponent of  'this  is  20  cos.  25°,  or  18-13  feet  per  second  per 
second  ;  and  the  easterly  component  is  20  sin.  25°,  or  8-452 
feet  per  second  per  second. 

If  a  force  of  30  Ibs.  is  in  a  northerly  direction,  its  com- 
ponent in  a  north-easterly  direction  is  30  cos.  45°,  or  21-21  Ibs. 

32.  The  resultant  of  two  or  more  forces  is  a  force  which 
might  be  substituted  for  them  without  changing  the  effect.    If 
two  strings  pull  a  small  body  with  forces  of  5  Ibs.  and  7  Ibs. 
(Fig.  9),  and  if  the  angle  between 

them  is  30°,  draw  o  P  equal   in 

length  to  5  inches,  and  make  the 

angle  QOP  equal  to  30°.      Make 

the  length  of  o  Q  7  inches.     Com- 

plete   the    parallelogram   Q  o  P  n, 

and  draw  the  diagonal  o  R.     Measure  o  R  in  inches  ;  we  find 

it  to  be   11  -6  inches,  so  that  the  resultant  of  the  two  forces 


Fig.  9. 


OF  THE 

UNIVERSITY 


30  APPLIED    MECHANICS. 

is   11 '6   Ibs.     One  string  acting  in  the  direction  OR  with  a 

pull  of  11 '6  Ibs.  will  produce  the  same  effect  at  o  as  the  two 

strings  did.     In  using  this  construction,  take  care  that  the 

arrow-heads  are  confluent— that  is,  that  they  all  point  away 

from  o,  or  they  all  point  towards  o.     Suppose  when  the  two 

strings  were  acting  we  had  found  by  experiment  that  a  third 

p        string  o  E  (Fig.  10)  would  just  prevent  the 

*f^       two  strings  from  causing  motion  at  o,  then 

E__£ — c)      > —       experiment  would   also  show  that  the  force 

Fig.  10.  in  o  E,  which  may  be  called  the  equilibrant 

of  o  P  and  o  Q,  is  exactly  equal  and  opposite 

to  the  resultant  of  o  P  and  o  Q. 

We  see  that  when  OQ  and  OP  are  given,  and  the  angle 
between  them,  we  may  use  the  above  principle,  called  the 
parallelogram  of  forces ;  or,  what  comes  to  the  same  thing,  the 
triangle  of  forces.  Draw  o  Q  one  of  the  forces,  draw  Q  R  the 
other,  and  let  their  arrow-heads  be  circuital ;  then  the  non- 
circuital  force  o  R  is  the  resultant ;  or  a  circuital  force  R  o 
would  be  the  equilibrant.  It  is  in  this  way  that  we  find  the 
resultant  or  vector  sum  of  any  two  vectors. 

In  vector  language,  OQ  +  QR  =  OR,  or  OQ  =  OR-QR. 
This  principle  is  very  easy  to  express ;  to  be  able  to  apply  it 
implies  a  considerable  experience.  We  mention  it  now  merely 
to  introduce  a  few  exercises. 

It  is  very  easy  to  solve  problems  graphically.  The  student 
must  work  some  numerical  exercises,  and  test  his  answers 
graphically. 

Example. — Two  forces,  o  P  and  o  Q,  of  5  Ibs.  and  7  Ibs. 
respectively,  act  a  point,  at  an  angle  of  56° ;  find  their 
resultant  by  calculation. 

In  Fig.  11  OP  and  o  Q  represent,  to  scale,  the  two  forces, 
the  angle  P  o  Q  being  56°.  Find  the  resolved  part  of  o  P 
(Art.  31)  in  the  direction  o  x,  say  ;  it  is  o  A  =  o  P  cos.  56°  = 
5  x  -5592  =  2-796.  Now  obtain  the  resolved  part  of  o  P 
in  the  direction  o  Y,  taken  at  right  angles  to  o  x ;  it  is 
0  B  =  o  P  .  cos.  P  o  B  =  5  sin.  56°  =  5  x  -829  =  4-145.  In- 
stead of  the  given  forces  we  now  have  o  A  and  o  Q  acting 
along  ox,  and  OB  along  o  Y.  Draw  OH  (Fig.  11)  equal  to 
OA  +  OQ,  that  is,  9796,  and  ov  =  4-145.  The  resultant  of 
these  is  o  R,  and  we  have 

OR2  =  OH*  +  ov2  =  (9-796)2  +  (4-145)2  =  24-996; 

.  • .  o  R=5  very  nearly 


APPLIED    MECHANICS.  31 

To  obtain  the  direction  of  o  R,  we  have 

RH        4*145         ,-,„, 
tan.  a  =  —  =  ^=^  =  '4231  : 
OH       9-796 

.  • .  a  =  23°  nearly. 

The  student  should  test  this  result  by  finding  o  R  graphi- 
cally by  the  method  explained  above.  He  should  also  observe 
that  if  the  angle  P  o  Q  had  been,  say,  130°,  OA  would  have  been 
directly  opposed  to  o  Q,  in  which  case  o  H  would  have  been 
obtained  by  subtracting  o  A  from  o  Q.  It  is  of  the  utmost 
importance  that  he  should  work  many  exercises  similar  to 
this  one,  so  as  to  become  familiar  with  the  method. 

It  will  be  shown  later  (Art.  94)  that  the  same  method 
is  employed  for  calculating  the  resultant  of  any  number  of 


forces.  In  every  case  we  resolve  the  forces  along  any  line  o  x, 
and  let  o  H  represent  their  resultant ;  then  resolve  along  o  Y 
taken  at  right  angles  to  o  x,  and  let  ov  represent  the  resultant 
of  these  resolved  parts,  o  R  and  a  are  then  easily  calculated. 

EXERCISES. 

1.  Force  of  20  Ibs.  at  an  angle  of  72°  with  the  horizontal;  what  are 
its  horizontal  and  vertical  components?  Ans,,  6-18  Ibs.,  19-02  Ibs. 

2.  A  man  walks  towards  the  north-north-east  at  4  miles  per  hour ;  at 
what  rate  is  he  getting  towards  the  east  ?    And  at  what  rate  towards  the 
north ?  Ans.,  T53  miles  an  hour;  3'69  miles  an  hour. 

3.  Forces  o  p  of  10  Ibs.  and  o  Q  of  7  Ibs.  at  an  angle  a  o  p  of  35° ; 
find  the  resultant.     Test  your  answer  by  working  the  problem  graph- 
ically. Ans.,  16-24  Ibs.  inclined  14°  18'  with  the  force  10  Ibs. 

4.  On  a  horizontal  surface  there  is  a  normal  pressure  of  4  tons  per 
square  inch  and  a  tangential  force  of  3  tons  per  square  inch  from  N.E.  to 
S.W.     What  is  the  total  force  per  square  inch  ? 

Ans.,  An  oblique  pressure  of  5  tons  per  square  inch,  making  an  angle 
tan  ~  J  |  with  the  N.E.  to  S.W.  direction  in  a  vertical  plane. 


32  APPLIED    MECHANICS. 

5.  An  anvil  is  carried  by  three  ropes,  which  make  angles  20°,  30°,  25° 
with  the  vertical ;  the  tensions  in  the  ropes  are  known  to  be  1,000  Ibs., 
700  Ibs.,  and  1,200  Ibs.  What  is  the  weight  of  the  anvil?  If  the  hori- 
zontal components  of  these  three  forces  are  drawn  as  balancing  one 
another,  find  the  azimuthal  angles  which  the  vertical  planes  through  the 
ropes  make  with  each  other — that  is,  find  the  angles  in  the  plan  of  the  ropes. 

Ans.,  2,633  Ibs. ;  85°  47',  137°  43',  136°  30'. 
33.  If  the  magnitude  of  OP  (Fig.  11)  is  called  p,  of  OQ  is  Q, 
of  o  R  is  R,  and  if  the  angle  Q  o  p  is  0,  it  can  be  proved,  that 

R  =  \/p2  +  Q2  +  2  P  Q  cos.  0. 
It  will  be  seen  from  Fig.  1 1  that 

OB  p  sin.  0 

OH          O  A   +   OQ          P.  COS.  0  +  Q* 

If  0  is  a  right  angle,  P  and  Q  are  the  resolved  parts  of  s.  in 
their  two  directions — 

R  =  V  p2  +  Q2,  and  tan.  R  o  Q  =  P/Q. 

EXERCISES. 

1.  P  =  3,  Q  =  4,  0  =  90°.     Then  R  =  5  and  a   is   an  angle  whose 
tangent  is  0'75.     [The  neat  way  of  making  this  statement  is  to  write 
«  =  tan.-1  0-75.] 

2.  P  =  3-045,  Q  =  7-462,  0  =  37° ;  find  R,  a.  Ans.,  R  =  10-06. 

a  =  10°30/. 

3.  p  =  3-045,  Q  =  7-462,  0  =  143° ;  find  R,  a.  Ans.,  R  =  5-353. 

a  =  20°. 

4.  p  =  12-06,  Q  =  1-002,  6  =  184° ;  find  R,  a.  Ans.,  R  =  13-05. 

a  =  187°  42'. 

34.  A  river  flows  at  1  mile  per  hour ;  a  swimmer  has  a 
velocity  of  2  miles  per  hour  relatively  to  the  water.  What 
is  his  velocity  relatively  to  the  bank  1  This  will  depend  upon 
the  direction  in  which  he  swims. 

Make  A  B  represent  1  mile  per  hour  down  the  river  (as 
Borne  students  cannot  get  out  of  their  heads  the  wrong  notion 
that  these  lines  represent  distance,  imagine 
the    drawing  to   be    infinitely  small    but 
greatly  magnified  merely  that  it  may  be 
c  examined) ;  make  B  c  represent  the  velocity 
of  swimming  to  scale,  the  direction  and 
sense  being  correct.     Take  care  that  the 
Pig.  12.  arrow-heads  are  circuital.       Then  A  c  is 

the  sum  of  the  two  velocities  possessed 
by  the  swimmer,  and  is  therefore  his  velocity  relatively  to  the 
bank.  Let  the  student  draw  B  c  in  all  sorts  of  directions 
and  reflect  upon  his  answers.  If  he  has  ever  swam  in  a  broad 
river  or  has  watched  a  swimming  dog  trying  to  reach  his 
master,  he  will  understand  his  answers  more  readily. 


APPLIED    MECHANICS.  33 

EXERCISES. 

1.  Given  the  above  velocities  and  that  the  stream  flows  due  south,  if 
the  absolute  motion  of  the  swimmer  is  to  be  south-east,  in  what  direction 
ought  he  to  swim  ?  Ans.,  24°  18'  S.  of  E. 

2.  A  steamer  moves  westward  at  10  feet  per  second ;  a  boy  throws  a 
ball  across  the  deck  northwards  at  4  feet  per  second.  What  is  the  velocity 
of  the  ball  relatively  to  the  water  ? 

Ans.,  10-77  feet  per  second,  21°  48'  N.  of  W. 

3.  A  steamer  has  a  velocity  of  14  knots  due  west ;  the  wind  blows  with 
a  velocity  of  7  knots  from  the  north.     What  will  be  the  apparent  velocity 
of  the  wind  to  a  person  on  board  the  steamer  ? 

Ans.,  15-7  knots  from  W.  26°  34'  N. 

4.  Velocity  of  a  ship  westwards  10  feet  per  second  ;  velocity  of  a  ball 
on  deck  5  feet  per  second  north-east  relatively  to  the  ship.     What  is  the 
total  velocity  of  the  ball  ?    Ans.,  7 '37  feet  per  second,  28°  40'  N.  of  W. 

5.  If  the  total  velocity  of  the  ball  is  12  feet  per  second  northwards, 
what  is  its  velocity  relatively  to  the  ship  ? 

Ans.,  15-62  feet  per  second,  50°  12'  N.  of  E. 

6.  A  railway  train  is  going  at  30  feet  per  second ;  how  must  a  man 
throw  a  stone  from  the  window  so  that  it  shall  leave  the  train  laterally  at 
1  foot  per  second,  but  have  no  velocity  in  the  direction  of  the  train's 
motion  ? 

Ans.,  30-02  feet  per  second ;  at  an  angle  of  178°  6'  with  direction  of 
motion  of  train. 

35.  A  bicyclist  is  ordered  to  travel  so  that  he  shall  be  more  to 
the  north  at  the  rate  of  3  miles  every  hour,  and  he  must  keep  to 
roads.     Notice  that  if  he  is  on  a  north  and  south  road  his  task  is 
very  easy.     If  his  road  is  directed  N.W.,  he  must  travel  at  4-242 
miles  per  hour.     If  his  road  is  W.N.W.,  he  must  travel  at  7'839 
miles  per  hour.     If  his  road  is  due  west,  his  task  is  an  impossible 
one.     If  his  road  makes  an  angle  6  with  due  north,  he  must  travel 
at  the  rate  of  3  -f-  cos.  0  miles  per  hour. 

36.  Water  flowing  from  the  inner  to  the   outer  part  of    a 
motionless  wheel  of  a  centrifugal  pump  is  guided  by  vanes  to 
follow  a  curved  path.     Suppose  its  radial  velocity  vr  known ;  show 
that  its  real  velocity  anywhere  is  vr  -r-  cos.  6,  if  0  is  the  angle 
which  the  vane  there  makes  with  the  radial  direction. 

A  student  will  find  this  an  excellent  graphical  exercise.    Draw 
circles  of   1   and   2   feet  radii  to  represent  the  inner  and  outer 
cylindric  surfaces    of    the  wheel    of   a 
pump.   Draw  any  shape  of  vane  connect- 
ing these,  but  you  had  better  take  a 
shape  from  an   actual  wheel   (see  Art. 
427).  Let  the  angle  at  A  with  the  tangent 
to  the  circle  there   (Fig.  13)  be  18 J°. 
Now  imagine  a  particle  of  water  travell- 
ing out  radially  at  0-1  foot  per  second.  p.     J8 
Imagine  it  to  take  10  seconds  to  get  to  Q. 

Mark  its  successive  positions  along  the   vane.     You  had  better 
also  trace  its  positions  backward  for  a  few  seconds  towards  A  from  s. 
Now  imagine  the  wheel  to  revolve  about  its  centre  so  that  A  moves 
C 


34 


APPLIED    MECHANICS. 


at  0-3  foot  per  second.  Map  out  the  real  positions  of  the  points 
which  you  found  on  the  vane,  and  therefore  the  real  path  of  a 
particle  of  water.  Try  to  follow  it  after  it  has  left  a,  assuming 
plenty  of  space  outside ;  but  this  is  a  problem  which  you  perhaps 
may  return  to  later. 

37.  Let  P  (Fig.  14)  be  a  point  on  the  rim  of  the  wheel  of  a 
centrifugal  pump.     Suppose  that  we  know  the  velocity  of  r,  and 
represent  it  by  the  distance  P  B  ;  also  that  we  know  the  velocity 
of  the  water  along  the  vane  of  the  wheel  at  p,  and  represent  it 
by  the    distance    A  P.      Now,   a    particle    of 
water  at  p    has    both    these  velocities ;    A  p 
relatively  to    the    wheel,   together    with  p  B 
because  the  wheel  is  in  motion.      Hence  the 
total  velocity  of  the  particle  of  water  is  repre- 
sented in  direction,  amount,  and  sense  by  the 
vector  sum  A  p  -\-  p  B,  and  we  can  either  use 
the    parallelogram    method    or    the    triangle 
method  to  find  it. 

Exercise. — The  rim  of  the  wheel  of  a  centrifugal  pump  goes  at  30  feet 
per  second ;  water  flows  radially  at  5  feet  per  second ;  the  vanes  are 
inclined  backward  at  an  angle  of  35°  to  the  rim.  What  is  the  absolute 
velocity  of  the  water?  What  is  the  component  of  this  parallel  to  the 
rim  ?  Ans.,  23'4  feet  per  second ;  22'8  feet  per  second. 

When  we  know  the  velocity  of  water  before  it  enters  a 
wheel  and  the  velocity  of  the  wheel  at  the  place,  and  we  wish 
the  water  to  enter  without  shock,  this  simply  means  that  the 
total  velocity  of  the  water  the  instant  after  it  enters  the  wheel 
shall  be  exactly  the  same  as  before  it  entered.  Thus  in  Fig.  15, 
if  the  velocity,  c  P,  of  water  is  known  before  it  enters  the  turbine 
wheel,  and  p  B  represents  the  velocity  of  the  wheel,  find  the 
velocity  which,  added  to  P  B,  will  have  c  P  for  a  resultant.  It  is 
p  A  if  P  Q,  represents  the  velocity  c  p  to  scale.  Make,  then,  p  A  the 
direction  of  the  vane  at  P.  The  water  will  flow  in  the  direction  p  A 
c  relatively  to  the  wheel,  and  it  has  also  the 

"X  velocity  p  B  because  it  moves  with  the  wheel ; 

its  total  velocity  is  the  same  as  before  it 
entered,  and  it  has  entered  without  shock. 
We  don't  much  care  how  the  vane  curves 
afterwards  so  long  as  it  curves  gradually ; 
it  is  its  direction  at  p  that  is  important. 

Exercise. — The  inner  circumference  of  a 

;  centrifugal  pump  wheel  goes  at  15  feet  per 

;  FJ(,  15  second  ;  water  approaches  it  radially  at  5  feet 

per  second.     What  is  the  angle  of  the  vanes 
if  the  water  is  to  enter  without  shock?  Ans.,  18°  26'. 

Note  that  the  angle  chosen  by  us  for  the  vane  at  A  (Fig.  13) 
enabled  the  water  to  enter  the  wheel  without  shock.  We  may  at 
once  say  that  it  is  only  the  angles  at  A  and  at  Q  that  are  of  real 


APPLIED   MECHANICS.  35 

importance  in  the  design  of  a  pump.  The  actual  shape  of  the  vane 
is  unimportant  so  long  as  the  curve  is  fairly  direct.  But  the 
exercise  ought  to  be  worked  carefully  throughout. 

We  made  the  tangent  of  the  angle  at  A  equal  to  the  radial 
velocity  •!,  divided  by  the  velocity  of  the  wheel  at  A  (see  Art.  33). 
We  always  endeavour  to  have  this  sort  of  relation  true  at  the  inner 
side  of  the  wheel  of  a  centrifugal  pump  or  turbine. 

38.  Usually  we  consider  the  earth  and  frame  of  a  machine 
to  be  fixed.  The  student  will  find  it  very  instructive  to  think 
of  some  link  or  wheel  as  fixed  (which  he  usually  thinks  of  as 
moving)  and  now  note  what  the  motions  are.  One  simple 
example  of  this  is  given  in  Art.  467,  and  we  there  show  that 
to  understand  the  relative  motions  in  a  four-link  mechanism 
is  to  understand  the  motions  in  a  very  great  number  of  other 
mechanisms. 

Train  A  is  passing  train  B.  Looked  at  by  an  observer  on 
the  ground,  how  very  different  is  their  appearance  from  what 
it  is  to  an  observer  in  either  train ! 

The  mathematics  of  the  subject  is  quite  easy.  It  is  simply 
this  :  If  a,  b,  c,  etc.,  are  lines  drawn  representing  what  may  be 
called  the  absolute  displacements  of  points  A,  B,  c,  etc.,  or  the 
rotations  of  bars  about  axes,  then  their  displacements  relatively 
to  a  frame  F,  whose  own  absolute  displacement  or  rotation  is 
/,  are  a  — /,  b  — f,c  — f,  etc.,  the  —  sign  meaning  vector 
subtraction. 

But  mathematics  does  not  satisfy  us  ;  we  want  the  instinct 
of  comprehending  easily  these  relative  motions.  That  we  do 
not  possess  it  is  evidenced  by  the  fact  that  all  the  mechanisms 
of  Art.  467  do  really  seem  to  us  different,  and  that  we  need  to 
give  the  name  epicyclic  train  to  a  train  of  wheels  when  the 
framework  which  connects  their  centres  is  allowed  to  move 
instead  of  remaining  at  rest,  as  it  does  in  our  usual  way  of 
studying  things.  Notice  that  this,  which  ought  to  be  an  easy 
subject,  follows  in  smaller  printing. 

39.  Thus,  for  example,  in  Fig.  16  we  have  a  train  of  three  wheels 
(the  student  ought  to  take  two  or  four  or 
more).  When  the  frame  F  is  at  rest  (that 
is,  we  take  speeds  relatively  to  the  frame) 
let  A,  B,  c  have  the  angular  velocities  o, 
ba,  co.  (evidently  in  the  figure  b  is  negative). 
Now  let  F  have  an  absolute  angular  velo- 
city, /,  and  the  absolute  velocities  of  the 
wheels  are  a  +  /,  ba  +  /,  ca  +  f. 


36  APPLIED   MECHANICS. 

1.  Suppose  A  is  at  rest  ;  o  +  /=  0,  a  =  —  /.  B'S  velocity  is 
(_  b  -f  I)/;  c's  velocity  is  (-  c  +  I)/.  Thus,  let  c  have  the 
same  number  of  teeth  as  A  ;  so  that  c  =  1,  c's  velocity  is  0. 
If  c  has  100  teeth,  and  alongside  it  on  what  is  practically  the  same 
spindle,  let  there  be  wheels  of  99  and  101  teeth  also  gearing  with  B. 
Evidently  the  absolute  speeds  of  the  three  will  be  (since  c  has  the 


the  arm  goes  round  and  the  motions  of  the  three  wheels  are 
observed,  we  call  this  Ferguson's  paradox.  If  we  could,  like  flies, 
move  with  the  arm,  there  would  be  nothing  paradoxical  about  it. 

Notice  that  A,  B,  and  c  may  be  bevel  wheels,  as  shown  in  Fig.  17. 

2.  Simpler  Case.  —  Wheels  A  and  B  connected  by  arm  F  ;   arm 


Fig.  17. 

rotates.  The  absolute  rotational  velocity  of  B  is  0.  What  is  A'S 
velocity  ? 

Here  ba  -f  /  =z  0.     Hence  a  =  —  fib  ;  so  that  A'S  velocity  is 

_  fjb  +f  or  /  fl  —  IY      Thus,  let   b  =  —  1,    as    it    is    in 

Watts'  sun  and  planet  motion;  A'S  velocity  is  2/.  In  this 
case,  however,  because  of  the  angularity  of  the  connecting- 
rod,  B  has  some  angular  velocity,  fluctuating  between, 
say,  -f-  £  and  —  £.  Hence  (since  b  =  —  1),  —  a  +/=  +  0, 
and  A'S  velocity  is  2/  +  j8. 

3.  In  Fig.  17  we  have  A,  B,  and  c,  three  bevel  wheels.  B  may 
be  carried  round  the  axes  of  the  others  on  a  spur  wheel  D.  Suppose 
we  are  looking  from  the  point  D  at  every  wheel,  and  the  numbers 
of  teeth  on  A  and  c  are  as  1  to  c,  then  relatively  to  D  the  speeds  of 
A  and  c  are  a  and  —  ac.  If  D  rotates  at  speed  /,  the  absolute 
velocities  of  A  and  c  are  a  +  /  and  -  ac  +  /. 

Example. — Let  c  =  1 ;  then,  if  A'S  speed  a  +/  is  called  a, 
so  that  o  =  a  -  f,  c's  speed  is  -  o  +  /  or  —  a  +  2/.  Thus, 


APPLIED    MECHANICS. 


37 


suppose  A  goes  at  20  revolutions  per  minute  or  a  =  20,  then  c's 
speed  is  —  20  +  2/;  so  that  if  /  is  gradually  changed  in  the 
following  way,  we  get  the  following  speeds  for  c  shown  in  the 
table  :— 

Again,  we  may  imagine  the 
speed  of  F  to  keep  constant  and 
that  of  A  to  vary,  and  we  get  the 
same  result. 

By  means  of  a  cam  we  may 
gradually  change  from  negative 
to  positive  velocities,  hut  here  we 
have  a  very  much  better  means 
by  ordinary  gearing. 

In  Fig.  17,  if  the  shaft  of  A  is 
rotated  by  coned  pulleys,  by  which 
it  may  be  given  varying  speeds, 
and  if  A  is  kept  rotating  at  a  fixed 
speed,  c  gets  speeds  which  may  be 
negative  or  positive  or  zero. 


Speed  of  F. 

Speed  of  C. 

4 

-  12 

6 

-  8 

8 

—  4 

10 

0 

12 

4 

14 

8 

16 

12 

18 

16 

20 

20 

Exercise. — An  epicyclic  gear  consists  of  an  annular  wheel  A  of  72  teeth, 
a  pinion  B,  and  a  spur  wheel  c  of  40  teeth  concentric  with  A.  The  arm 
which  carries  the  axis  of  B  makes  30  revolutions  per  minute.  (1)  If  A 
be  a  dead  wheel,  find  the  revolutions  of  c.  (2)  If  c  be  a  dead  wheel,  find 
the  revolutions  of  A.  Ans.,  (1)  84  ;  (2)  46|. 

Again,  if  A  makes  4  revolutions  and  c  6  revolutions  in  the  same 
direction,  find  the  revolutions  of  the,  arm.  Ans. ,  4|. 

Exercise. — In  a  horse-gear  for  driving  a  chaff-cutter,  the  bracket 
that  holds  the  pole  supports  also  a  short  horizontal  shaft  carrying  a  bevel 
wheel  of  31  teeth  and  a  bevel  pinion  of  16  teeth.  The  pinion  gears  into 
a  horizontal  bevel  ring  of  80  teeth  that  is  stationary,  and  forms  part  of 
the  framing.  The  bevel  wheel  of  31  teeth  also  gears  with  a  bevel  pinion 
of  22  teeth  which  is  loose  on  the  central  vertical  axis,  and  this  pinion 
carries  with  it  a  bevel  wheel  of  60  teeth  that  gears  with  a  pinion  of 
16  teeth  on  the  high  speed  horizontal  shaft.  Find  the  number  of  revolu- 
tions of  the  high  speed  shaft  for  each  circuit  of  the  horse. 


38 
CHAPTER    III. 

WORK     AND     ENERGY, 

40.  Work. — To  do  work  it  is  necessary  to  exert  a  force 
through  a  certain  distance  in  the  direction  of  the  force.  Thus,  if 
we  exert  a  force  of  20  Ibs.  through  a  distance  of  6  feet,  we  do 
20  x  6,  or  1 20  foot-pounds  of  work.  If  a  body  of  5  Ibs.  weight 
changes  its  level  by  the  amount  of  10  feet,  whether  it  does  this 
by  a  direct  vertical  fall  or  rise,  or  is  moved  up  or  down  an 
inclined  plane  or  curved  surface,  so  long  as  there  is  no  friction, 
the  amount  of  work  given  out  by  the  body  in  falling  or  given 
to  it  to  make  it  rise  is  always  the  same,  5  x  10,  or  50  foot- 
pounds. 

Example. — The  weight  in  a  certain  clock  is  20  Ibs.,  and 
after  being  wound  up  it  can  fall  through  a  distance  of  40  feet. 
Suppose  we  wish  to  alter  this  height,  making  it  10  feet ;  what 
weight  must  we  use  1  Evidently  the  work  given  out  by  the 
new  weight  in  falling  10  feet  must  be  equal  to  that  given  out  by 
the  old  weight,  or  800  foot-pounds.  In  fact,  the  new  weight 
must  be  80  Ibs.  Of  course  we  must  apply  this  weight  to  the 
clock  by  means  of  a  block  and  pulleys,  or  we  must  reduce  the 
diameter  of  the  drum  proportionately ;  and  if  in  applying  it  we 
introduce  more  friction  than  there  used  to  be  in  the  clock,  we 
must  further  increase  the  weight,  so  as  to  be  able  to  overcome 
this  friction. 

The  work  done  by  a  force  is  well  illustrated  by  the  pulling 
of  a  tramcar.  If  the  pulling  force  P  Ibs.  is  not  directly  along 
the  track,  but  makes  an  angle  6  with  it,  the  effective  force,  or 
the  resolved  part  of  p  in  the  direction  of  motion  is  P  cos.  0, 
and  this,  multiplied  by  the  distance  moved  through  in  feet,  is 
the  work  done  in  foot-pounds. 

When  a  body  is  pulled  up  a  curve  the  work  done  in  over 
coming  the  force  of  gravity  (we  are  neglecting  work  spent  in 
overcoming  friction)  is  simply  the  weight  of  the  body  multiplied 
by  the  difference  in  level. 

Thus  in  Fig.  18  the  work  done  in  moving  a  body  of  weight  vr 
from  A  to  B  along  the  curve  is  simply  wy,  where  y  is  the  difference 
in  level  of  A  and  B. 

Proof : — Let  the  co-ordinates  of  any  point  p  be  x,  y  ;  and  of  <a,  a 
point  indefinitely  near  to  P,  x  -f  8#,  y  +  5y. 


APPLIED  MECHANICS. 


The  weight  w  resolved  in  the  direction  of  the  tangent  at  P  is  w 
bin.  0,  and  this  multiplied  by  the  distance  P  a  (we  are  supposing 
that  the  tangent  at  Q,  is  parallel  to  the  tangent  at  P,  as  Q  is 
supposed  to  be  indefinitely  near  to  P),  or  P  Q  .  w  sin.  0  is  the  work 
done  against  gravity  in  pulling  the  body  from  P  to  a.  Therefore 
the  whole  work  done  in  pulling  the  body  from  A  to  B  is  the  sum 


A  N 

Fig.  18. 

of  all  such  terms  as  i>  a  .  w  sin.  0.     But  P  a  sin.  0  is  Q  B,  or  by,  or 
in  the  limit  dy.     Therefore  the  whole  work  done  is 


f 

J  C 


•w  dy  or  wy. 


Gravity  does  this  work  when  the  body  falls.  But,  both 
when  a  body  is  moved  up  and  down,  energy  is  wasted  or  con- 
verted into  heat  in  overcoming  friction.  Hence,  when  a 
weight,  w  Ibs.,  is  lifted  in  level  h  feet,  the  useful  work  done  is 
\\h,  but  there  is  energy  wasted.  Again,  when  w  falls, 
gravity  does  the  work  wA,  but  there  is  energy  wasted.  If 
we  are  depending  upon  the  total  store  wA  to  drive  machinery, 
the  useful  work  done  is  less  than  wA,  the  difference  being 
wasted,  or  rather  converted  into  heat,  by  friction. 

41.  Horse-power.  —  One  horse-power  is  the  work  of 
33,000  foot-pounds  done  in  one  minute.  Power  means  not 
merely  work,  but  work  done  in  a  certain  time  ;  the  time 
rate  of  doing  work.  The  work  done  in  one  minute  by  any 
agent  divided  by  33,000  is  the  horse-power  of  that  agent. 
In  a  steam-engine  the  piston  travels  four  times  the  length 
of  the  crank  in  one  revolution,  and  all  this  time  it  is 
being  acted  upon  by  the  pressure  of  steam.  If  the  mean  or 
average  pressure  urging  the  piston  is  60  Ibs.  per  square  inch, 
and  the  area  of  the  piston  is  150  square  inches,  then  the  total 
average  force  urging  the  piston  is  150  x  60,  or  9,000  Ibs.  If  the 
crank,  whose  length  is  0'9  foot,  makes  70  revolutions  per 


40  APPLIED   MECHANICS. 

minute,  then  the  piston  travels  4  times  0'9  x  70,  or  252  feet 
per  minute,  so  that  the  work  done  in  one  minute  is  9,000  x  252, 
or  2,268.000  foot-pounds.  Dividing  this  by  33,000,  we  find 
the  hors-e-power  of  the  steam-engine  to  be  68 '7.  The  mean 
pressure  is  best  found  by  the  use  of  an  indicator  which  draws 
for  us  an  indicator  diagram.  Measuring  the  pressures  at  ten 
equidistant  places  on  this  diagram,  adding  them  together,  and 
dividing  by  ten,  gives  the  average  pressure.  Or,  measure  the 
area  by  means  of  a  planimeter  in  square  inches  ;  divide  by  the 
extreme  length  of  the  diagram  parallel  to  the  atmospheric  line; 
this  gives  the  average  breadth,  and  therefore  the  average 
pressure  to  scale. 

As  the  pressure  of  steam  is  usually  given  per  square  inch, 
it  is  usual  to  take  the  diameter  of  the  cylinder  in  inches,  but 
distances  passed  through  by  the  piston  are  evidently  to  be 
measured  in  feet. 

Example. — We  find  by  a  spring  balance  that  some  horses 
or  a  steam-engine  have  been  pulling  a  carriage  with  an  average 
pull  of  120  Ibs,  during  one  minute,  the  space  passed  over  in 
the  minute  being  500  feet ;  what  is  the  horse-power  expended 
on  the  carriage  ?  Here  120  Ibs.  act  through  the  distance  of 
500  feet,  and  the  work  done  in  one  minute  is  evidently  500  x 
1 20,  or  60,000.  Dividing  by  33,000,  we  find  the  horse-power  to 
be  1-818. 

42.  Energy  is  the  capability  of  doing  work.  When  a  weight 
is  able  to  fall,  it  possesses  potential  energy  equal  to  the  weight 
in  Ibs.  multiplied  by  the  change  of  level  in  feet  through  which 
it  can  fall.  When  a  body  is  in  motion,  it  possesses  kinetic 
energy  equal  to  half  its  mass  (its  weight  in  London  in  pounds 
divided  by  32-2  is  its  inertia,  which  is  usually,  but  we  think 
unwisely,  called  its  mass),  multiplied  by  the  square  of  its 
velocity  in  feet  per  second.  (See  Art.  190.) 

Example. — A  body  of  60  Ibs.  is  100  feet  above  the  ground, 
and  has  a  velocity  of  150  feet  per  second;  what  is  its  total 
amount  of  mechanical  energy? — that  is,  what  energy  can  it 
give  out  before  it  reaches  the  ground,  and  becomes  motionless  ? 
Here  the  potential  energy  is  60x100,  or  6,000  foot-pounds. 
Its  kinetic  energy  is  60  x  150  x  150 -f- 64-4,  or  20,963  foot- 
pounds. So  that  the  total  amount  is  26,963  foot-pounds. 

Suppose  this  body  to  lose  no  energy  through  friction  with 
the  air,  and  suppose  that,  after  a  time,  it  is  at  a  height  of  20  feet 
above  the  ground ;  find  its  velocity.  Answer  :  Its  potential 


APPLIED    MECHANICS.  41 

energy  is  now  60x20,  or  1,200  foot-pounds,  therefore  its 
kinetic  energy  must  be  25,763 ;  and  evidently  this,  multiplied 
by  64-4  and  divided  by  60,  gives  27,652*29,  the  square  of  the 
new  velocity.  Its  velocity  is  therefore  166-3  feet  per  second. 
In  such  a  question  we  are  not  concerned  with  the  direction  in 
which  the  body  is  moving.  It  may  be  a  cannon-ball,  or  a  falling 
or  rising  stone;  or  the  bob  of  a  pendulum.  Given  its  velocity 
and  height  at  any  instant,  we  can  find  for  any  other  height 
what  its  velocity  must  be,  or  for  any  other  velocity  what  its 
height  must  be. 

On  a  Switchback  railway,  the  loss  of  energy  (roughly  pro- 
portional to  distance  travelled)  every  journey  is  represented 
by  the  lift  of  a  few  feet  which  is  effected  by  the  attendant 
at  each  end  before  making  a  fresh  start.  Neglecting  this  loss, 
it  is  easy  to  calculate  the  velocity  at  any  place  if  we  know  the 
vertical  depth  of  the  place  below  the  starting  point.  The 
deepest  places  on  the  line  are  places  of  greatest  velocity; 
the  highest  places  are  those  of  least  velocity.  In  a  bob  of  a 
pendulum  we  see  a  continual  conversion  of  kinetic  into 
potential  and  of  potential  into  kinetic  energy,  the  total  store 
remaining  constant,  except  in  so  far  as  friction  is  converting 
mechanical  energy  into  heat.  (See  Art.  192.) 

The  energy  of  a  strained  body — for  example,  a  strained 
spring — is  another  store  of  mechanical  energy.  It  is  an  excellent 
laboratory  experiment  to  measure  how  much  energy  a  spring 
can  store  without  getting  permanently  hurt  in  shape  or  broken. 
This  store  of  energy  is  called  its  resilience.  As  the  elongation 
of  a  spring  is  proportional  to  the  load,  if  we  gradually  increase 
the  load  from  0  to  w  Ibs.,  the  •  elongation  increases  from  0  to 
x  feet,  so  that  the  average  force  for  the  whole  length  being 
^  w,  the  energy  stored  is  f  w  x  foot-pounds. 

A  weight  vibrating  vertically  at  the  end  of  a  spiral  spring 
gives  a  good  example  of  the  continual  conversion  of  the  three 
kinds  of  mechanical  energy  into  one  another.  The  total  store 
gradually  diminishes,  partly  by  friction  in  the  atmosphere,  but 
greatly  also  by  an  internal  frictional  resistance  or  viscosity  in 
the  material  of  the  spring.  The  student  will  find  it  interesting 
to  compare  the  behaviour  of  a  spring  of  steel  and  a  spring  of 
indiarubber  in  this  respect ;  the  viscosity  of  the  indiarubber 
damps  the  vibrations  with  great  rapidity.  Many  hours  may 
be  well  spent  in  studying  these  phenomena,  making  quantita- 
tive measurements. 


42  APPLIED   MECHANICS. 

Air  and  other  fluids  in  a  compressed  condition  contain 
stores  of  energy. 

43.  A  student  may  employ  his  leisure  in  calculating  the 
possible  stores  of  energy  in  1  lb.  of  various  materials :  1  Ib.  of 
hydrogen,  4'8  x  107  foot-pounds;  kerosene,  2  x  107  foot-pounds; 
coal,   12  x  106  foot-pounds  (the    weight  of  oxygen  or  air  for 
combustion  is  not  counted  in) ;    1  lb.  of  cast  iron  in  rim  of 
pulley  at  highest  speed  to   produce  greatest  working  stress, 
1,000  foot-pounds  ;   steel  spring  of  the  best  kind,  270  foot- 
pounds ;  usual  spiral  spring  of  round  wire,  135  foot-pounds. 

The  heat  given  to  1  lb.  of  water  to  raise  it  from  0°  to  lc  C. 
or  from  99°  to  100°  0.  in  temperature,  is  nearly  the  same. 
This  is  a  unit  of  heat  energy.  Joule  showed  that  this  is 
equivalent  to  1,400  foot-pounds.*  When  energy  is  in  the  heat 
form,  a  heat-engine  may  be  used  to  convert  part  of  it  into 
mechanical  energy ;  unfortunately  the  amount  convertible, 
depends  upon  how  high  is  the  temperature  of  the  stuff  contain- 
ing the  heat  energy  above  the  temperature  of  the  refrigerator 
or  exhaust.  Hence  it  is  that  in  the  very  best  steam-engines  we 
seldom  convert  more  than  one-seventh  of  the  heat  energy  of 
the  steam  into  mechanical  energy,  the  other  six-sevenths  being 
degraded  in  temperature,  and  going  to  the  refrigerator  useless 
for  our  purposes.  In  discussing  heat-engines  we  express  all 
energy  in  foot-pounds. 

The  energy  store  is  most  intense  as  to  volume  (we  do  not 
include  the  weight  of  the  air  needed  for  combustion)  in  a  pound 
of  kerosene,  being  about  half  as  much  again  as  in  a  pound  of 
coal.  We  look  forward  to  the  time  when  no  heat-engine 
will  be  needed  in  the  conversion,  and  we  may  then  be  able  to 
convert  over  90  per  cent,  of  the  energy  of  a  pound  of  kerosene 
into  the  mechanical  form,  as  at  the  present  time  the  chemical 
energy  of  zinc  is  convertible  in  a  battery  or  electric  motor,  or 
the  chemical  energy  of  oats  and  other  food  is  convertible  in  the 
animal  machine,  which  is  probably  a  gas  battery  and  electric 
motor. 

44.  When  the  engine  of  the  Finsbury  College  is  working 
mainly  for  the  electric  light  and  possibly  one  electric  motor,  the 
students  have  sometimes,  during  a  long-continued  measurement, 
been  able  to  trace  what  becomes  of  the  energy  of  one  pound 

*  The  latest  determination  for  the  mean  specific  heat  of  water  from  0°  C. 
to  100°  C.  is  1399  foot-pounds,  or  for  1  gramme  it  is  0'995  calorie  ;  1  calorie, 
the  heat  to  raise  1  gramme  of  water  from  10°  C.  to  11°  C.  is  4 '2  Joules  or 
4'2  X  107  ergs.  The  heat  from  20°  C.  to  21°  C.  is  l-30th  of  1  per  cent.  le^s. 


APPLIED    MECHANICS.  43 

of  coal.  They  had  previously  measured  the  chemical  energy 
in  a  pound  of  coal,  and  tested  all  sorts  of  instruments  and 
ideas  used  in  their  measurements. 

The  energy  of  1  Ib.of  coal  is,  say,  12,000,000  foot-pounds; 
of  this  4,000,000  go  up  the  chimney  or  get  wasted  by  radiation 
as  heat,  and  8,000,000  foot-pounds  of  heat  energy  reach  the 
engine  in  steam ;  of  this  only  one-thirteenth,  or  600,000  foot- 
pounds is  converted  into  mechanical  energy  and  given  to  the 
piston ;  the  other  twelve-thirteenths  go  off  to  the  condenser  and 
are  wasted.  Only  500,000  are  given  out  by  the  engine  to  the 
long  shaft  which  drives  the  dynamo  machine,  about  400,000 
foot-pounds  leave  the  dynamo  as  electric  energy,  and  part  is 
wasted  by  conversion  into  heat  in  the  conductors  to  the  lamps. 
At  the  lamps  we  have  about  370,000  foot-pounds  of  electric 
energy  converted  into  heat  and  light.  If  any  of  the  electric 
energy  or  all  of  it  is  given  to  an  electric  motor,  perhaps  more 
than  90  per  cent,  of  it  will  be  converted  into  mechanical 
power. 

Men  who  make  measurements  of  this  kind  get  to  have  very 
clear  ideas  as  to  the  various  forms  of  energy  and  the  fact  that 
it  is  indestructible,  and  cannot  be  wasted,  although  it  may 
alter  into  forms  in  which  it  may  not  be  available  for  use ;  and 
therefore  we  say  that  it  is  wasted. 

The  very  best  steam-engines  use  more  than  1J  Ibs.  of  coal 
per  hour  for  each  horse-power  given  out.  We  cannot  hope  for 
much  improvement;  this  is  about  one  useful  for  nine  total. 
Gas-engines  using  Dowson  gas  already  give  out  one  horse- 
power for  1  Ib.  of  coal  consumed  per  hour ;  we  hope  for  con- 
siderable improvement.  Oil-engines  give  out  one  horse-power 
for  less  than  0-9  Ib.  of  kerosene  per  hour;  we  hope  for  very 
considerable  improvement. 

45.  Students  ought  to  work  many  numerical  exercises  on  me- 
chanical energy :  w  Ib.  of  water  raised  vertically  h  feet,  the  energy 
is  w/i  foot-pounds.  If  time  is  given,  find  the  work  done  per 
minute  and  divide  by  33,000;  this  is  useful  horse-power.  Divide 
this  by  the  efficiency  of  a  pump,  and  we  have  the  actual  horse- 
power which  must  be  supplied  to  the  pump.  Or,  if  the  w  Ibs. 
of  water  fall  per  minute  through  a  turbine,  the  head  available 
being  h  feet,  we  have  wh  ~  33,000  as  the  total  horse-power,  and 
the  turbine  will  probably  give  out  70  per  cent,  of  this  usefully, 
0'70  or  70  per  cent,  being  the  efficiency  of  a  good  turbine. 

If  there  were  no  friction,  a  waggon  of  weight  w  requires  to 


44  APPLIED    MECHANICS. 

be  pulled  with  a  force  of  w-j-s  up  a  road  which  rises  1  foot  in 
every  s  feet  of  its  length.  The  frictional  resistance  to  motion 
of  a  vehicle  on  a  level  road  is  usually  stated  as  so  many  pounds 
per  ton  weight  of  the  vehicle.  If  there  is  also  an  incline  wo 
add  the  two  tractive  forces  together — one  for  the  incline  without 
friction,  the  other  to  overcome  friction  on  the  level.  The 
resistance  in  pounds  per  ton  of  a  moving  train  (including 
engine)  on  the  level  is  found  roughly  by  adding  2  to  one- 
quarter  of  the  speed  in  miles  per  hour.  This  is  for  speeds 
greater  than  20  miles  per  hour.  At  less  speeds  there  is  quite 
a  different  law,  which  may  for  some  trains  and  permanent 
ways  be  indicated  by  the  following  figures  : — 


Speed  in  miles  per      j 
hour.                        ) 
Resistance  in  pounds  } 
per  ton.                    ( 

0 

20 

If 
10 

2 

7 

5 

5 

10 
6 

A  curved  line  adds  1 2  per  cent,  to  the  resistance  on  the  average 
English  railways.  The  tractive  force  of  heavy  waggons  on 
macadamised  roads  may  be  taken  as  50  Ibs.  per  ton,  on  paved 
roads  30  Ibs.  per  ton,  and  on  gravel  roads  as  150  Ibs.  per  ton. 
These  rules  are  good  enough  for  academic  exercise  work. 

If  the  pull  on  a  tramcar  is  recorded  as  the  ordinate  of  a 
diagram,  of  which  the  abscissa  represents  distance  along  the 
track,  the  area  of  the  diagram  represents  to  some  scale  the 
work  done.  The  average  height  of  the  diagram  represents  the 
average  pull  p.  This  average  pull  p,  multiplied  by  the  whole 
length  of  the  track,  is  the  whole  work  done.  In  most  practical 
cases  the  average  can  only  be  obtained  by  making  the  diagram, 
finding  its  area,  and  dividing  by  its  length,  just  as  we  do  with 
an  indicator  diagram  of  an  engine.  But  there  are  cases  where 
we  can  calculate  an  average. 

Example. — A  chain  of  length  I  and  weight  wl,  with  a 
weight  w  at  the  end  of  it,  is  to  be  wound  up  by  a  capstan ; 
what  work  will  be  done  ?  Obviously  the  average  pull  is  w-f- 
half  the  weight  of  the  chain,  or  w  + J  wl,  and  the  total  distance 
is  I,  so  that  the  work  done  is  wl  + 1  wl2.  As  a  rule,  it  is  wise 
to  plot  the  varying  pull  as  the  ordinate  of  a  curve  on  squared 
paper,  as,  without  the  aid  of  the  calculus,  one  is  apt  to  state  what 
is  the  average  pull  without  due  thought.  Thus,  if  the  above- 


APPLIED    MECHANICS.  45 

mentioned  chain  varies  in  heaviness  per  foot,  the  average  pull 
due  to  it  is  not  half  its  weight. 

Observe  that  we  ought  to  call  the  above  the  space  average 
of  a  force.  The  space  average  of  a  force,  multiplied  by  the 
distance,  is  the  work  done. 

A  little  consideration  will  show  a  student  that  a  time 
average  is  a  very  different  thing. 

46.  According  to  the  results  of  some  experiments  by  Prof. 
R.  H.  Smith  on  the  cutting  of  metal  in  the  lathe  without 
water  or  oil,  the  force  on  the  tool  is  not  much  affected  by 
speed. 

For  both  thin  and  moderately  thick  shavings  at  all  speeds, 
feeds,  and  depths  of  cut,  we  may  roughly  take  it  that  forged 
steel  takes  twice  as  much  power  to  cut  it  as  does  cast  iron; 
wrought  iron  takes  one  to  one  and  a  half  times  as  much  as  cast 
iron. 

For  broad  thin  shavings,  cast  iron  required  more  cutting 
force  than  wrought  iron. 

The  force  is  neither  proportional  to  the  breadth  of  the 
shaving  nor  the  depth,  but  it  is  more  nearly  proportional  to 
depth  than  breadth. 

It  is  interesting  to  note  that  before  these  experiments  it 
was  usual  in  books  to  follow  Weisbach  in  saying  that  for  iron 
P  =fbd  where  f  was  50,000  Ibs.  per  square  inch.  Smith's 

P 
experiments  show  that  this  rule  is  not  true,  and  that  —  varies 

Id 

from  92,000  to  239,000.  Possibly  when  it  is  discovered  that 
there  are  other  things  to  be  done  in  college  engineering  labora- 
tories than  to  break  endless  numbers  of  specimens  of  metal 
with  a  100-  or  200-ton  testing-machine,  we  may  have  further 
experimental  results  on  which  practical  engineers  may  rely. 

Professor  Smith  tried  various  depths,  d  inches,  and  breadths, 
b  inches,  of  shaving  from  cast  iron,  wrought  iron,  and  forged 
steel,  in  every  case  measuring  P,  the  pressure  on  the  tool  in 
pounds,  at  the  cutting  edge.  In  almost  every  case  we  find  from 
his  numbers  that  the  energy  E  usefully  spent  per  cubic  inch 
of  metal  removed,  diminished  about  30  per  cent,  as  b  was 
increased  from  -03  to  '05,  being  about  its  minimum  value  when 
b  was  -05  ;  but  probably  it  does  not  increase  much  for  greater 
values  of  b.  The  minimum  values  of  E  for  various  depths  of 
cut  were  as  follow  :  — 


APPLIED   MECHANICS. 


d 

b 

P 

E 



•05 
•135 

•056 
•056 

288 
690 

8,700 
7,700 

'  Cast  iron. 

•03 
•06 
•14 

•056 
•06 
•056 

215 
570 
810 

10,700 
13,000 
8,700 

)  Wrought 
I      ifon. 

•02 

•056 

250 

18,700 

~\ 

•04 
•06 

•056 
•056 

480 
705 

18,000 
17,700 

V  Steel. 

With  cast  iron  if  b  =  '05,  and  probably  for  greater 'values  of 
b,  E  is  much  the  same  for  cuts  of  the  depths  '05  and  '135  inch, 
being  about  9,000  foot-pounds  per  cubic  inch  of  metal  removed. 

With  steel,  if  b  ='05,  and  probably  for  greater  values  of  b, 
E  is  much  the  same  for  cuts  of  the  depths  '02,  -04,  and 
•06  inch,  being  about  18,000  foot-pounds  per  cubic  inch  of 
metal  removed. 

With  steel,  if  b  ='05,  and  probably  for  greater  values  of  6, 
E  is  10,700  foot-pounds  for  a  cut  of  -03  inch  depth,  13,000 
foot-pounds  for  a  cut  of  '06  inch  depth,  and  is  practically  the 
same  as  cast  iron  for  a  cut  of  '14  inch  depth. 

EXERCISES. 

1.  Two  tons  of  rock  can  fall  to  a  depth  of  320  feet;  find  the  work 
which  it  may  do.  Ans.,  1,433,600  foot-lbs. 

2.  In  lifting  an  anchor  of  1^  tons  from  a  depth  of  15  fathoms  in  six 
minutes,  what  is  the  useful  man-power,  if  a  man-power  is  defined  as 
3,500  foot-lhs.  per  minute?  Ans.,  14-4. 

3.  The  pull  on  a  tramcar  is  200  Ibs.  at  an  angle  of  25°  with  the  track ; 
what  is  the  component  in  the  direction  of  the  track  ?     "What  work  is  done 
in  a  distance  of  10  feet  along  the  track  ?      If  the  speed  is  4  feet  per 
second,  what  is  the  usefully  expended  horse-power  ? 

Ans.,  181-26  Ibs.  ;  1812-6  foot-lbs  ;  1-318. 

4.  What  horse-power  is  involved  in  lowering  by  2  feet  the  level  of  the 
surface  of  a  lake  2  square  miles  in  area  in  300  hours,  the  water  being 
lifted  to  an  average  height  of  5  feet  ?  Ans.,  58'5. 

5.  Taking  the  average  power  of  a  man  as  TVth  of  a  horse-power,  and 
the  efficiency  of  the  pump  used  as  0-4,  in  what  time  will  ten  men  empty  a 
tank  of  50  feet  x  30  feet  x  6  feet  filled  with  water,  the  lift  being  an 
average  height  of  30  feet  ?  Ans.,  21  hours  14  minutes. 

6.  The  diameter  of  a  steam-engine  cylinder  is  9  inches,  the  length  of 
crank  9  inches,  the  number  of  revolutions  per  minute  110,  and  mean 
effective  pressure  of  the  steam  35  Ibs.  per  square  inch ;  find  the  indicated 
horse-power. 


APPLIED    MECHANICS.  47 

7.  One  gas-engine  uses  24  cubic  feet  of  coal-gas,  and  another  98  cubic 
feet  of  Dowson  gas  per  hour  per  useful  horse-power ;    what  are   their 
efficiencies  ?     The  calorific  powers  of  coal-gas  and  of  Dowson  gas  per  cubic 
footare  520,000  and  123,000  foot-pounds  respectively.     Ans.,  0'159;  0-164. 

8.  What  would  be  the  indicated  horse-power  of  an  Otto  gas  engine 
which  has  a  piston  12  inches  in  diameter  and  a  crank  8  inches  long? 
The  engine  works  at  150  revolutions  a  minute,  and  there  is  an  explosion 
every  2  revolutions,  the  mean  effective  pressure  in  the  cylinder  during  a 
cycle  being  62  Ibs.  per  square  inch.  Ans.,  21. 

9.  The  average  breadth  of  an  indicator  diagram  for  one  side  of  the 
piston  is  1-58  inches,  and  for  the  other  side  it  is  T42  inches,  and  1  inch 
represents  32  Ibs.  per  square  inch.     Piston,  12  inches  diameter;  crank,  1 
foot ;  110  revolutions  per  minute.     What  is  the  indicated  horse-power  ? 

Ans.,  72-38. 

10.  What  must  be  the  effective  horse-power  of  a  locomotive  which 
moves  at  the  steady  speed  of  35  miles  an  hour  on  level  rails,  the  weight 
of  engine  and  train  being  120  tons,  and  the  resistances  16  Ibs.  per  ton? 
What  additional  horse-power  would  be  necessary  if  the  rails  were  laid 
along  a  gradient  of  1  in  112  ?  Ans.,  179-2 ;  224. 

11.  In  10  find  in  each  case  how  far  the  train  would  move  after  steam 
was  shut  off,  assuming  the  above  constant  resistance  and  neglecting  rota- 
tory motions.     Find  also  the  speed  of  the  train  after  the  latter  had  moved 
?ver  a  distance  of  1,000  feet  from  the  point  where  steam  was  shut  off. 

Ans.,  5,728  feet;  2,545-8  feet ;  3 1-8  miles  per  hour;  27 -3  miles  per  hour. 

12.  A  flywheel  weighs  2|  tons,  and  its  mean  rim  has  a  velocity  of  40 
feet  per  second ;  what  is  its  kinetic  energy  ?    If  the  velocity  be  reduced 
3  per  cent.,  what  is  the  reduction  in  the  kinetic  energy  ?    If  the  kinetic 
energy  be  reduced  by  10,000  foot-lbs.,  by  how  much  is  the  velocity 
reduced  ?    In  estimating  the  latter,  why  would  it  be  wrong  to  subtract 
from  40  feet  per  second  the  velocity  which  corresponds  to  10,000  foot-lbs. 
of  energy  in  this  flywheel  ? 

Ans.  139,130  foot-lbs. :  130,900  foot-lbs. ;  1-5  ft.  per  second. 

13.  A  machine  discharges  n  projectiles  per  minute,  each  of  w  Ibs. 
moving  with  the  velocity  of  v  feet  per  second ;  what  is  the  actual  horse- 
power ?    If  the  efficiency  of  the  machine  is  e,  and  it  is  driven  by  a  steam- 
engine  which  uses  w  Ibs.  of  steam  per  hour  per  horse-power  given  out  by 
it,  what  is  the  total  steam  per  hour?    If  the  engine  is  governed  by 
throttling,  and  if  the  total  steam  per  hour  follows  the  rule :  —steam  per 
hour  =  a-\-  b  x  brake  horse-power  (where  a  and  b  are  known  to  us),  and 
if  for  half  an  hour  n^  projectiles  are  discharged  per  minute,  and  if  for  the 
next  half  hour  n2  projectiles  are  discharged  per  minute,  how  much  steam 
is  used  during  the  hour  ?       .  nwv2    .    wnw2  .         b(n^  +  M2)wp2 

m">  gx  66000  '  66000"^  '  '          132000^    ' 

14.  Town  refuse  is  about  £  a  ton  per  unit  of  the  population  per  year. 
In  the  careful  burning  of  1  Ib.  of  refuse,  0-5  Ib.  of  water  at  212°  F.  may 
be  converted  into  steam  at  212°  F.     If  1  Ib.  of  coal  is  able  to  evaporate  in 
a  good  boiler  10  Ibs.  of  water,  how  many  tons  of  coal  per  year  would  pro- 
duce the  same  amount  of  steam  as  the  refuse  from  5,000,000  inhabitants  ? 

Ans.,  5,000,000  x  i  x  0-4 -^  10,  or  100,000  tons  of  coal  per  year. 
If  we  get  one  actual  horse-power  for  40  Ibs.  of  refuse  per  hour ;  if  the 
engines  drive  pumps  of  90  per  cent,  efficiency,  pumping  water  to  a  reservoir ; 
if  the  water  drives  motors  in  the  city  with  a  total  efficiency  of  30  per  cent., 


4&  APPLIED    MECHAfrWS. 

working  on  an  average  2  hours  per  working  day ;  what  is  the  average 

horse-power  supplied  to  each  house  if  there  are  500,000  houses  in  the  city  ? 

Ans.,    5,000,000x1^x2240-7-40  or   42xl07  is  the  actual  energy  in 

horse-power  hours.     The  pumped  energy  is  3 '78  x  108  horse-power 

hours.     The  supplied  energy  is  I  134  x  108  horse-power  hours.     The 

average  total  horse-power  (2  hours  a  day  for  313  days)  is  1*8  x  105, 

or  180,000,  and  per  house  it  is  0'3G  horse-power. 

15.  The  section  of  a  stream  is  12  square  feet,  the  average  velocity  of 
the  water  is  2  feet  per  second ;  there  is  an  available  fall  of  25  feet ;  what 
is  the  horse-power  available  ?     A  turbine  drives  a  dynamo  machine  which 
sends  electric  power  to  a  motor  at  a  distance.      The   efficiency  of  the 
turbine  is  70  per  cent.;  of  the  dynamo,  87  per  cent.;  ten  per  cent,  of  the 
energy  from  the  dynamo  is  wasted  in  transmission,  and  the  efficiency  of 
the  motor  is  72  per  cent.  ;  how  much  power  is  given  out  by  the  motor  ? 
The  voltage  at  the  dynamo  is  102  ;  what  is  the  current  in  amperes  ? 

Ans.,  68;  26 '8  ;  303. 

16.  Electric  lamps  giving  1  candle-power  for  4  watts;  how  many  10- 
or  how  many  16 -candle  lamps  may  be  worked  per  electric  horse-power? 
The  combined  efficiency  of  engine,  dynamo,  and  gearing  being  70  per 
cent.,  what  is   the   candle-power  available  for  every  indicated  horse- 
power? Am,,  18;  11;  130'55. 

17.  On  a  switchback  the  carriage  is  966  Ibs.,  neglecting  friction;  find 
its  kinetic  energies  when  it  is  5,  10,  and  15  feet  below  its  starting-point. 
And  if  the  starting-point  is  20  feet  above  datum  level,  write  out  in  two 
columns  its  two  kinds  of  energy  at  each  point.     If  the  above  points  are 
O'l,  0-4,  and  0'7  of  the  distance  along  the  track,  and  if  loss  of  energy  by 
friction  is  proportional  merely  to  distance  along  the  track,  and  if  the 
carriage  has  to  be  lifted  T6  feet  at  the  end  of  each  journey,  find  the 
correction  in  the  kinetic  energy  at  each  place. 

Ans.,  Potential  energies,  14,490,  9,660,  4,830  foot-lbs. ;  kinetic 
energies,  4,830,  9,660,  14,490  foot-lbs. ;  corrected  kinetic  energies,  4,675-4, 
9,041-6,  13,407  foot-lbs. 

18.  The  calorific  powers  of  1  Ib.  of  each  of  the  following  fuels  are 
given  in  centigrade-pound  heat  units ;  convert  into  foot-lbs.  if  1  heat  unit 
=  1,400  foot-lbs. :— charcoal,  7,000;  coke,  7,000;  coal,  8,800  to  7,330; 
wood,  4,200  ;  and  kerosene,  12,200. 

.  19.  The  pull  on  a  tramcar  was  registered  when  the  car  was  at  the 
following  distances  along  the  track : — 0,  200  Ibs.;  10  feet,  150  Ibs.; 
25  feet,  160  Ibs.;  32  feet,  156  Ibs.;  41  feet,  163  Ibs.;  56  feet,  170  Ibs.; 
60  feet,  165  Ibs. ;  73  feet,  160  Ibs. ;  what  is  the  average  (space)  pull  on 
the  car,  and  what  is  the  effective  work  done  in  pulling  the  car  through 
the  distance  of  73  feet?  Ans.,  161  Ibs.,  11,800  foot-lbs.  about. 

20.  A  chain  hanging  vertically  520  feet  long,  weighing  20  Ibs.  per 
foot,  is  wound  up  ;  what  work  is  done?  Ans.,  2,704, OuO  foot-lbs. 

21.  Four  cwt.  of  material  are  drawn  from  a  depth  of  80  fathoms  by  a 
rope  weighing  1-15  Ibs.  per  linear  foot ;  how  much  work  is  done  altogether, 
and  how  much  per  cent,  is  done,  in  lifting  the  rope  ?     What  horse-power 
wculd  be  required  to  raise  the  material  in  four  and  a  half  minutes  ? 

Ans.,  347,520  foot-lbs  ;  38  ;  2'34. 

22.  (a)  A  cut  of  -06  inch  depth  is  being  made  on  a  4-inch  wrought- 
iron  shaft  revolving  at  10  revolutions  per  minute ;  the  traverse  feed  is  0'03 
incb  per  revolution;  the  pressure  on  the  tool  is  found  to  be  435  Ibs., 


APPLIED    MECHANICS.  49 

what  is   the  horse-power  expended  at  the  tool  ?    How  much  metal   is 
removed  per  hour  per  horse-power  ? 

Ans.,  -1381 ;  98-28  cubic  inches. 

(b)  When  the  traverse  feed  is  -06  inch  per  revolution,  the  pressure  on 
the  tool  is  found  to  be  570  Ibs. ;    find  the  horse-power,  and    the   metal 
removed  per  hour  per  horse-power.  Ans.,  '181 ;  150  cubic  inches. 

(c)  If  the  above  horse-power  is  called  the  useful  power  TJ,  and  it  is 
found  that  the  actual  horse-power  given  to  the  lathe  is  0*1  + 1-5  u,  find 
actual  horse-power  and  metal  removed  per  hour  per  actual  horse-power. 

Ans.,  '3071,  44-2  cubic  inches;  '3715,  73'1  cubic  inches. 

23.  What  is  the  kinetic  energy  of  a  tramcar  moving  at  6  miles  per 
hour,  laden  with  36  passengers,  each  of  the  average  weight  of  11  stones  ? 
Weight  of  car,  2|  tons.     What  is  its  momentum?     If  stopped  in  2  sees., 
what  is  the  average  force  ?     If  the  force  is  constant,  this  must  also  be  the 
space  average  force  ;  find  the  distance  of  stopping  if  the  force  is  constant. 

Ans.,  13,400  foot-lbs;  3045-565  ;  1522-8  Ibs.  ;  8-8  feet. 

24.  A  ball  weighing  5  ounces,  and  moving  at  1,000  feet  per  second, 
pierces  a  shield,  and  moves  on  with  a  velocity  of  400  feet  per  second ; 
what  energy  is  lost  in  piercing  the  shield  ?  Ans,,  4,076  foot-lbs. 

25.  A  fire-engine  pump  is  provided  with  a  nozzle,  the  section  area  of 
which  is  1  square  inch,  and  the  water  is  projected  through  the  nozzle 
with  an  average  normal  velocity  of    130  feet  per  second;   find  (1)  the 
number  of  cubic  feet  discharged  per  second,    (2)  the  weight  of  water 
discharged  per  minute,  (3)  the  kinetic  energy  of  each  pound  of  water  as 
it  leaves  the  nozzle,  (4)  the  horse-power  of  the  engine  required  to  drive 
the  pump,  assuming  the  efficiency  to  be  70  per  cent. 

Ans.,  (1)  -9  cubic  feet ;  (2)  1-51  tons;  (3)  262-3  foot-lbs. ;  (4)  38'3. 

47.  Bicycle  problems. — When  a  man  says  that  his  bicycle  is 
geared  to  D  inches,  he  means  that  he  advances  IT  D  inches  for 
one  turn  of  his  pedals.  Let  the  diameter  of  the  pedal  circle 
be  d  inches.  Let  W0  be  weight  in  Ibs.  of  rider,  WQ  of  machine, 
W0  +  w0=w.  Let  w  be  the  uniform  vertical  force  which  the 
rider  applies  to  each  pedal  alternately;  if  w  is  negative,  it 
means  that  he  is  back-pedalling.  Let  F  be  the  force  in  Ibs. 
which  would  pull  the  bicycle  along  at  the  velocity  of  v  miles 
per  hour ;  2  w  d  is  the  work  done  in  inch-pounds  by  the  rider 
in  one  revolution,  and  this  is  equal  to  ?r  D  F. 

Ort*  =  ~F....(l) 

60  x  £3  QOO'  °r  FV  "*"  ^"^'  ^e  horse-power  usefully  expended.... (2) 

12  x  v  x  5,280       336  v 

— = =  »,  the  number  of  revolutions  of  a  pedal  per 

IT  D  x  60  D 

minute....  (3). 

[It  is  useful  to  remember  that  a  machine*  geared  to  56  inches 
goes  at  10  miles  an  hour  when  the  pedals  make  one  turn  per 
second.] 


50  APPLIED    MECHANICS. 

The  vertical  force  is  not  by  any  means  constant  in  practice, 
nor  indeed  ought  it  to  be ;  it  is  and  ought  to  be  greatest  when 
the  crank  is  horizontal.  With  proper  ankle  action  the  force 
is  always  somewhat  in  the  direction  of  the  circular  path  of  the 
pedal,  but  for  exercise  work  there  is  no  harm  in  assuming  a 
uniform  vertical  force  w  in  the  down-stroke  and  no  force  in  the 
up-stroke.  In  practice  it  is  difficult  to  avoid  pressing  on  the 
pedal  in  its  up-stroke. 

If  P0  is  the  value  of  F  on  a  level  road,  then  on  a  rising 
slope  of  1  in  s  we  have 

F  =  FO  +  y  . . . .  (4). 

If  it  is  a  descending  slope,  —  is  negative.     FO  may  usually  be 

taken  as  proportional  to  w. 

The  following  data  were  roughly  measured  by  myself. 
They  are  good  enough  for  academic  problems.  They  suit  my 
own  bicycle  on  a  good  road.  Students  ought  to  obtain  data  of 
their  own  by  careful  measurement. 

A  rider  weighing  127  Ibs.  on  "a  cycle  weighing  33  Ibs.  (or 
w=160  Ibs.)  finds  that  on  a  descent  of  1  in  80,  with  his  feet  off 
the  pedals,  he  is  just  able  to  get  on  very  slowly  but  steadily ; 
on  a  long  slope  of  1  in  40  his  steady  speed  (feet  off  pedals)  was 
9J  miles  per  hour ;  on  a  long  slope  of  1  in  20  his  steady  speed 
(feet  off  pedals)  was  20  miles  per  hour.  In  these  three  cases 
the  values  of  FO  were  : — 

8-0  OT  2>    TO  °r  *'    2-0  M  8' 

We  need  careful  experiments ;  and  it  is  not  of  much  use 
speculating  on  the  probable  law  of  resistance  of  a  safety  bicycle 
with  pneumatic  tyres  on  a  certain  kind  of  road.  A  constant 
term  for  quasi-solid  friction,  and  a  term  (the  most  important  at 
high  speeds)  proportional  to  the  square  of  the  speed  relatively 
to  that  of  the  atmosphere  for  air  resistance  :  these  we  ought  to 
have.  Resistance  due  to  unevenness  of  the  ground  would  be 
constant  if  a  certain  kind  of  unevenness  were  to  repeat  itself 
at  intervals  so  far  apart  on  the  road  that  the  vibration  due  to 
each  had  time  to  die  away  before  the  next ;  but  speculation  is 
vague,  especially  as  the  kind  of  vibration  will  often  depend  on 
the  velocity.  In  our  present  state  of  knowledge  we  cannot  be 
far  wrong  in  assuming 

F4  =  w  (a  +  bv  -f  cv*). 


MECHANICS.  51 

If  the  above  values  of  PO  are  correct  at  the  given  speeds, 
—  0,  9|,  and  20  miles  per  hour,  we  find 


as  a  formula  which    represents  our    experimental  data  well 
enough  for  exercise  purposes. 

Hence,  going  up  a  slope  of  1  in  s  we  have 

F_  w   /1+   v    ,_*L  +  80\ 
~  80  V     r  20  ^  200         *   / 

for  this  bicycle.     When  w=  160  Ibs. 


Example  1.  —  What  horse-power  is  expended  in  going  at 

160 
12  miles  an  hour  on  a  level  road  1     Here  -  =0. 


Th3  speed  is  88  v  or  1,056  feet  per  minute,  and  the  horse-power 


Example  2.  —  At  12  miles  an  hour,  going  up  or  down  an 
incline  of  1  in  60,  what  is  the  useful  horse-power  ? 

F  =  4-64  +  "?         —  7-31  or  1-97  Ib. 
—  60 

Horse-power  up       =  l>056  X  7'31  =  0-234. 
33,000 

Horse-power  down  =  1>0^  *  *'97  =  0-033. 
33,000 

EXERCISES. 

1.  In  Examples  1  and  2,  what  force,  w,  does  the  rider  exert  upon 
his  pedal  if  his  bicycle  is  geared  to  D  =  60  inches  and  the  diameter  of  the 
pedal  circle  is  13  inches  ? 

Ans.,  On  the  level,  w  =  33-6  Ibs.  ;  going  up,  w  =  63  Ibs.  ;  going 
down,  w  =  14-3  Ibs. 

2.  On  what  slope  downward  would  the  velocity  of  12  miles  an  hour  be 
steadily  maintained,  feet  off  pedals  ?  Am.  ,  1  in  34£. 

3.  Going  down  a  slope  of  1  in  30  at  8  miles  per  hour,  what  is  the  force 
on  the  pedal?  Ans.t  -12-30  Ibs. 

The  minus  sign  means  that  the  rider  must  back-pedal. 

4.  If  a  rider  whose  weight  is  157  Ibs.,  back-pedals  on  this  same  bicycle 
with  a  force  of  only  10  Ibs.,  down  a  slope  of  1  in  25,  what  is  his  velocity  f 

Ans.t  13  '6  miles  per  hour. 


52  APPLIED    MECHANICS. 

The  following  example  is  for  students  who  can  integrate  :  — 
The  resistance  to  motion  being  F  r=  a  -f-  bv  -f  cv2  on  a  road 
which  does  not  alter  in  character  for  a  whole  journey,  compare  the 
work  done  in  going  over  a  certain  distance,  I  —  first,  at  a  constant 
speed,  v0  ;  second,  at  the  same  average  speed,  but  varying  according 
to  the  law  v  =  v0  -f-  /  sin.  qt.  In  both  cases  the  time  is  the  same 
if  the  journey  is  just  so  long  as  to  be  finished  by  the  rider  in  the 
same  state  as  to  speed,  etc.,  with  which  he  starts. 

f  yds  =  f  F  *  dt  =  \rvdt  =  f  («r  +  W  +  ws)  dt, 

the  work  done.  If  we  calculate  this  for  a  time,  T,  where  q  =.  2irr, 
it  is  just  as  good  as  for  any  number  of  such  periods.  The  value 
divided  by  T  gives  the  average  work  per  second,  or 


whereas,  if  /  is  0  and  the  speed  is  constant,  we  have  the  average 
work  per  second  av0  +  bv02  +  cv03.  Hence  the  fractional  increase  of 
work  is  (*/  -f  3cv0f2)  /  2  (av0  +  bv<?  +  cvQ3).  Thus,  taking  our 
values  of  Art.  47  for  a  weight  of  160  Ibs., 

' 


If  the  road  has  everywhere  an  upward  slope,  1  in  *,  we  take 


but  we  had  better  take  a  level  road,  so  that  a  =  2.     In  any  case, 

b  —  -1,        c=  -01. 
Taking  an  average  speed  of  10  miles  per  hour  and 

v  =  10  +  3  sin.  qt, 

so  that  the  speed  fluctuates  between  13  and  7  miles  per  hour,  wo 
have  a  fractional  waste  of  power 

8_          b  -f  9Qg  3  . 

20  a  +  104  +  lOOc         80  ' 

that  is,  the  power  at  constant  speed  being  80,  the  average  power  at 
varying  speed  is  83. 

48.  Much  of  what  we  have  given  may  be  said  to  be  mere  exer- 
cise work  in  the  use  of  formulae.  But  we  hope  that  it  is  much 
more  than  that,  and  that  students  are  getting  to  understand 
how  the  formulae  are  derived.  Much  of  this  book  is  devoted 
to  the  explanation  of  how  these  formulae  are  derived,  and 
somewhat  similar  exercises  will  be  given  later.  Our  object 
has  been  to  familiarise  students  with  the  notion  of  the 
quantitative  transformation  of  energy.  The  subject  of  this 
book  is  almost  altogether  the  study  of  energy  and  momentum, 
We  close  this  introductory  part  of  our  subject  with  a  few 
problems  011  the  hydraulic  transmission  of  power  and  the 
propulsion  of  ships. 


APPLIED    MECHANICS. 


53 


For  academic  exercises  it  is  sufficiently  correct  to  say  (see  Art.  69) 
that  the  energy  wasted  per  pound  of  water  flowing  in  a  pipe  is 
experimentally  found  to  he  k  times  its  kinetic  energy,  where  k  has  the 
following  values: — In  I  feet  of  pipe  of  diameter  d  feet,  £  =  -0232  l/d; 
entrance  or  exit  hy  cylindric  pipe  to  or  from  a  reservoir,  k  =  0-5  ; 

bend  in  a  pipe,  k  =  j  -131  +  1'847  (  d-^^  }  a,  where  a  is  the 

fraction  of  two  right  angles  through  which  the  hend  extends, 
D  the  diameter  of  the  circle  of  which  the  centre  line  of  the  pipe 
is  part  and  d  the  pipe's  diameter.  Probably  only  the  first  of 
these  values  of  k  is  fairly  correct. 

Exercise. — Prove  that  if  the  horse-power,  H,  enters  a  straight 
pipe  as  pressure  water,  the  waste  power,  w,  is  '00374  I  H3/p3^5, 
where  p  is  the  pressure  at  the  entering  end  in  pounds  per  square 
inch.  Notice  that  the  fractional  loss  of  power  is  the  fractional 
loss  of  pressure. 

If  v  is  the  voltage  and  A  the  current  in  amperes,  the  horse- 
power delivered  electrically  to  a  conductor  is  v  A  watts  (746  watts 
are  equal  to  1  horse-power).  The  loss  in  the  conductor  is  A2  R 
watts,  if  R  is  its  resistance  in  ohms.  Hence,  if  H  is  the  horse- 
power sent  in,  the  wasted  power  is  w  =  746  H2  R/v2. 

As  R  =  -   X  -044  if  the  copper  conductor  is  n  miles  long  (going 

a 
and  coming,  so  that  the  distance  is  ^  n  miles)  and  a  square  inches 

in  cross-section,  w  =  32*7  ~2  -.     Notice  that  the  fractional  loss 

of  power  is  the  fractional  loss  of  voltage. 

Exercise. — Prove  that  at  an  entering  pressure  of  700  Ibs.  per 
square  inch,  if  we  admit  the  following  amounts  of  hydraulic 
power,  we  have  the  following  amounts  of  waste  power.  Also  find 
the  values  in  the  table  for  a  conductor. 


H 

The  Power 
sent  in. 

At  pressure  of  700  Ibs.  per 
square  inch. 
Horse-power  lost  in  one  mile. 

At  700  volts. 
Horse-power  lost  in  one  mile. 

Conductor 
0*25  square  inch 
in  cross-section. 

Conductor 
0'125  square  inch 
in  cross-section. 

6-inch  pipe. 

3-inch  pipe. 

20 
50 
100 
200 
300 

0-23 

1-84 
14-72 
117-8 

0-9 
7-4 
59 

0-67 
2-67 
10-68 
24-03 

0-34 
1-34 
5-34 
21-36 
48-06 

Up  to  the  highest  usual  speeds  of  commercial  ships  we 
may  assume  without  great  error  that,  for  vessels  not  dissimilar 


54  APPLIED   MECHANICS. 

in  form  and  character  and  going  at  the  usual  speeds,  the 
indicated  horse-power  is  H  =  D|  v3  -h  c,  where  D  is  the  dis- 
placement in  tons  and  v  is  the  speed  in  knots  and  c  is  a 
constant,  which  for  many  classes  of  vessel  may  be  taken  as 
not  very  different  from  240. 

Exercise  1. — What  is  the  indicated  horse-power  of  a  vessel  of  1,330 
tons  moving  at  a  speed  of  12  knots,  if  it  obeys  the  above  rule  ? 

Ana.,  871. 

Exercise  2. — If  a  vessel  of  1,720  tons  moves  at  10  knots  when 
its  indicated  horse-power  is  655,  what  is  the  value  of  c  in  such  a  class  of 
vessel?  Ana.,  219. 

The  resistance  to  the  motion  of  a  ship  is  considered  to  be  made 
up  of  two  parts.  1.  The  skin  friction  in  pounds,  s  =/A  v  n,  where 
v  is  the  speed  in  knots,  n  is  .1*83  for  varnished  or  painted  wooden 
models  or  clean  iron  ships,  A  is  the  wetted  area  in  square  feet, 
/  is  -009  for  ships  of  over  200  feet  long,  and  -012,  -0106,  -0096 
for  ship  lengths  of  8,  20,  and  50  feet.  At  speeds  of  6  to  8  knots  in 
ordinary  vessels  this  skin  resistance  is  about  80  or  90  per  cent,  of 
the  whole  ;  at  high  speeds  it  is  about  half  the  whole. 

2.  A.  residuary  resistance  due  to  the  fact  that  eddies  (the 
smaller  part)  and  waves  are  produced.  Eddy  resistance  is  thought 
not  to  be  more  than  8  per  cent,  of  the  skin  resistance  even  at  high 
speeds.  It  is  mainly  caused  by  bluntness  of  the  stern  of  a  vessel. 
In  two  perfectly  similar  ships,  similarly  loaded,  of  lengths  I  and 
L,  at  speeds  v  and  v  v/Z/J,  which  are  said  to  be  the  corresponding 
speeds,  the  residuary  resistances  are  proportional  to  I3  and  L3. 

The  skin  resistances  81  and  »x  of  the  ship  and  its  model  can  be 
calculated  from  Froude's  numbers  given  above.  Hence,  if  R  is 
the  resistance  in  pounds  of  a  ship  L  feet  long,  A  its  wetted  area  in 
square  feet,  v  its  speed  in  knots,  and  if  r  and  I  are  the  resistance 
and  length  of  a  model  which  is  exactly  similar  and  of  similar 
draught  when  the  model  is  drawn  at  the  corresponding  speed 
v  knots,  where  v :  v  ::  */  L  :  */  I,  prove  that  it  follows  from  the 
above  that 

R  =^r—  -009  Av1'8 

if  the  ship  is  more  than  200  feet  long,  and  the  model  is  from  8  to 
30  feet  long. 

Example. — Before  building  a  vessel  400  feet  long,  of  wetted 
surface  26,000  square  feet,  we  wish  to  know  R,  its  resistance,  at 
v  =  12  knots.  _A  model  is  made  10  feet  long,  it  is  drawn  at  a 
speed  of  12\/40  or  1-9  knots  in  the  tank,  and  its  resistance  r 
is  found  to  be  0'9  Ib.  We  find  R  to  be  39,720  Ibs. 

Prove  that  R  in  pounds  x  v  in  knots  -?•  325  =  utilised  horse- 
power. In  this  case  we  find  1,463  horse-power.  The  indicated 
power  will  probably  be  more  than  3,000. 

The  vagueness  of  our  knowledge  as  to  the  probable  loss  of 
power  by  friction,  makes  any  attempt  to  calculate  u  for  the  above 


APPLIED   MECHANICS.  55 

purpose  rather  useless,  and  the  better  use  of  the  tank  would  there- 
fore seem  to  lie  in  helping  to  improve  a  particular  class  of  vessel. 

The  following  great  simplification  has  recently  been  tried  by 
Colonel  English.  Suppose  an  existing  vessel  to  be  run  at  various 
speeds  and  its  indicated  horse-power  noted.  Now,  assume  that  the 
effective  horse-power  in  a  new  ship  will  be  the  same  fraction  of  the 
indicated  that  we  take  it  to  be  in  the  existing  ship  —  say  one-half. 
Find  the  resistance  of  the  existing  ship  at  the  speed  vr  We  wish 
to  know  the  resistance  of  the  new  ship  at  the  speed  v?.  We  only 
need  to  compare  the  wave  and  eddy  resistances,  which  we  shall 
call  w,  and  w2.  Make  two  models,  one  of  the  existing  ship  and 
one  of  the  ship  being-  designed.  Let  the  values  of  v,  D,  s 
L,  w  for  the  two  ships  and  the  two  models  be  indicated  by 
capital  and  small  letters,  the  existing  ship  and  its  model  having 
the  affixes  x.  s  is  skin  friction  ;  D  is  displacement,  which  in  similar 
ships  is  proportional  to  the  cubes  of  the  lengths. 


Let    vl  =  vl*,     *2  =  v26,     and    let    ^  =  *„  ; 

\D1  /  \D2/ 

that  is,  make  the  models  of  such  sizes  that  v2  and  %  fts  well  as  Vj 
and  Vj,  are  "  corresponding  speeds,"  and  yet  that  the  speeds  of  the 

two  models  shall  be  the  same.     In  fact,  -r  =  —  (  —  )  •     Now  let 

'  <*!       DJ  Vv2/ 

the  two  models  be  towed  from  the  two  arms  of  a  lever  whose 
fulcrum  may  be  adjusted  and  the  ratio  of  the  resistances,  n,  of 
the  second  to  the  first  may  be  measured.  Note  that  we  need  only 
find  this  ratio  —  a  much  easier  thing  to  do  than  to  find  either 
resistance.  Show  that  the  total  resistance  of  the  new  ship  is 


56 


CHAPTER    IV. 

FRICTION. 

49.  WE  have  said  that  the  mechanical  principles  which  must 
be  studied  by  the  young  engineer  are  few  in  number,  but  they 
must  be  very  familiar  to  him.  It  is  not  well  to  say  that  any 
one  method  of  study  is  more  important  than  another,  the  fact 
being  that  a  student  must  not  only  study  in  the  workshops 
and  drawing-office,  but  he  must  read,  work  numerical  exercises, 
and  make  a  great  many  quantitative  laboratory  experiments 
to  illustrate  these  principles.  Our  aim  is  to  get  students  to  think, 
and  it  is  astonishing  how  difficult  it  is  to  effect  this  object. 
We  cannot  easily  get  students  to  wrangle  over  these  subjects. 
We  have  few  pretty  lecture  experiments.  Even  in  chemistry 
and  experimental  physics,  pretty  lecture  experiments  are  not 
very  effective  in  causing  students  to  think.  Students  will 
think  about  things  that  they  do.  Hence  it  is  that  boys  should 
be  allowed  to  chip  and  file  metals,  and  to  pare  and  cut  wood. 
Merely  in  learning  how  to  hold  a  chipping-chisel  or  in  setting 
a  plane  iron,  a  student  must  think  about  the  properties  of 
materials  and  forces.  Country  boys  who  make  their  own  things 
have  a  great  advantage  over  town  boys  who  buy  their  things 
in  shops.  In  the  mechanical  laboratory,  I  find  that  even  the 
dullest  student  begins  to  think  for  himself  if  he  is  not  too 
much  spoon-fed  ;  and  if  his  difficulties  are  not  cleared  away  by 
some  wretched  routine  system  of  laboratory  work  being  adopted 
by  cheap  laboratory  instructors,  the  fundamental  principles  of 
mechanics  will  become  part  of  his  mental  machinery. 

It  is  not  necessary  to  illustrate  everything ;  a  few  things 
carefully  done  in  the  laboratory  are  better  than  many.  For 
example,  let  the  triangle  of  forces  be  illustrated  in  its  simplest 
form.  The  principle  is : — If  three  forces  act  on  a  small  body 
and  just  keep  it  at  rest,  then  if  we  draw  on  a  sheet  of  paper 
three  straight  lines  parallel  to  the  directions  of  the  three  forces, 
and  let  them  form  a  triangle  in  such  a  way  that  arrow-heads 
representing  the  directions  of  the  forces  go  round  the  triangle 
circuitally,  it  will  be  found  that  the  lengths  of  the  sides 
of  the  triangle  are  proportional  to  the  amounts  of  the  forces, 
^ig.  19  shows  how  the  strings,  pulleys,  and  the  smooth  ring  p 


APPLIED    MECHANICS. 


57 


Fig.  19. 


are  used.     I  am  in  the  habit  of 
using    three    scale  -  pans    with 
weights  in  them,  and  I  measure 
beforehand  the   weights  of   the 
scale-pans  themselves.      Put  al- 
most any  weights  in  at  random 
(only  you  will  find  that  any  two 
must  be  greater  than  the  third), 
and   let    P   find   its   position  of 
equilibrium,  and   you  will    find 
the  rule  to  be  nearly  true  every 
time.     Also  for  the  same  set  of 
weights  you  will  find  that  there 
is  a  small  region  within  which,  anywhere,  the 
centre  of  the  ring  P  may  be  placed  without  dis- 
turbing the  state  of  equilibrium ;  this  is  owing 
to  the  friction  of  the  pulleys.     The  polygon  of 
forces  is  also  easily  illustrated.      (See  Fig.  20.) 

50.  Our  one  Theory  insufficient. — No  man  can  make  these 
trials  without  finding  that  there  is  a  great  deal  to  be  observed 
beyond  what  his  f 
teacher  or  his 
book  has  taught 
him.  A  force 
has  been  repre- 
sented by  the 
pull  in  a  string 
passing  over  a 
little  pulley  with 
a  weight  at  its 
end.  He  finds 
that  as  his  pul- 
ley works  more 
easily,  and  as  its 
pivots  are  better 
oiled,  his  illus- 
tration of  the 
law  is  better  and 
better ;  in  fact, 
he  finds  that  the 
pull  in  a  string  is 
not  exactly  the 


Fig-  20. 


APPLIED   MECHANICS. 


Fig.  21. 


same  on  the  two  sides  of  a  pulley.  If  he  takes 
one  pulley  and  one  string  and  two  weights, 
called  A  and  B,  Fig.  21,  at  its  ends,  he  will 
find  that  there  is  equilibrium  even  when 
the  two  weights  are  not  exactly  equal.  If 
A  is  slightly  greater  than  B,  and  he  further 
increases  the  weight  of  A  till  it  is  just  able 
to  overcome  B,  then  the  difference  between 
the  weights  represents,  in  some  fashion, 
what  may  be  called  the  friction  of  the 
pulley.  If  now  he  increases  both  the  weights 
which  he  uses,  he  will  find  that  the  fric- 
tion is  proportionately  increased,  and  he 
will  get  to  understand  why  we  so  generally 
find  in  machinery  that  there  is  a  law,  "fric- 
tion is  proportional  to  load."  This  law 
is  not  quite  true,  but  it  is  sufficiently  true 
to  be  of  great  value  to  engineers.  Again, 
he  sees  that  this  friction,  which  is  a 

resistance  experienced   in   the   rubbing   together  of  any  two 

surfaces,  is  a  force  which  always  opposes  motion,  always  acts 

against  the  stronger  influence.  Suppose, 

for   example,    that   he    found    that   a 

weight  of  5'1  ounces  was  just  able  to 

overcome   a  weight  of  5   ounces ;    he 

will  find  that  a  weight  of   about  4*9 

ounces   will    just   be   overcome   by   a 

weight  of  5  ounces,  and  that  there  is 

equilibrium   with    5    ounces   and   any 

weight  varying  from  5*1  to  4 '9.     Fric- 
tion always  helps  the  weaker  forces  to 

produce  a  balance. 

51.    Law     of    Work. — Take    any 

machine,  from  a  simple  pulley  to  the 

most  complicated   mechanism.     Let  a 

weight,  A,  hung  from  a  cord  round  a 

grooved  pulley  or  axle  in  one  part  of 

the  mechanism,  balance  another  weight, 

B,  hung  from  a  cord  round  another  axle 

or  pulley  somewhere  else.     In   Fig.  22 

we  have  imagined  that  the  mechanism 

is  enclosed  in  a  box,  and  only  the  two 


Fig.  22. 


APPLIED   MECHANICS.  59 

axles  in  question  make  their  appearance.  Now  move  the 
mechanism  so  that  A  falls  and  B  rises  steadily.  Suppose 
that  when  A  falls  1  foot  B  rises  20  feet,  then  if  there  were  no 
friction  in  the  machine  a  weight  at  A  is  exactly  balanced  by 
one-twentieth  of  this  weight  at  B.  This  is  the  law  which  you 
will  find  proved  in  books  on  mechanics.  The  reason  why  it 
is  true  is  this.  The  work  or  mechanical  energy  given  out  by  a 
body  in  falling  is  measured  by  the  weight  of  the  body  multiplied 
into  the  distance  through  which  it  falls.  It  is  in  this  way  that 
we  get  the  energy  derivable  from  the  fall  of  a  certain  quantity 
of  water  down  a  waterfall,  and  it  is  in  this  way  that  we  find 
out  whether  a  certain  waterfall  gives  out  enough  power  to 

'  drive  a  mill.  Similarly  the  energy  given  to  a  body  when  we 
raise  it,  is  measured  by  the  weight  of  the  body  multiplied  by  the 
vertical  height  through  which  it  is  raised.  Now,  every  experi- 
ment we  can  make  shows  that  energy  is  indestructible,  and 
consequently,  if  I  give  energy  to  a  machine,  and  find  that  none 
remains  in  it,  that  there  are  no  means  there  of  converting 
mechanical  energy  into  heat  by  friction  or  into  any  other  form 
such  as  electrical  energy,  or  vice  versd,  and  no  storage  or 
unstorage  of  energy,  as  by  lifted  weights  or  the  coiling  of 
springs,  or  by  increasing  or  diminishing  the  kinetic  energy  of 
anything  (this  is  why  I  pre-suppose  uniform  velocity  in  A  and 
B),  then  all  the  energy  given  to  the  machine  must  be  given  out 
by  it.  This  is  often  what  people  really  mean  when  they  say 
that  their  machine  is  supposed  to  have  no  friction. 

\  Therefore  the  energy  given  out  by  A  in  falling  must  be 
equal  to  the  energy  received  by  B  in  rising;  and  as  A  falls  1  foot 
when  B  rises  20  feet,  the  weight  of  A  must  be  twenty  times  the 
weight  of  B.  If,  then,  there  were  no  friction  in  the  machine, 
and  if  a  weight  of  20  Ibs.  were  hung  at  A  and  a  weight  of  1  Ib. 
at  B,  we  should  find  that  if  we  start  A  downwards  or  upwards 
there  will  be  a  steady  motion  produced.  Any  excess  at  A  will 
cause  it  to  overcome  B,  the  weights  moving  more  and  more 
quickly  as  the  motion  continues. 

Now,  in  our  machine,  Fig.  22,  we  can  always  find  by  trial 
what  is  the  velocity  ratio — that  is,  the  speed  of  B  as  compared 
with  the  speed  of  A — and  this  is  usually  called  the  mechanical 
advantage  when  there  is  no  friction.  I  have  chosen  a  machine 
in  which  I  suppose  that  if  A  has  a  uniform  motion,  so  has  B. 
But  if  when  A  is  uniform,  B  is  not  uniform  in  its  motion,  then 
the  velocity  ratio  for  any  particular  position  must  be  measured 


60  APPLIED   MECHANICS. 

during  exceedingly  small  motions,  as,  after  a  little  motion  every- 
thing alters.  Let  us  continue  to  suppose  that  the  velocity 
ratio  does  not  alter.  Now,  when  we  try  to  balance  a  weight 
at  B  by  a  weight  at  A,  we  find  that  the  above  relation  is  quite 
untrue.  Hang  a  weight  of  1  Ib.  at  B,  hang  a  weight  of  20  Ibs. 
at  A,  there  is  certainly  a  balance ;  but  when  we  have  somewhat 
less  or  more  than  20  Ibs.  at  A  there  still  is  balance.  The 
reason  for  this  is  that  there  is  friction  in  the  mechanism,  and 
this  friction  always  tends  to  resist  motion,  always  acts  against 
the  stronger  influence. 

52.  Effect  of  Friction. — We  proceed  to  find  out  in  what  way 
friction  modifies  the  law  given  in  books.  You  must  make 
actual  experiments  with  some  machine  if  you  are  to  get  any 
good  from  your  reading.  Hang  on  a  weight,  B,  and  find  the 
weight,  A,  which  will  just  cause  a  slow,  steady  motion.  Do 
this  every  time  when  a  number  of  different  weights  are 
placed  at  B.  Now  suppose  you  have  measured  the  velocity 
ratio — that  is,  suppose  you  find  that  B  rises  four  times  as 
rapidly  as  A  falls.  Then,  according  to  the  books,  there  would 
be  an  exact  balance  if  A  were  four  times  the  weight  cf  B. 
On  actual  trial,  however,  I  find  in  a  special  case  the  following 
table  of  values  : — 

A  overcomes  B  when 

A  is  23-4  ounces  and  B  is  5  ounces 

»     447        „         „        10       „ 


65-4  „        15      „ 


86-8 
107-5 
128-8 
149-6 
171-0 


20 
25 
30 
35 
40 


But  if  there  had  been  no  friction  in  the  first  experiment,  A 
would  have  been  20  ounces  instead  of  23-4,  hence  the  friction 
is  represented  by  this  3-4  ounces.  For  every  experiment  let 
this  be  done,  subtract  four  times  B  from  A  and  call  this 
difference  the  friction.  Now  how  shall  we  compare  this 
friction  with  the  corresponding  load  ? 

53.  The  Use  of  Squared  Paper. — And  here  we  come  to  a 
matter  of  the  greatest  importance  to  the  practical  man.  How 
do  we  practically  compare  two  things  whose  values  depend  on 
one  another  1  How  do  we  find  out  the  law  of  their  de- 
pendence ?  It  is  a  strange  fact  that  there  should  be  a  class  in 


APPLIED   MECHANICS. 


6.1 


Fig.  23. 


the  community  who  have  a  little  difficulty  in  manipulating 
decimals  in  arithmetic,  but  it  is  almost  a  stranger  evidence  of 
neglected  education  that  so  many  people  should  be  ignorant 
of  the  great  uses  to  which  a  sheet  of  squared  paper  may  be  put. 
A  sheet  of  squared  paper  can  be  bought  very  cheaply.  It  hag 
a  great  number  of  horizontal  lines  at  equal  distances  apart,  and 


62  APPLIED    MECHANICS. 

these  are  crossed  by  a  great  number  of  vertical  lines  of  the  same 
kind,  so  that  the  sheet  is  covered  with  little  squares.  This 
sheet  will  enable  me  first  of  all  to  correct  for  errors  of  observa- 
tion in  the  above  series  of  experiments ;  and,  secondly,  to 
discover  the  law  which  I  am  in  search  of.  A  miniature 
drawing  is  shown  in  Fig.  23,  many  lines  being  left  out  because 
of  the  difficulties  of  wood-cutting.  At  the  bottom  left-hand 
corner  I  place  the  figure  0,  and  I  write  10,  20,  etc.,  to  indicate 
the  number  of  squares  along  the  line,  0  A.  Instead  of  10,  20, 
etc.,  I  might  write  1,  2,  etc.,  or  100,  200,  etc.,  according  to  the 
scale  I  am  going  to  use.  Indeed,  on  account  of  the  friction 
being  so  much  less  than  the  weight  A  with  which  it  is  to  be 
compared,  I  number  the  squares  along  the  vertical  line  0  P  by 
1,  2,  etc.,  instead  of  10,  20,  etc.  We  can  employ  any  scale 
we  please  in  representing  any  of  the  things  to  be  compared, 
and  it  is  usual  to  multiply  all  the  numbers  of  one  kind  by  some 
number,  so  as  to  represent  all  our  experiments  on  one  sheet  of 
paper,  and  on  as  much  of  this  sheet  as  possible.  Having  sub- 
tracted four  times  B  from  A,  I  find  the  following  numbers  : — 


A.  Friction. 

23-4  .         .         .         .3-4 

44-7  ....    47 

65-4  .        .        .         .5-4 
86-8  .        .         .6-8 


A.  Friction. 

107-5  .  .  .  .7-5 

128-8  .  .  .  .8-8 

149-6  .  .  .  .9-6 

171-0  .  .  .  .  11-0 


I  now  find  on  my  sheet  of  paper  the  point  P,  which  is  23-4  hori- 
zontally and  3  -4  vertically,  and  mark  it  with  a  cross  in  pencil. 
Q  is  44-7  horizontally  and  4*7  vertically,  and  so  for  the  others. 
The  last  point,  w,  is  171  horizontally  and  11-0  vertically.  We 
guess  at  the  decimal  part  of  a  small  square.  The  point  P 
represents  the  two  numbers  of  my  first  experiment,  and  every 
other  point  represents  the  two  observations  made  in  one 
experiment.  Now  we  are  certain  that  if  there  is  any  simple 
law  connecting  load  and  friction,  the  points  p,  Q,  to  w,  lie  in  a 
simple  curve  or  in  a  straight  line.  You  see  that  in  this  case  110 
curve  is  needed  to  suit  the  points ;  we  assume  that  they  would 
lie  in  a  straight  line,  only  that  we  made  some  errors  of  observa- 
tion. You  must  now  find  what  straight  line  lies  most  evenly 
among  all  the  points  ;  this  you  can  do  by  means  of  a  fine 
stretched  string,  and  the  line  M  N  seems  to  me  to  answer  best. 
It  tells  me,  for  instance,  that  when  A  is  44 '7  the  friction  is 
really  4*5,  instead  of  4-7.  Take  any  point  in  the  line,  its 


APPLIED    MECHANICS.  63 

vertical  measurement  gives  me  the  true  friction  corresponding 
to  a  load  represented  by  its  horizontal  measurement.  Thus, 
for  instance,  you  see  that  friction  5*0  corresponds  to 
load  55. 

This  is  a  simple  way  of  correcting  errors  made  in  experi- 
ments, but  you  cannot  hope  to  understand  much  about  it  till 
you  actually  make  experiments,  and  use  the  squared  paper. 
You  will  find  the  matter  all  very  simple  when  you  try  for 
yourself ;  my  description  of  it  is  as  complicated  as  if  I  were 
teaching  you  by  mere  words  to  walk,  or  to  bicycle. 

If  at  any  time  you  make  a  number  of  measurements  of 
two  variable  things  which  have  some  relation  to  one  another, 
plot  them  on  a  sheet  of  squared  paper,  and  correct  by  using  a 
flexible  strip  of  wood  or  a  ruler,  to  draw  an  easy  curve  or  a 
straight  line  so  that  it  passes  nearly  through  all  the  points. 
If  the  line  is  straight,  the  law  connecting  the  two  things  will 
prove  to  be  a  very  simple  one.  In  the  present  case  it  means 
that  any  increase  in  the  load  is  accompanied  by  a  proportionate 
increase  in  the  amount  of  friction.  Thus,  when  the  load  is  0, 
the  friction  is  2'3 ;  when  the  load  is  100,  the  friction  is  7 '2. 
That  is,  when  the  load  increases  by  100,  the  friction  increases 
by  4-9,  so  that  the  increased  friction  is  always  the  fraction, 
•049,  of  the  increased  load.  In  fact,  it  is  evident  that  we  can 
calculate  the  friction  at  any  time  from  the  rule 

Friction  =  2-3  +  '049  A. 

That  is,  multiply  the  load  A  in  ounces  by  -049,  and  add  2-3, 
the  answer  is  the  friction. 

54.  Law  of  Friction. — Our  result  is  that  the  total  friction  is 
equal  to  the  friction  2 '3  of  the  machine  unloaded,  together 
with  a  constant  fraction,  '049,  of  the  load.  Now  when  a 
similar  series  of  experiments  is  tried  on  any  machine,  be  it  a 
watch  or  clock,  or  be  it  a  great  steam-engine,  we  always  find 
this  sort  of  simple  law. 

If  you  clean  all  the  bearings  or  pivots,  or  if  you  use  a 
different  kind  of  lubricator,  you  will  get  other  values  for  the 
two  numbers  in  the  above  rule,  but  the  law  will  remain  of  the 
same  simple  kind.  I  find  it  nearly  impossible  to  get  rny  pupils 
to  believe  that  rough  and  rusty  old  machines,  such  as  screw- 
jacks  or  hydraulic  jacks,  which  have  been  long  in  use,  are 
far  more  instructive  to  study  than  beautiful,  specially  made, 
frictionless  machines.  In  my  laboratory,  now,  at  Finsbury, 


64        .  APPLIED    MECHANICS. 

there  is  an  ideal  screw-jack ;  the  weight  p  (Fig.  40)  is  not  a 
single  weight,  but  two  equal  ones  at  the  ends  of  two  cords 
which  pass  round  two  equal  grooves  and  produce  a  true  couple 
in  turning  the  screw.  I  prefer  the  rough  old  thing  previously 
in  use,  just  as  I  prefer  the  cheap  Attwood's  machine  (Fig.  163), 
which  used  to  be  employed  in  my  laboratory,  to  either  of  the 
two  elaborate  machines  which  are  now  in  use.  The  pulley  of 
Fig.  21,  which  I  use,  is  out  of  balance  and  badly  made;  it 
gives  ever  so  much  more  instruction  than  if  it  were  so 
expensively  and  correctly  made  that  the  friction  which  one 
wants  to  measure  had  almost  disappeared.  My  laboratory 
crane  is  a  model  crane,  much  too  carefully  constructed ;  but 
my  hydraulic  jack  and  differential  pulley  block  are  the  real 
things,  made  "for  human  nature's  daily  use."  Our  object  is 
not  to  find  out  how  to  make  a  machine  with  the  most  friction- 
less  bearings;  else  we  should  find  it  instructive  enough  to 
work  experimentally  with  ball  bearings,  such  as  are  used  in 
cycles  (Art.  70),  and  with  friction  wheel  bearings.  Again, 
a  student  is  told  to  judge  with  his  eye  as  to  whether  a  weight 
A  is  falling  steadily,  with  a  uniform  velocity.  Let  him  find 
out  for  himself  how  much  of  the  steadiness  is  due  to  irregu- 
larities in  the  rubbing  surfaces  of  the  machine,  and  how  much 
can  be  altered  by  altering  the  weight.  Do  not  spoon-feed  him. 
Let  it  be  a  discovery  of  his  own  that  his  eye  is  somewhat 
defective  as  a  speed-measurer ;  he  will  be  led  to  suggest  plans 
for  more  accurate  working  if  you  refrain  from  forcing  upon 
his  attention  your  elaborate  plans.  Indeed,  your  electric  and 
other  contrivances  for  measuring  velocity  may  be  so  elaborate 
as  to  hide  altogether  from  a  student  the  main  object  of  his 
experiments;  just  as  when  a  young  student  works  with  a  200- 
ton  testing-machine,  it  is  almost  impossible  for  him  to  think 
of  the  little  specimen  of  material  which  is  being  tested,  the 
testing-machine  itself  takes  up  so  much  space. 

55.  Force  of  Friction. — I  have  in  all  this  used  the  term  "fric- 
tion," or  the  term  "  effect  of  friction,"  to  mean  the  difference 
between  the  weight  which  would  balance  another  through  the 
mechanism  if  there  were  no  resistance  to  the  rubbing  of 
surfaces,  and  the  weight  which  will  just  overcome  the  other 
when  there  is  such  resistance.  Observe  that  there  is  a  great 
difference  between  the  cases,  A  overcoming  B  (Fig.  21),  and  B 
overcoming  A.  What  we  have  called  friction  is  due  to  the 
rubbing  at  all  sorts  of  surfaces  in  all  sorts  of  directions,  at  all 


APPLIED    MECHANICS. 


65 


Pig.  24. 


sorts  of  velocities  under  all  sorts  of  pressures,  and  we  are  led 
to  study  it  in  its  simplest  form,  where  at  one  part  of  a  pair  of 
surfaces  the  rubbing  is  exactly  of  the  same  kind  and  in  the 
same  direction  as  at  another  part ;  so  that  we  may  speak  of  the 
resultant  force  which  resists  motion  as  the  force  of  friction. 
Experiments  may  be  made  upon  the  apparatus  shown  in  Fig.  24, 
where  A  B  represents  a  table,  the  upper  level  surface  of  which 
is  wood,  iron,  brass,  or  other  material  to  be  experimented 
upon.  We  usually  experiment  on  smooth  surfaces,  c  is  a 
little  slide  made  of  any  material  whose  coefficient  of  friction 
with  the  table  we  wish  to  find.  Different  weights  may  be 
placed  on  it.  The  weight  of  the  slide,  together  with  the 
weight  lying  upon  it,  is  the  total  force  (R)  pressing  the  two 
surfaces  together,  c  is  pulled  by  the  weight,  w,  hung  from  a 
string,  passing  over  a  pulley  working  on  very  frictionless 
pivots.  The  weight,  w,  which  will  just  cause  the  slide  to  keep 
up  a  steady  motion  on  the  table,  is  taken  as  a  measure  of  the 
friction.  Of  course,  however,  it  really  includes  the  resistance 
of  the  pulley,  but  this  is  usually  neglected,  as  we  know  from 
previous  experiment  that  it  is  small.  It  is  found  necessary  to 
start  the  slide  by  giving  a  little  jerk  to  the  arrangement,  as 
the  friction  when  the  slide  is  motionless  is  found  to  be  some- 
what greater  than  when  it  is  moving.  This  is  one  of  the  most 
instructive  experiments  which  can  be  made  in  mechanics,  and 
I  hope  that  every  reader  will  make  a  series  of  observations. 
Let  him  correct  his  results  by  means  of  squared  paper,  and  he 
will  find  it  nearly  true  that  the  friction  is  a*  constant  fraction 
of  the  force  pressing  the  surfaces  together.  This  fraction  is 
called  the  coefficient  of  friction  and  usually  denoted  by  p.  I 


66  APPLIED    MECHANICS. 

give  its  value  for  a  few  surfaces,  but  a  student  had  better 
depend  upon  the  values  which  he  himself  arrives  at ;  they  will 
not  be  the  same  ;  they  may  differ  greatly  from  these. 

Oak  on  oak,  fibres  parallel  to  direction  of  motion  .  .  0'48 

,,               „      perpendicular     ,,               „  .  .  0'34 

„  ,,      endwise  „  ,,  0*19 

Metals  on  oak  „     parallel               „          ^    „  .  0-5  to  0'6 

Wrought  iron  on  wrought  iron,  wrought  iron  on  cast 

iron t  •  0-18 

Cast  iron  on  cast  iron       .        .        .        .         .  .  .  0'16 


Fig.  25. 

A  layer  of  oil  or  other  lubricant  between  the  surfaces  will 
greatly  reduce  the  friction.  Figures  given  for  the  coefficient 
of  friction  when  a  lubricant  is  used  are,  however,  very  greatly 
misleading,  but  for  student's  exercise  work  p.  may  be  taken  any- 
thing between  '04  and  '01  for  sperm  oil,  between  not  too  gener- 
ously but  continuously  lubricated  surfaces,  being  twice  as  much 
for  greases  as  for  sperm  oil. 

It  is  very  interesting,  after  determining  the  coefficient  in 
the  case  of  a  certain  pair  of  materials,  to  diminish  the  size  of 


APPLIED    MECHANICS.  67 

the  slide.  You  will  find  that  unless  you  diminish  it  so  much 
that  the  pressure  actually  alters  the  surfaces  in  contact  from 
being  quite  plane  you  will  get  pretty  much  the  same  result. 
You  will  also  find  that,  whether  the  motion  of  the  slide  is 
quick  or  slow,  if  the  weight  maintains  the  motion  steady 
when  it  is  slow  it  will  also  maintain  it  steady  when  quick. 

If,  instead  of  using  a  cord  and  weight,  we  move  the  slide 
by  tilting  the  table  more  and  more  from  the  horizontal  (see 
Fig.  25),  the  slide  getting  an  occasional  shove  to  start  it,  let  the 
inclination  of  the  table  be  found  in  degrees  when  the  weight 
of  the  body  itself  is  just  able  to  keep  up  a  steady  motion.  The 
tangent  of  the  angle  of  inclination  of  the  table  when  this 
occurs  can  be  found  in  a  book  of  mathematical  tables ;  it  proves 
to  be  equal  to  the  coefficient  of  friction.  This  method  of  ex- 
perimenting is  much  easier  and  is  more  exact  than  the  other, 
but  it  is  not  so  instructive  for  a  beginner.  (See  Art.  90.) 

56.  Loss  of  Energy  due  to  Friction. — In  the  simple  case 
with  which  we  began  Art.  50,  the  difference  of  pull  in  a  cord 
on  the  two  sides  of  a  pulley  was  what  we  called  the  friction 
of  the  arrangement,  whereas  we  see  that  the  friction  takes 
place  at  every  point  where  rubbing  occurs,  not  only  at  the 
pivot  but  even  in  the  fibres  of  the  cord  itself,  and  the  force 
at  one  point  may  be  very  different  from  the  force  at  another 
point.  Again,  any  force  acting  on  the  cord  has  a  greater 
leverage  about  the  axis  than  any  of  the  forces  of  friction  has. 
The  real  connection  between  the  two  things  is,  then,  this : 
what  we  have  generally  called  "  the  effect  of  friction,"  or  "  the 
friction  of  the  arrangement, "  multiplied  by  the  velocity  of  the 
cord  on  which  it  is  measured,  is  equal  to  the  sum  of  all  such 
products  as  the  friction  at  any  point  in  a  rubbing  surface, 
multiplied  by  the  velocity  of  rubbing.  In  fact,  if  the  weight  A 
in  falling  causes  the  weight  B  to  rise,  the  work  done  by  A  is 
greater  than  the  work  done  on  B  by  an  amount  which  is  called 
the  work  lost  in  friction,  and  this  is  the  work  done  against  the 
forces  of  friction  at  all  the  rubbing  surfaces. 

If  we  know  the  force  of  friction  at  any  place,  in  pounds, 
and  the  distance,  in  feet,  through  which  this  force  is  overcome — 
that  is,  the  distance  through  which  rubbing  has  occurred — the 
product  of  force  by  distance  measures  the  work  or  energy  spent 
in  overcoming  friction,  in  foot-pounds.  This  energy  is  all 
wasted,  or  rather,  it  is  all  changed  into  heat  and  does  not  come 
out  of  the  machine  as  mechanical  work,  the  shape  in  which  it 


68  APPLIED    MECHANICS. 

was  when  we  put  it  into  the  machine.  And  inasmuch  as  no 
machine  can  be  constructed  which  will  move  without  friction, 
we  never  get  out  of  a  machine  as  much  mechanical  work  as  we 
put  into  it. 

57.  Friction  at  Bearings  of  Shafts. — At  almost  every 
rubbing  surface  which  you  can  consider,  the  force  of  friction  is 
different  at  every  point  of  the  surface,  and  it  is  generally  acting 
in  different  directions  at  different  points.  Consider,  for  example, 
a  horizontal  shaft  and  its  bearing  (Fig.  26).      The  force  of 
friction  at  c,  per  square  inch  of  area  of 
rubbing  surface,  is  probably  not  the  same 
A  >        i||         as  at  A.     A  very  little  difference   in   the 
*r~    sizes  of   the  journal  and  step  will   cause 
a  considerable  difference  in  the  pressure  per 
square    inch    at   c   or   at    A.       Now    the 
force   of  friction   at  c,   multiplied   by  the 
Fig.  26.  velocity   of    rubbing,    gives    the   work    or 

energy  lost  per  second  in  friction  at  c  \ 
and  this,  added  to  the  energy  lost  at  every  other  place 
where  rubbing  occurs,  gives  the  total  loss  of  energy  per 
second  at  all  the  points.  It  is  not,  then,  a  simple  matter  to 
investigate  the  force  of  friction  at  every  point  of  such  a 
bearing ;  and  the  rigidity  of  the  metal,  and  a  number  of  other 
important  matters,  not  to  speak  of  the  nature  of  the  lubricant, 
must  be  taken  into  account  in  investigating  the  force  of 
friction  everywhere  when  the  shaft  is  transmitting  different 
amounts  of  power.  As  we  have  already  seen,  however,  experi- 
ment shows  that  the  energy  lost  in  friction  for  a  certain  amount 
of  motion  increases  proportionately  with  the  energy  actually 
transmitted  by  the  shaft.  Keeping  in  mind,  then,  the  general 
law — "the  force  of  friction  is  proportional  to  load" — it  is  easy 
to  see  how  to  reduce  the  frictional  loss  in  any  machine.  For 
instance,  when  a  wheel  is  transmitting  power,  the  load  on  the 
rubbing  surfaces  of  its  bearings  or  pivots  depends  on  the  power 
transmitted.  Now,  the  actual  force  of  friction  at  the  rubbing 
surface  is  about  the  same,  whatever  be  the  size  of  the  bearing  \ 
but  the  distance  through  which  rubbing  occurs  when  the  wheel 
makes  one  revolution  is  less  as  we  have  a  less  diameter  of  bearing ; 
in  fact,  the  force  of  friction,  multiplied  by  the  circumference  of 
a  cylindric  bearing,  is  the  energy  in  foot-pounds  lost  in  one 
revolution.  Our  rule  is,  then,  to  make  this  diameter  as  small 
as  possible,  consistently  with  sufficient  strength.  The  wheel  of 


APPLIED    MECHANICS.  69 

a  carriage  is  made  large,  and  the  axle,  where  rubbing  occurs, 
is  made  as  small  as  possible,  because  in  this  way  the  carriage 
moves  over  a  great  distance  for  a  small  amount  of  rubbing. 
There  is  another  reason,  however,  for  the  use  of  large  wheels 
in  carriages  on  common  roads  —  namely,  their  requiring  a  less 
tractive  force  to  get  over  obstacles,  such  as  stones.  In  some 
machines,  where  it  is  important  that  there  should  be  very  little 
friction  at  the  bearings  of  axles,  the  axles  are  made  to  lie  at 
each  end  in  the  angle  formed  by  two  wheels  with  plain  rims. 
The  main  axle  rolls  on  these  wheels,  and  it  is  only  at  the  axles 
of  the  wheels  that  there  is  rubbing.  This  rubbing  is  a  very 
slow  motion,  and  as  the  force  of  friction  is  but  little  increased 
in  consequence  of  the  weights  of  the  friction  wheels,  the  energy 
lost  in  friction  may  be  made  very  small  in  this  way.  Every 
curious  student  is  aware  of  the  way  in  which  rolling  takes  the 
place  of  sliding  in  the  ball-bearings  of  cycles.  This  kind  of 
bearing  will  probably  be  greatly  used  in  ordinary  machinery. 
The  resistance  is  as  if  we  had  a  co-efficient  of  friction  inversely 
proportional  to  the  diameter  of  the  ball  or  friction  roller, 
because  of  indentation  of  the  rolling  surfaces. 

If  you  compare  Watt's  parallel  motion  which  is  still  used 
in  some  pumping  engines  to  cause  the  piston-rod  to  move  in  a 
straight  line,  with  the  slide  which  is  now  so  common,  you  will 
see  that  there  is  very  much  less  loss  of  energy  by  friction  when 
the  parallel  motion  is  employed,  because,  whereas  in  the  slide 
the  rubbing  motion  is  as  much  as  the  motion  of  the  piston,  in 
the  parallel  motion  rubbing  only  occurs  at  the  pins  of  the 
arrangement.  Unfortunately,  this  arrangement  does  not  allow 
the  piston-rod  end  to  move  exactly  in  a  straight  line,  and 
produces  some  friction  between  the  piston  and  its  cylinder,  and 
between  the  piston-rod  and  stuffing-box  ;  and  it  is  also  much 
more  costly  and  less  compact  than  slides.  Hence  slides  are 
now  in  general  use. 

58.  In  a  journal  of  length  I  and  diameter  d,  if  w  is  the  load  and 
IJLW  the  force  of  friction,  at  n  turns  per  minute,  the  velocity  being 
proportional  to  nd,  the  rate  per  second  at  which  heat  is  developed 


per  square  inch  of  area  is  proportional  to    -ry-  .     Calling  •••-  the 

pressure  p,  we  see  that  pnd  ought  to  he  constant  in  all  journals 
if  they  are  to  have  about  the  same  rise  of  temperature  above 
surrounding  objects,  because  the  giving  out  of  heat  by  a  surface 
per  square  inch  may  be  taken  as  proportional  to  this  rise  of 
temperature.  This  rule  is  found  to  be  somewhat  misleading  for 


70  APPLIED    MECHANICS. 

lubricated  bearings,  because  /u  is  not  by  any  means  constant.  The 
values  of  pnd  found  in  practice  are  1,000,000  to  1,500,000  in 
locomotive  crank  pins,  calculated  from  full  pressure  and  speed ; 
250,000  in  marine-engine  crank  pins;  60,000  to  200,000  in  stationary 
engine  crank  pins;  crank  shaft  bearings,  36,000  ;  railway  carriage 
axles,  300,000.  When  what  is  understood  to  be  a  constant  in 
one's  theory  varies  between  30,000  and  1,500,000,  it  strains  one's 
sense  of  humour  to  maintain  the  gravity  necessary  in  the  writer  of 
a  text  book.  It  is  evident  that  we  must  pay  more  attention  to 
mere  pressure ;  and  it  may  be  said  that  in  practice  it  is  the  rule 
not  to  greatly  exceed  200  Ibs.  per  square  inch  in  shafting,  unless 
there  is  bath  lubrication,  and  then  the  limit  is  500  Ibs.  per  square 
inch;  600  Ibs.  per  square  inch  in  crank  pins;  1,000  Ibs.  per  square 
inch  in  crosshead  pins. 

The  practical  engineer  has  by  processes  of  success  and  failure 
arrived  at  dimensions  in  machine  design  which  we  have  always  the 
desire  to  see  reasons  for  in  our  theory.  It  is,  however,  sometimes 
forgotten  that  a  complete  theory  must  be  a  very  complicated  one, 
and  attempts  to  deduce  (find  reasons  for)  certain  veryguseful  rules 
(sometimes  impertinently  called  rules  of  tlmmb)  from  very  im- 
perfect theory,  do  not  always  succeed. 

It  would  be  interesting  to  find  out  why  it  is  the  universal 
practice  of  good  engineers  to  make  the  ratio  of  length  I  of  a 
bearing  to  its  diameter  d  increase  nearly  in  proportion  to  the 
number  of  revolutions  per  minute.  It  has  usually  been  lost  sight 
of  that  these  bearings  never  occur  in  long  lengths  of  shafting — 
only  in  separate  machines  like  fans,  centrifugal  pumps,  and  dynamo 
machines.  In  long  lengths  of  shafting,  bearings,  as  their  name 
implies,  are  mainly  used  as  mere  supports ;  but  in  separate 
machines  they  not  only  carry  weight — their  function  is,  especially 
in  light  machines  running  at  great  speeds,  to  keep  the  shaft  fixed 
in  direction.  If  then  there  is  any  bending  moment  M  in  the 
spindle,  due  to  centrifugal  force  through  want  of  balance,  or  due 
to  other  causes,  it  is  easy  to  show  that  the  pressure  per  square  inch 
of  bearing,  besides  what  is  due  to  steady  load,  is  proportional  to 
M/^H  and  this  multiplied  by  n  is  supposed  to  be  kept  constant.  If  M 
is  proportional  to  the  twisting  moment,  or  to  the  horse-power,  or 
d3n,  we  have  a  rule  l/d  so  n  which  agrees  with  the  practical  one.  I 
have  myself  worked  out  such  a  rule  for  M  as  a  very  likely  one  in 
certain  kinds  of  dynamo  machines.  In  all  probability  the  rule  for 
any  quick-speed  machine  would  turn  out  to  be  this :  that  if 
questions  of  cost  of  construction  and  space  did  not  intervene,  the 
ratio  of  I  to  d  ought  not  only  to  increase  with  n,  but  also  with  L, 
the  distance  between  the  main  bearings  of  the  machine.  Where 
there  is  a  possibility  of  error  in  the  allineation  of  the  two  bearings, 
we  have  a  reason  for  the  ratio  of  I  to  d  increasing  as  the  square 
root  of  the  speed. 

Exercises. — 1.  Find  the  horse-power  necessary  to  turn  a  shaft  9" 
diameter,  and  making  75  revolutions  per  minute,  if  the  total  load  on  it 
is  12  tons  and  <£,  the  angle  of  friction,  is  such  that  sin.  <p  =  -015.  Re- 
member  that  tan.  <(>  =  >*.  Ans.,  2-16 


ALLIED    MECHANICS. 


2.  find  the  horse-power  absorbed  in  overcoming  the  friction  of  a 
foot-step  bearing  4"  diameter,  the  total  load  being  1|  tons,  the  number 
of  revolutions  100  per  minute,  and  the  average  co-efficient  of  friction  *07. 

Am.,  0-5  nearly. 

3.  If  it  be  assumed  that  the  power  wasted  between  the  end  of  a  flat 
pivot  and  its  step  is  proportional  at  each  point  to  the  product  of  the 
velocity  and  pressure,  what  horse-power  will  be  absorbed  by  such  a  pivot, 
3"  diameter,  when  running  at  120  revolutions  per  minute,  the  load  on 
the  pivot  being  1\  tons,  and  the  average  co-efficient  of  friction  -06  ?    Inte- 
gration gives  the  total  energy  wasted  per  second  as  f  a  p  w  n,  where  a  is 
the  angular  velocity  in  radians  per  second,  w  is  the  total  load,  K  is  outside 
radius  of  pivot,  and  p.  is  the  co- efficient  of  friction. 

Ans.,  -639. 

4.  The  length  of  a  journal  is  9"  and  its  diameter  6" ;  it  carries  a  load 
of  3  tons.     What  horse-power  is  absorbed  in  friction  when  making  100 
revolutions  per  minute,  the  average  co- efficient  of  friction  being -015? 
What  number  of  thermal  units  per  minute  will  be  conducted  away  per 
square  inch  of  the  brass  ?  Ans.,  0*48  ;  0'2  centigrade  heat  units. 

5.  A  shaft  makes  50  revolutions  per  minute.     If  the  load  on  the  bear- 
ing be  8  tons,  and  the  dia- 
meter of  the  bearing  7  inches, 

at  what  rate  is  heat  being 
generated,  the  average  co- 
efficient of  friction  being 
•05? 

If  3  thermal  units  escape 
per  minute  when  the  temper- 
ature of  the  bearing  is  1°  C . 
higher  than  that  of  surround- 
ing objects,  what  will  be  the 
increase  in  temperature  caused 
by  the  heat  produced  at  the 
bearing?  Ans.,  19°-6  C. 


59.  Friction  and 
Speed.— You  will  find 
it  instructive  to  experi- 
ment with  such  a  piece 
of  apparatus  as  is  re- 
presented in  Fig.  28, 
designed  to  measure  the 
friction  between  sliders 
of  different  materials 
and  the  cast-iron  sur- 
face P.  Here  we  have 
a  pulley  with  a  broad, 
smooth  outer  surface. 
On  this  surface  lies  a 
slide  made  slightly  Fig.  27. 

concave,  to  fit  the  rim 
of  the  pulley.     On  this 

slide  we  can  hang  different  loads  w  by  the  arrangement  shown 
in  the  figure,  and  the  slide  can  only  move  a  small  distance  in 


72  APPLIED    MECHANICS. 

any  direction  on  account  of  stops.  There  is  a  fly-wheel  to  give 
steadiness  of  motion  when  the  apparatus  is  worked  by  hand.  Sup- 
pose, now,  that  P  rotates  in  the  direction  of  the  arrow.  Friction 
causes  the  slide  to  move  in  the  direction  of  motion  until  it  is 
brought  up  by  a  stop.  Now  let  weights  be  placed  in  the  scalo 


Pig.  28. 

pan  w  until  the  slide  is  held  in  a  position  half  way  between  the 
stops.  Evidently  the  force  of  friction  between  the  slide  and  r  is 
just  balanced  by  the  weight  in  the  scale-pan.  With  this  appar- 
atus you  can  not  only  find  the  co-efficient  of  friction  for  two 
rubbing  surfaces  easily  at  any  speed,  but  you  can  very  quickly 
vary  your  experiments,  s  is  a  speed  counter. 

To  what  extent  I  ought  to  be  ashamed  of  the  following  facts  I 
don't  know.  They  are  instructive.  Successive  generations  of 
etudents  at  Finsbury  obtained  results  from  an  apparatus  like  Fig.  28. 


APPLIED    MECHANICS.  73 

It  was  arranged  to  be  driven  by  the  college  engine  at  many 
different  speeds ;  and  there  was  a  speed  counter.  The  slide  had  a 
much  longer  arc  of  contact  than  is  shown  in  the  figure,  because  a 
rocking  motion,  instead  of  sliding,  was  apt  to  be  set  up.  The  load 
was  applied  at  a  single  point  in  the  centre  of  the  top  of  the  slider, 
In  every  case  when  the  load  was  kept  constant  the  friction  was 
greatest  at  low  speeds ;  it  got  less  and  less  as  the  speed  increased, 
and  reached  a  minimum  value ;  after  which  it  increased  steadily  as 
the  speed  increased,  the  rate  of  increase  of  friction  with  speed 
getting  less  towards  our  highest  speeds.  I  thought  these  curves 
obtained  by  students  well  worthy  of  special  study,  and  several 
times  projected  an  investigation  of  my  own,  and  urged  others  to 
take  it  up ;  but  I  was  too  busy  with  other  matters  to  give  it  much 
attention.  Now,  from  Prof.  Osborne  Reynolds's  explanation  of 
the  results  of  Mr.  Beauchamp  Tower's  experiments  one  sees  how 
very  different  the  phenomenon  is  from  what  occurs  between  flat 
surfaces.  Air  is  being  pumped  by  friction  into  the  space  between 
the  slider  and  pulley,  and  the  pressure  underneath  the  slider  is 
greater  than  the  atmospheric  pressure,  and  varies  from  point  to 
point,  as  we  have  proved  by  inserting  little  pressure-gauges. 

We  are  about  to  use  now  another  piece  of  apparatus  (Fig.  29), 
the  slider  being  flat,  and  lying  on  a  circular,  flat  horizontal 
plate,  which  may  be  kept  rotating  at  any  speed.  The  slider  is 
prevented  from  moving  much  by  stops,  and  friction  is  balanced, 
as  in  the  above  case,  by  a  scale-pan.  The  apparatus  is  only 
now  (July,  1896)  being  run  for  the  first  time.  What  sorts  of  results 
will  be  obtained  from  it  I  do  not  know. 

60.  Although,  on  the  whole,  in  any  machine  the  average  forces 
of  friction  do  not  seem  to  depend  much  upon  speed,  and  they  increase 
in  proportion  to  the  load,  and  in  other  ways  seem  to  follow  the 
laws  set  down  in  Art.  55,  when  we  make  experiments  on  the 
friction  at  any  one  place  in  a  machine  we  obtain  inconsistent 
results.  So  long  as  our  theory  of  an  action  is  wrong,  our  experi- 
ments give  rise  to  what  are  called  inconsistent  results.  On  the 
hypothesis  of  Art.  55  the  mathematicians  have  built  a  science,  and 
thousands  of  examples  and  exercises  have  been  invented  to  illus- 
trate it.  The  exercises  and  examples  are  valuable  to  the  engineer ; 
but  he  must  remember  that  they  have  been  invented  by  mathema- 
ticians for  the  training  of  mathematicians,  and  he  must  exercise 
caution  in  using  the  results.  I  give  some  examples  in  Art.  58. 
Two  substances,  when  they  really  touch,  get  welded  together ;  and 
this  seizing  seems  to  occur  in  some  journals  and  footsteps  under 
heavy  loads.  Bodies  said  to  touch  or  rub  on  one  another  are 
really  separated  by  a  layer  of  air  or  other  fluid.  A  slider  like  c 
(Fig.  24)  is  separated  from  the  table  A  B  by  a  layer  of  air ;  and 
the  greater  the  load,  the  less  is  the  thickness  of  air. 

Students  will  find  it  very  interesting  to  study  the  friction 
between  two  scraped  surfaces  in  the  workshop.  If  one  plate  is 
laid  down  on  the  other,  there  is  usually  very -little  friction,  because 
there  is  a  thick  layer  of  air  separating  the  surfaces.  By  putting 
on  a  load,  and  giving  small  sliding  motions,  we  can  make 
the  layer  of  air  very  thin — so  thin,  indeed,  that  when  the  top 


74 


APPLIED    MECHANICS. 


plate  is  lifted  the  "bottom  one  sticks  to  it,  and  lifts  also,  partly 
because  there  is  a  partial  vacuum  between  them,  partly  also,  prob- 
ably, because  of  molecular  attraction  seeing  that  it  occurs  even  in 
a  good  vacuum.  Now,  when,  through  the  one  plate  having  lain 
on  the  other  a  considerable  time,  or  through  pressure,  we  get  the 
layer  of  air  very  thin,  it  is  found  that  there  is  considerable  resistance 
to  sliding.  In  fact  in  this  case,  where  we  might  expect  to  find  the 
phenomena  of  friction  assuming  their  simplest  form,  we  find  what 
seem  to  be  the  most  inconsistent  results.  When  c  (Fig.  24)  slides, 
it  seems  as  if  fresh  air  were  being  carried  into  the  space  between 
the  surfaces,  keeping  them  apart,  and  that  the  greater  the  velocity 


Fig.  29. 


of  rubbing  the  more  air  is  carried  in,  so  that  the  separation 
between  the  surfaces  is  proportional  to  the  velocity  of  rubbing. 
When  we  come  to  discuss  fluid  friction,  and  reflect  that  all  the 
friction  we  know  of  is  really  fluid  friction,  we  shall  not  be  aston- 
ished that  the  laboratory  results  from  the  rubbing  of  solids  are 
sometimes  inconsistent-looking — we  shall  wonder  greatly  that  the 
above-mentioned  law  should  be  even  approximately  true.  But  I 
cannot  think  the  effect  merely  one  of  the  fluid,  there  is  also 
molecular  attraction.  In  any  lubricated  bearing  which  is  not  kept 
flooded  with  oil,  the  rules  called  the  laws  of  solid  friction  are  found 
to  be  approximately  true — that  is,  we  may  take  the  load  on  the 
bearing  in  pounds,  multiplied  by  a  coefficient  p,  as  representing  a 
force  of  friction ;  and  this,  multiplied  by  the  distance  of  rubbing 
in  feet,  is  the  mechanical  energy  converted  into  heat  by  friction. 
ft  is  less  as  the  temperature  is  greater,  partly  because  the  viscosity 


APPLIED    MECHANICS.  75 

of  a  fluid  diminishes  with  temperature  (see  Art.  63) ;  but  it  is  not 
only  because  of  this,  for  the  body  of  the  lubricant  also  alters  with 
temperature — that  is,  it  tends  more  to  get  squeezed  out  of  place. 
Of  course  the  friction  depends  greatly  upon  the  nature  of  the 
lubricant,  and  the  phenomena  are  really  so  very  complicated  that 
in  the  present  state  of  our  knowledge  the  reader  must  perforce  be¥ 
satisfied  with  the  rough  general  law  that  I  have  mentioned.  That' 
the  supply  of  oil,  however  small,  shall  be  continuous,  and  not  inter- 
mittent, is  regarded  as  the  most  important  condition  in  the  lubrica- 
tion of  bearings.  The  lubricant  ought  to  suit  the  nature  of  the 
load.  Thus  great  body  is  necessary  when  the  loads  are  great,  so 
that  the  oil  may  not  be  squeezed  out;  and  greases  and  solid 
lubricants,  such  as  soapstone  and  plumbago,  must  be  used  for  very 
heavy  loads.  It  is  also  usual  to  cast  plugs  of  white  metal  and 
other  soft  alloys  in  recesses  of  the  step,  and  in  some  cases  to  line 
the  whole  step  with  such  a  soft  alloy.  As  for  the  detailed 
construction  of  pedestals,  hangers,  A  frames,  and  other  supports  for 
shafting,  this  is  to  be  learnt  in  the  drawing- office  and  shops,  and  it 
would  be  useless  to  refer  to  it  here. 

61.  As  it  has  been  found  that  with  some  kinds  of  material  the 
statical  friction — that  is,  the  friction  which  resists  motion 
from  rest — is  somewhat  greater  than  the  friction  of  the  surfaces 
when  actually  moving,  experiments  have  been  made  to  deter- 
mine whether,  at  very  small  velocities  indeed,  with  such 
materials,  there  is  not  a  gradual  increase  in  the  friction.  It  is 
known  that  at  ordinary  velocities  the  friction  is  much  the 
same  as  at  a  velocity  of  *01  foot  per  second.  We  have  reason 
to  believe  that  with  metals  on  metals  and  air  between,  there  is 
the  same  friction  at  all  velocities,  even  down  to  one-five- 
thousandth  of  a  foot  per  second,  whereas  with  metals  on  wood 
the  friction  increases  gradually  as  the  velocity  diminishes,  until 
when  the  velocity  is  0,  the  friction  is  what  we  call  static 
friction.  Again,  at  very  high  velocities  it  has  been  found  that 
there  is  a  very  decided  diminution  of  the  coefficient  of  friction 
between  a  cast  iron  railway  brake  and  the  wrought  iron  tyre 
of  a  wheel.  The  coefficient  was  '33  for  very  slow  motion,  '19 
for  a  speed  of  29  feet  per  second,  and  '127  for  a  speed  of  66 
feet  per  second.  It  has  also  been  observed  in  these  railway 
brake  experiments  that  when  a  certain  pressure  is  applied  for 
a  short  space  of  time  the  friction  diminishes.  All  such  results 
as  these,  however  interesting  they  may  be  to  the  railway 
engineer,  tell  us  nothing  about  what  I  have  hitherto  called 
friction,  because  I  have  supposed  the  rubbing  surfaces  to 
remain  unaltered,  whereas  these  railway  brakes  are  rapidly 
worn  away,  and  the  effects  of  abrasion  and  polishing  are  of  an 


76  APPLIED    MECHANICS. 

utterly  different  kind  from  the  effects  of  friction  of  which  I 
have  hitherto  been  speaking. 

62.  We  must  remember  that  although  friction  leads  to  waste 
of  energy,  all  the  energy  spent  in  overcoming  friction  being 
converted  into  another  form  of  energy  called  heat,  still  the 
force  of  friction  is  very  useful.      The  weight  resting  on  the 
driving-wheels  of  a  locomotive  engine  multiplied  by  the  co- 
efficient of  friction  between  the  wheels  and  rails  represents  the 
greatest  pull  which  the  engine  can  exert  upon  a  train.    Suppose 
the  weight  on  the  driving-wheels  to  be  15  tons,  and  that  the  co- 
efficient of  friction  of  wrought  iron  on  wrought  iron  is  about  0-2, 
the  greatest  pull  which  the  locomotive  can  exert  is  15  x  0*2,  or 
3  tons.  If  the  train,  including  the  locomotive  itself,  resists  with  a 
greater  force  than  this,  the  driving-wheels  must  slip ;  if  the  train 
resists  with  a  less  force  than  this,  there  is  no  slipping,  the  wheels 
simply  roll  on  the  rails.     Again,  it  is  the  friction  between 
the  soles  of  our  feet  and  the  ground  that  enables  us  to  walk ; 
friction  enables  us  to  handle  objects ;  friction  enables  a  nail  to 
remain  in  wood ;  friction  keeps  mountains  from  rolling  down. 

63.  Fluid  Friction. — I  have  been  considering  the  friction 
between  solid  bodies  only.     The  friction  between  liquids  and 
solids  or  between  liquids  and  liquids  is  of  a  very  different  kind. 
If  a  man  attempts  to  dive  into  water  unskilfully,  and  falls  prone, 
you  know  that  the  water  offers  a  very  considerable  resistance 
to  a  change  of  shape.     Now  this  is  mainly  the  resistance  that 
any  body  offers  to  being  rapidly  set  in  motion.     If  you  came 
colliding  against  the  end  of  the  most  frictionless  carriage,  you 
would  also  experience  its  resistance  to  being  suddenly  set  in 
motion;    whereas  the   constant  steady  resistance   to  motion 
which  the  carriage  experiences  when  moving  with  a  uniform 
velocity  is  called  friction.     What  I  wish  rather  to  refer  to  is 
the  resistance  to  the  motion  of  water  in  a  pipe,  the  resistance 
to  the  steady  motion  of  a  ship. 

In  nearly  all  ordinary  cases  the  motion  is  complicated  and 
difficult  to  study.  The  simplest  motion  is  in  plane  parallel 
layers.  Imagine  two  infinite  plane  parallel  boundaries  with 
the  fluid  between  :  one  of  the  boundaries  at  rest,  the  other 
moving  with  uniform  velocity  v  in  its  own  plane.  Imagine 
the  fluid  to  stick  to  each  boundary.  If  b  is  the  distance 
between  them,  the  tangential  force  per  unit  area  required  to 
keep  up  the  motion  is  p.  v  -f  b  if  p.  is  the  coefficient  of 
viscosity.  Theory  shows  that  p.  ought  to  be  constant  if  the 


APPLIED    MECHANICS. 


77 


Fig.  30. 


motion  is  truly  in  plane  layers.  As  we  cannot  experiment 
with  infinite  surfaces,  I  thought  that  I  could  approach  the 
condition  most  nearly  with  the  apparatus  shown  in  Fig.  30. 
F  is  a  hollow  cylindric  body  supported  so  that  it  cannot  move 
sidewise,  and  yet  so  that 
its  only  resistance  to 
turning  is  due  to  the 
twist  it  would  give  the 
suspension  wire,  A.  c  c  is 
water  or  other  liquid  fill- 
ing the  annular  space  be- 
tween the  cylindric  sur- 
faces D  D  and  E  E,  and 
wetting  both  sides  of  F. 
When  the  vessel  D  D,  E  E 
is  rotated,  the  water  mov- 
ing past  the  surfaces  of  F 
tends  to  make  F  turn 
round,  and  this  frictional 
torque  is  resisted  by  the 
twist  which  is  given  to  the  wire.  The  amount  of  twist  in  the 
wire  gives  us,  then,  a  measurement  of  the  viscosity  of  liquids, 
and  investigations  may  be  made  under  very  different  conditions. 
The  above  apparatus  was  designed  and  partly  constructed 
in  Japan  in  1876.  Experiments  made  with  it  by  Finsbury 
students  on  olive  oil  are  described  in  the  Proceedings  of  the 
Physical  Society  of  London,  March,  1893.  At  constant 
temperature  below  a  certain  critical  speed,  I  found  that  the 
friction  was  proportional  to  the  velocity,  so  that  p  could  be 
found.  At  that  critical  speed  I  found  that  there  was  a  sudden 
change  in  the  law,  and  above  that  speed  the  friction  is  pro- 
portional to  a  higher  power  of  the  speed  than  1.  We  know 
that  above  the  critical  speed  the  plane  motion  which  I 
described  above  would  become  unstable,  and  eddies  would  be 
formed.  From  a  theoretical  point  of  view  it  is  curious*  that 

*  Some  experiments  of  Mr.  D.  Baxandall,  not  yet  published,  show  that 
the  forces  of  friction — that  is,  the  resultant  forces  applied  to  the  solid  bodies 
which  form  the  boundaries  of  a  mass  of  fluid  to  maintain  relative  motion — are 
strictly  proportional  to  the  relative  velocity  at  small  speeds,  and  there  is 
always  a  critical  speed  above  which  the  friction  is  proportional  to  a  higher 
power  of  the  velocity.  We  have  tried  surfaces  arranged  like  those  of  churns, 
with  paddles  and  curiously  shaped  vanes,  and  the  law  is  always  true.  With 
mixtures  of  glycerine  and  water,  the  higher  power  of  the  velocity  above 
referred  to  depends  on  the  proportions  of  glycerine  and  water. 


78  APPLIED    MECHANICS. 

the  phenomenon  should  have  been  so  marked  between  my 
cylindric  surfaces,  because  even  at  slow  speeds  cylindric  motion 
ought  to  be  unstable  inside  a  fixed  cylindric  surface — that  is,  in 
the  inner  part  of  my  trough. 

At  speeds  below  the  critical,  I  measured  p.  at  many 
different  temperatures,  and  noted  the  rapid  decrease  in  it  as 
the  temperature  increased. 

Yery  interesting  observations  may  be  made  at  small  speeds 
by  immersing  similar  and  equal  heavy  discs  of  brass  in  air, 
water,  and  oil,  suspending  them  by  fine  steel  wires.  (See  Fig.  30.) 
When  the  suspension  wires  are  twisted  and  let  go,  the  bodies 
vibrate  like  the  balance  of  a  watch.  But  it  is  only  the  one  which 
vibrates  in  air  that  goes  on  vibrating  for  a  long  time  ;  the  one 
in  water  keeps  up  its  motion  longer,  however,  than  the  one  in 
oil,  showing  that  there  is  more  frictional  resistance  in  oil  than 
in  water,  and  more  in  water  than  in  air.  The  rates  of  diminu- 
tion of  swing  or  the  stilling  of  the  vibrations  tell  us  the  relative 
viscosities  of  the  fluids.  If,  by  means  of  a  pointer  or  mirror 
attached  to  the  wire,  you  observe  the  various  angular  displace- 
ments, noting  the  time  for  each,  and  then  plot  your  observa- 
tions on  squared  paper  (as  in  Art.  53),  you  will  find  what  is 
very  nearly  a  curve  of  sines  for  the  vibrations  in  air  ;  and  for 
the  different  liquids  damping  curves,  which  show  the  effect  of 
friction  in  the  liquids.  Similarly,  the  rate  of  diminution  of 
swing  of  the  vibrating  fluids  in  U  tubes,  one  containing  water 
and  the  other  oil,  tells  us  about  the  relative  co-efficients  of 
viscosity  of  the  liquids. 

64.  The  motions  in  these  cases  are  not  so  simple  as  in  the  case 
which  I  considered  experimentally.  The  question  of  the  resist- 
ance to  the  passage  of  fluids  through  pipes  is  one  which  has 
attracted  much  attention,  and  the  results  of  experiments  seemed 
very  inconsistent  until  Professor  Osborne  Reynolds  considered 
the  problem.  It  was  known  that  the  pressure  difference  at  the 
ends  of  a  level  uniform  pipe  necessary  to  produce  a  certain 
flow  was  proportional  to  the  length  of  the  pipe,  and  it  was  usual 
to  say  that  the  force  of  friction  was  proportional,  as  in  all 
other  cases  of  fluid  friction,  to  the  wetted  area  ;  it  is  quite 
independent  of  the  pressure ;  it  is  proportional  to  the  velocity 
of  the  water  when  the  velocity  is  small,  but  at  high  speeds  it 
increases  much  more  quickly  than  the  speed.  Thus,  as  I  said 
in  the  first  edition  of  this  book,  of  water  flowing  in  a  certain 
pipe,  "  at  the  velocities  of  1,  2,  3,  etc.  inches  per  second,  the 


APPLIED    MECHANICS. 


79 


friction  is  proportional  to  the  numbers  1,  2,  3,  etc.,  whereas 
at  the  velocities  of  1,  2,  3  yards  per  second  the  friction  is  pro- 
portional to  the  numbers  1,  4,  9,  etc."  At  small  velocities, 
three  times  the  speed  means  three  times  the  friction ;  whereas 
at  great  velocities,  such  as  those  of  ships,  three  times  the  speed 
means  nine  or  more  times  the  friction.  We  see,  then,  that 
friction  in  fluids  is  proportional  to  the  speed  when  the  speed  is 
small,  to  the  square  of  the  speed  when  the  speed  is  greater,  and 
at  still  greater  speeds  the  friction  increases  more  rapidly  than 
the  square  of  the  speed.  The  resistance  to  motion  of  a  rifle 
bullet  is  proportional  to  the  square  root  of  the  fifth  power 
of  the  speed ;  that  is,  a  bullet  going  at  four  times  the 
velocity  meets  with  thirty-two  times  the  frictional  resistance 
from  the  atmosphere.  (See  Art.  68.)  Again,  it  has  been 
found  that  the  friction  is  much  the  same  whatever  be  the 
pressure.  Thus  it  is  found  that  when  the  disc  and  liquid 
apparatus  is  placed  in  a  partial  vacuum  or  under  considerable 
pressure,  there  is  exactly  the  same  stilling  of  the  vibrations. 

This  fact  is  illustrated  by  the  apparatus,  Fig.  31.  Water 
tends  to  flow  from  vessel  A  to  vessel  B,  through  the  long  tube. 
Whether  the  tube  is  in  the  position  shown  in  Fig.  31,  or  in 
the  position  Fig.  32,  or  is  acting  as  a  syphon,  we  find  the  same 
flow  through  it ;  the  same  quantity  of  water  passes  through  it 
per  second,  although  the  pressure  of  the  water  in  the  tube  in 


Pig.  81. 

the  position  Fig.  32  is  very  much 

greater  than  in  the  position  Fig.  31, 

or  again  when  the  tube  is  a  syphon. 

In  the  apparatus  actually  used  by  Fig.  32. 

me,    there    is   a   stopcock    in    the 

middle  of   the   tube,    and   by  nearly    closing   it   one  is   sure 

that  the  friction  occurs  at  the   place  where  the    pressure    is 


80 


APPLIED    MECHANICS. 


greatest  in  Fig.  32.  The  comparison  is  most  readily  made 
by  observing  how  long  it  takes  for  a  certain  change  of  levels 
to  take  place  in  the  two  vessels,  repeating  this  several  times 
with  the  tube  in  various  positions,  beginning  and  ending  each 
experiment  with  the  same  difference  of  levels.  Again,  fluid 
friction,  for  even  considerable  velocities,  does  not  seem  to 
depend  much  on  the  roughness  of  the  solid  boundary.  This 
seems  to  be  due  to  the  fact  that  a  layer  of  fluid  adheres  to 
the  solid  surface  and  moves  with  it.  Even  when  the  disc  of 
Art.  63  is  indented,  or  when  large  grooves  are  cut  in  it  we 
find  practically  the  same  frictional  resistance. 

Comparison  of  the  Laws  of  Fluid  and  Solid  Friction. 


Friction  between  Solids. 


Fluid  Friction. 


1.  The    force    of    friction    does 
not  much  depend  on  the  velocity, 
but  is   certainly   greatest  at  slow 
speeds. 

2.  The  force  of  friction  is  pro- 
portional   to    the    total    pressure 
between  two  surfaces. 

3.  The  force  of  friction  is  in- 
dependent   of    the    areas    of    the 
rubbing  surfaces. 

4.  The  force  of  friction  depends 
very  much   on  the  nature  of   the 
rubbing  surfaces,  their  roughness, 
etc. 


1.  The    force    of    friction   very 
much  depends  on  the  velocity,  and 
is  indefinitely  small  when  the  speed 
is  very  slow. 

2.  The  force  of  friction  does  not 
depend  on  the  pressure. 

3.  The  force  of  friction  is  pro- 
portional to  the  area  of  the  wetted 
surface. 

4.  The    force     of     friction     at 
moderate    speeds    does    not  much 
depend  on  the  nature  of  the  wetted 
surfaces. 


65.  Molecular  theory  gives  us  the  cause  of  fluid  friction  in  such 
a  fluid  as  air.  Layers  of  fluid  at  different  velocities  are  continually 
interchanging  molecules  by  ordinary  diffusion  ;  consequently,  the 
relative  motion  is  being  destroyed,  the  rate  of  loss  of  momentum 
by  one  layer  and  gain  of  it  by  the  other  enabling  us  to  state  that 
the  force  required  to  maintain  the  motion  is  proportional  to  the 
surface  of  contact  and  to  the  relative  velocity.  In  regard  to 
friction  in  gases,  the  explanation  is  complete,  the  greater  diff  usivity 
at  higher  temperatures  causing  the  viscosity  to  be  greater  also. 
Indeed,  viscosity  is  proportional  to  the  square  of  the  absolute  tem- 
perature. In  the  same  way,  if  two  trains  were  passing  one  another  and 
the  same  number  of  passengers  jumped  from  each  train  to  the  other, 
the  trains  would  become  more  equal  in  speed  ;  there  would  seem 
to  be  a  mutual  frictional  force  between  them  proportional  to  the 


APPLIED    MECHANICS.  81 

rate  of  loss  or  gain  of  momentum  per  second.  But  in  liquids  there 
is  less  viscosity  at  higher  temperatures,  although  there  is  greater 
diffusivity.  This  is  probably  due  to  the  fact  that  mere  diffusivity 
is  all-important  in  gases,  the  molecules  of  which  exert  no  forces 
upon  one  another  except  by  collision  ;  wheroas  in  liquids,  although 
the  greater  diffusivity  at  higher  temperatures  would  tend  to  make 
them  behave  like  gases,  in  regard  to  viscosity,  forces  are  always 
acting  between  the  molecules  which  resist  shearing  strain,  and 
these  forces  get  less  as  the  temperature  increases. 

Now,  in  any  case  of  relative  motion  between  the  bounding 
surfaces  of  a  fluid,  beyond  a  certain  velocity,  motion  in  plane 
layers  becomes  unstable  and  sinuous  motion  sets  in.  This  means 
that  the  surfaces  across  which  interchange  of  momentum  by 
diffusion  may  take  place  become  greater  in  area ;  so  that  above 
a  certain  critical  speed  I  take  it  that  we  may  expect  almost  any 
law  connecting  friction  and  speed.  Theory  shows  that  for  any 
given  shape  of  surface  the  critical  speed  will  be  less  as  the  density 
of  the  fluid  is  greater,  and  it  is  less  as  JJL  is  less.  A  very  friction- 
less  fluid  is  very  unstable.  (See  Appendix.) 

I  believe  that  all  friction  said  to  be  between  solid  surfaces  is  to 
be  regarded  as  taking  place  in  the  fluid  which  always  separates 
such  surfaces.  This  statement  seems  a  mere  truism.  It  is  like 
many  another  yet  to  be  made  by  discoverers  in  applied  physics. 
As  a  matter  of  fact,  the  above  table  showing  the  utter  difference 
in  character  between  the  phenomena  of  solid  and  fluid  friction 
quite  hid  from  everybody's  view  the  fact  that  all  friction  must  be  a 
fluid  friction,  until  Professor  0.  Reynolds  opened  our  eyes.  He  has 
given  us  in  his  lectures  at  the  Royal  Institution  and  in  his  paper 
published  in  the  Transactions  of  the  Royal  Society  the  suggestion 
that  it  is  to  some  extent  in  the  solution  of  hydrodynamic  problems 
we  must  look  for  an  explanation  of  the  curious  phenomena  of  solid 
friction.  I  have  already  mentioned  a  curious  phenomenon  often 
brought  to  my  notice  in  connection  with  the  use  of  the  apparatus 
shown  in  Fig.  27.  Let  me  now  describe  some  experiments  made 
on  the  friction  of  journals.  Probably  everybody  has  been  occa- 
sionally interested  in  curious  results  obtained  when  testing  oils  with 
the  Thurston  oil-tester  (Fig.  33).  Many  of  these  will  be  found 
published  in  Mr.  Thurston's  book  on  "The  Materials  of  Engi- 
neering," Part  I. ;  but  every  mechanical  laboratory  ought  to 
be  provided  with  the  apparatus,  that  students  may  study  the 
phenomena  for  themselves.  In  Hirn's  experiments,  made  in  1855, 
he  found  that  the  force  of  friction  was  proportional  to  the  square 
root  of  the  product  of  load  and  velocity.  In  the  experiments  of 
Mr.  Beauchamp  Tower  upon  a  steel  journal  with  a  gun-metal  cap 
only,  the  cap  being  loaded,  the  lubrication  being  practically  an 
oil  bath,  the  friction  was  found  to  be  practically  independent  of  the 
load  for  loads  so  excessive  as  from  100  to  520  Ibs.  to  the  square 
inch  (the  diameter  being  4  inches  and  length  6  inches,  the 
pressure  is  taken  as  the  whole  load  divided  by  24),  and  in  all  cases 
to  be  practically  proportional  to  the  square  root  of  the  velocity. 
If,  instead  of  such  excessive  lubrication  as  we  have  in  an  oil  bath, 
f.hore  was  only  the  lubrication  due  to  an  oily  pad  pressed  again?t 


APPLIED    MECHANICS. 


Fig.  S3. 

the  journal  below,  the  ordinary  law  assumed  for  the  friction  of 
solids  was  found  to  he  approximately  followed.  In  collar  hearings, 
as  in  the  thrust  hearings  of  a  propeller  shaft,  he  again  found  that 
the  ordinary  laws  of  solid  friction  are  fairly  well  followed,  and 
that  only  very  much  less  pressures  (75  Ibs.  per  square  inch  at  high 
speeds  and  90  Ihs.  per  square  inch  at  low  speeds)  were  possible 
without  seizing.  These  curious  phenomena  have  "been  completely 
explained  hy  Professor  O.  Reynolds.  They  depend  upon  the 


APPLIED   MECHANICS.  83 

carrying  of  the  lubricant  into  the  space  "between  the  step  and  the 
journal,  and  if  there  is  not  an  oil  bath  and  the  motion  is  all 
in  one  direction,  the  lubricant  leaves  the  place  where  it  is  most 
wanted  and  the  journal  seizes  at  comparatively  low  pressures; 
whereas  if  there  is  such  an  irregularity  of  motion  or  reversal  of 
motion  as  helps  the  lubricant  to  maintain  its  place,  very  great 
pressures  may  be  employed.  Thus  in  crank  pins  and  railway 
axles  pressures  as  high  as  400  Ibs.  per  square  inch  with  sperm,  oil 
and  600  Ibs.  with  mineral  grease  have  been  used ;  and,  indeed,  in 
slow-moving  steam  engines  nearly  double  these  pressures  have 
been  used.  It  is  well  to  remember  that  at  the  place  where  the 
journal  most  nearly  approaches  the  step  the  pressure  in  the  oil 
becomes  very  great,  and  if  there  is  an  opening  there  the  oil  is 
forced  out.  *It  is  now  quite  common  to  employ  a  force-pump  to 
pump  oil  at  great  pressure  into  these  parts  of  the  bearings,  and  in 
consequence  much  higher  loads  on  bearings  are  possible  than  used 
to  be  the  case. 

In  any  case,  we  must  try  to  understand  the  distribution  of 
pressure  in  the  bearing,  so  that  in  our  endeavours  to  utilise  an 
ordinary  syphon-lubricator  for  example,  we  shall  not  be  attempting 
impossible  things. 

66.  It  is  in  the  drawing- office  and  shops  that  students  will  become 
acquainted  with  the  methods  in  actual  use  for  supporting  horizontal 
shafts.     Footsteps  for  vertical  shafts,  which  give  endless  trouble  in 
many  high  factories,  are  now  in  many  cases  water-  or  oil-borne, 
being  converted  into  the  rams  of  hydraulic  presses  having  only 
a  very  small  range  of  vertical  motion,  or,  rather,  so  arranged  that 
the  lifting  force  of  the  fluid  shall  always  be  less  than  the  weight  of 
the  shaft. 

67.  It  is  when  great  forces  have  to  be  overcome  slowly,  and 
particularly  with  a  long  translational  motion,  that  water-pressure 
machinery  shows  itself  most  greatly  superior  to  other  machinery, 
for  friction  seems  to  be  nearly  independent  of  pressure.     But  if  in 
any  place  the  water  is  set  in  rapid  motion  there  is  internal  friction 
and  waste   of  energy.     Where  fluids  move  so   slowly  that  the 
friction  is  proportional  to  the  velocity,  we  seldom  consider  it  in 
our  engineering  work.     At  the  valves  of  pumps  and  in  pipes  it 
is  usually,  on  the  whole,  economical  to  let  energy  be  wasted  in 
water  friction.     There  is  a  certain  relationship  between  velocity,  t;, 
and  diameter,  d,  of  pipe  for  a  particular  fluid  which  causes  a  certain 
v  to  be  critical.     Below  that  value  of  v  the  water  flows  in  straight 
streams ;  at  that  critical  value  the  beautiful  straight  lines  which 
Professor  Reynolds  shows  coloured  with  aniline  dye  suddenly  break 
up  into  confused,  smoke-like  eddying  cloud.     Below  the  critical 
velocity  the  total  pressure  difference — or  friction,  as  we  may  call 
it — required  to  keep  up  the  flow  is  proportional  to  the  velocity. 
Above  that  critical  velocity  the  friction  is  proportional  to  a  power 
of  the  velocity  which  varies  from  1'7  to.  2,  depending  upon  the 
nature  of  the  material  of  the  pipe. 

68.  The  mathematical  investigation  of  the  resistance  to  the  pas- 
sage of  a  body  through  a  viscous  fluid  is  so  difficult  that  we  have 
almost  no  results  which  may  be  relied  upon.    Without  viscosity 


84  APPLIED    MECHANICS. 

there  would  be  no  resistance  to  steady  motion,  whatever  tho  shape  of 
the  object.  It  is  difficult  to  imagine  that  there  would  he  no  propelling 
force  on  a  sailing-boat  if  the  air  were  Motionless,  and  yet  this  is  so. 
Even  in  the  case  of  a  ship,  experiments  on  which  have  been  going 
on  continuously  since  ships  were  first  built,  our  knowledge  is  very 
incomplete.  Roughly,  we  may  take  it  that  resistance  is  generally 
proportional  to  square  of  speed.  In  the  case  of  shot  this  law  holds, 
probably  up  to  speeds  of  300  feet  per  second ;  from  400  to  1,000 
feet  per  second  the  resistance  is  possibly  proportional  to  the  2^ 
power  of  the  speed.  Beyond  1,100  feet  per  second  we  may  take, 
p  being  in  pounds,  d  the  diameter  of  a  shot  in  feet,  v  the  velocity 
in  feet  per  second,  P  =  fd2  (v  —  800),  where  /—  3  for  spherical 
and  2  for  elongated  shots  with  ogee-shaped  heads.  The  velocity  is 
greater  than  that  of  sound,  and  probably  it  is  to  this  that  the 
change  of  law  is  due.  The  fact  that  even  in  the  steadiest  winds 
there  is  pulsation,  causes  scientific  speculation  about  wind  pressure 
to  be  difficult. 

69.  Reynolds  has  deduced  from  hydrodynamics  the  rational 
formula 

LBn   f2-n  Dn-3  yn  j  A  ....    (1) 

as  the  loss  of  energy  per  pound  of  fluid  passing  through  a 
pipe  of  length  L  feet  and  diameter  D  feet  at  v  feet  per  second. 
(I  have  reduced  his  numbers  to  suit  the  foot  as  the  unit  of  length.) 
The  index  n  is  1  for  velocities  below  the  critical  velocity,  vc  =.  '039 
P/D,  and  n  varies  from  1/7  to  2  at  higher  velocities  than  the  critical. 
p  is  proportional  to  the  co-efficient  of  viscosity,  which  changes 
with  temperature.  In  the  case  of  water  he  takes 

p  =  1  -5-  (1  -  -0336  6  +  -000221  02)  .  .  .  .  (2), 
where  6  is  temperature  Centigrade. 

A  =  1-917  x  106;         B  =  36-8. 

Note  that  the  critical  velocity  depends  upon  the  temperature 
and  size  of  pipe.*  Thus,  for  a  tube  T\jth  of  an  inch  in  diameter,  the 
critical  velocity  is  4-65  feet  per  second ;  for  a  pipe  1  inch  in 
diameter  the  critical  velocity  is  -465  feet  per  second ;  for  a  6-inch 
pipe  (d  =  0-5)  the  critical  velocity  is  -077  feet  per  second.  In  all 
practical  hydraulic  cases  the  critical  velocity  is  exceeded,  and  for  a 
general  rule,  with  cast-iron  pipes  in  actual  use,  we  usually  take 
n  =  2.  In  this  case,  in  (1),  the  influence  of  p,  the  temperature 
term,  is  unfelt ;  that  is,  in  practical  hydraulic  work,  temperature 
has  no  important  influence.  The  formula  now  becomes 

LB2V2/AD  Or  '0007  Ltf2/D    .  .   .  .   (3). 

As  a  mnemonic*  for  this  simple  formula,  let  the  student  imagine  thai 
a  solid  prism  of  water  of  length  L  is  moved  along  a  pipe  rubbing 
all  round  its  perimeter,  the  friction  being  proportional  to  the  square 
of  the  velocity  and  to  the  area  of  the  rubbing  surface.  Thus,  if  s 
is  the  wetted  perimeter,  LS  is  the  area  and  the  force  of  friction  is  LSV*  ; 
that  is,  if  p  is  the  pressure  difference  which  produces  the  motion 
and  A  is  the  area  of  cross-section,  pA  aLsv2.  The  loss  of  energy  per 

pound  being  proportional  to  p,  this  <x  Lv2  — .     Now,  A/*  is  called 
*  See  Appendix. 


APPLIED    MECHANICS.  85 

the  hydraulic  mean  depth,  w,  of  any  channel,  and  we  find  loss  of 
energy  per  pound  =  <?Lt>2/m  .  .  .  .  (4). 

In  the  case  of  a  round  channel  full  of  water 

m  =  -D2  -T-  ir  D  =  _  D. 

Comparing  (4)  with  (3),  the  formula  of  Osborne  Reynolds  corre- 
sponds to  c  =  -000175. 

The  constant  in  the  formula  (3)  agrees  with  that  of  D'Arcy 
for  small  pipes.  Reynolds  has  compared  (1)  with  D'Arcy's  experi- 
ments as  well  as  with  his  own,  from  the  smallest  sizes  of  pipes  to 
20  inches  diameter,  and  finds  that  there  is  practical  agree- 
ment. D'Arcy's  pipes  had  joints  which  somewhat  vitiated  the 
results.  Reynolds  gives  for  n  the  values  :  —  Lead-  jointed  pipes, 
1-79;  varnished,  T82  ;  glass,  1'79  ;  new  cast-iron,  1'88;  incrusted 
pipe,  2'0  ;  cleaned  pipe,  1*91.  This  formula  of  Reynolds  is  rational, 
and  suits  every  imaginable  size  of  pipe,  and  I  prefer  it.  But  that 
of  D'Arcy  is  more  commonly  used  for  pipes  of  from  3  inches  to 
2  feet  in  diameter,  and  it  is  often  used  in  academic  exercises,  in 
which,  indeed,  almost  any  loss  of  energy  per  pound  of  water 
(usually  called  loss  of  head)  is  expressed  as  /  x  the  kinetic  energy 

per  pound  of  water,  or  /  x  —  .     D'Arcy  gives 


for  a  straight  pipe  of  length  L  feet  and  diameter  D  feet. 

70.  Resistance  to  Rolling.  —  When  one  wheel  or  c*ylindric 
body  rolls  upon  another  there  is  some  conversion  of  mechanical 
energy  into  heat.  The  power  lost  seems,  roughly,  to  be  propor- 
tional to  the  force  pressing  the  two  bodies  together,  to  the 
velocity  of  rolling,  and  to  the  curvature  of  the  smaller  of  the 
two.  We  have  very  little  experimental  knowledge  of  the  subject. 
In  all  probability  the  power  wasted  is  proportional  to  the 
strain  energy  per  second  stored  in  the  material,  the  waste 
being  due  to  viscosity.  When  the  velocity  is  very  great,  as  at 
the  driving-wheels  of  locomotives,  secondary  effects  are  pro- 
duced, waves  of  compression  and  extension  travelling  in  the 
rim  of  the  wheel  and  in  the  rail,  with  very  curious  results.  In 
some  experiments  which  I  have  made  with  great  pressures 
between  hard  cast  iron  wheels  rolling  upon  one  another,  there 
seems  to  have  been  much  local  heating  just  at  the  surfaces. 
At  the  end  of  some  months  of  work  a  quantity  of  black  dust 
had  been  produced,  and  each  particle,  when  examined  by  the 
microscope,  looked  like  a  piece  of  slag. 

Besides  energy  wasted  by  changing  strain  in  the  material, 
there  is  slipping  at  the  surfaces  in  contact.  A  student  who 


86  APPLIED    MECHANICS. 

remembers  that  when  a  strut  is  compressed  it  swells,  and  when 
a  tie  bar  is  lengthened  it  gets  thinner,  can  study  the  "  creep  " 
which  occurs  both  here  'and  in  belting,  for  himself.  Imagine 
points  one  inch  apart  upon  the  rim  of  an  iron  wheel,  and 
another  set  upon  an  unstrained  plane  indiarubber  surface. 
Now  draw  the  wheel  as  it  indents  the  surface.  As  in  Fig.  34, 

points  1,  2,  3,  and  4 
are  further  apart,  and 
points  6,  7,  8,  9  are 
nearer  together  than 
in  the  unstrained  con- 
dition, and  hence  the 
metal  and  indiarnb- 
ber  surfaces  slide 
upon  one  another.  No 
ordinary  material 
Fig.  84.  coating  seems  to  have 

much  effect  in  pre- 
venting the  sliding.  A  cast  iron  wheel  on  planes  of  cast 
iron,  boxwood,  and  on  indiarubber,  seems  to  have  frictional 
resistances  to  rolling  in  the  proportion  of  1  :  2  :  8. 

The  loss  of  energy  in.  belting  is  partly  due  to  this,  partly 
due  to  energy  wasted  in  bending  and  unbending  the  belt. 
Both  the  lessened  distance  of  rolling  and  the  slip  of  a  belt 
seem  to  be  proportional  to  the  power  transmitted.  M.  Raffard 
has  actually  used  a  dynamometer  on  this  principle.  He 
transmits  his  power,  to  be  measured,  by  means  of  a  thick 
indiarubber  belt  through  two  equal  pulleys;  the  difference 
of  speed  of  these  pulleys  is  taken  to  be  a  measure  of  the 
power.  The  slip  is  quite  noticeable  when  speed  cones  are  used 
in  driving  machines  at  various  speeds  with  variable  power,  for 
the  actual  speed  has  to  be  carefully  measured ;  calculation 
from  the  known  sizes  of  the  steps  of  the  cones  giving  in- 
accurate results. 

When  pressures  are  not  too  great,  as  in  the  ball  bear- 
ings of  cycles  and  some  machine  tools,  there  can  be  no  doubt 
whatever  of  the  ease  of  running.  Fig.  35  shows  the  ordinary 
adjustable  ball  bearing  used  in  bicycles.  D  is  the  fork  and  H  the 
hub  of  a  wheel.  The  spindle  A  is  fixed  to  the  fork  D.  One  of 
the  hard  steel  cones  c  is  tight  against  a  shoulder  v ;  the  other 
c'  is  tightened  just  enough-  to  let  the  wheel  revolve  easily,  and 
then  it  is  locked  by  the  lock-nut  K.  The  linings  of  the  shaped 


APPLIED    MECHANICS.  87 

ends  of  c  are  hardened  steel,  and  a  number  of  hard  steel  balls 
are  placed  between.     Fig.  36  is  an  enlarged  drawing  of  the 


Fig.  86.  Fig.  36. 

ball  and  the  linings,  showing  that  the  radii  of  curvature  of  the 
ball,  cone,  and  cup  are  different;  the  friction  of  the  bearing 
will  be  much  less  than  if  the  radii  of  curvature  were  nearly 
the  same.  A  very  little  oil  getting  inside  the  hub  finds  its 
way  to  the  balls. 

Some  experiments  (Proc.  I.  C.  E.,  Vol.  119,  p.  456)  on  rollers  be- 
tween flat  cast-iron  plates,  give  as  the  resistance  in  pounds  per  Ib.  to 
rolling  c/v/r  where  r  is  radius  in  inches  and  c  =  '0063  for  cast-iron, 
•0120  for  wrought  iron,  '0073  for  steel.  These  are  13  per  cent, 
greater  for  wrought  iron  plates  and  13  per  cent,  less  for  steel  plates. 
The  crushing  load  in  pounds  on  a  wrought  iron  roller  seems  to  be 
444  r  per  inch  of  its  length. 

Several  experimenters  are  now  engaged  in  procuring  for  us 
more  exact  information  on  rolling  friction. 

In  using  roller  bearings  on  carriages,  it  has  been  found  that 
there  is  a  diminution  of  from  23  (on  gradients  of  1  in  20)  to  60  per 
cent,  (on  gradients  of  1  in  140)  of  the  tractive  effort  required  with 
ordinary  bearings.  In  ordinary  machinery,  the  loss  of  energy  by 
friction  has  been  found  (in  one  experiment)  to  be  less  than  one- 
third  of  what  it  is  with  good  ordinary  bearings.  Oil  is  only  needed 
to  prevent  rusting.  (See  Appendix.) 


88 


CHAPTER    V. 

EFFICIENCY. 

71.  Mechanical  Advantage. — In  books  on  mechanics  you 
will  usually  find  that  when  simple  machines  are  described,  they 
are   only  considered  in  relation   to   their 
Mechanical  Advantage.     That  is,  suppose  a 
small   weight    E,    now  usually  called   the 
effort,  is  able  by  means  of  the  mechanism 
to  cause  a  larger  weight,  R,  usually  called 
the  resistance,  to  rise,  the  ratio  of  R  to  E 
is  called  the  mechanical  advantage.    Now, 
in   nearly   all   cases   you    will   find   that, 
when  there  is  a  mathematical  investigation 
of  a  machine,  the  assumption  is  made  that 
there  is  no  friction.  I  have  already 
shown    you   that   the    problem    of 
taking   friction   into    account  is  a 
very  difficult  one.    But,  as  we  have 
seen,  a   practical  man  can  experi- 
ment on  the  effect  of 
friction;  and,  happily 
for  us,  he  obtains  re- 
sults which  are  gener- 
ally very  simple.     Let 
the  reader  make  a  few 
experiments     himself, 
or  let  him  by  means  of 
squared  paper  find  the 
relation  between  E  and 
R  from  the  following 
results,  taken  from  a 
crane,  Fig.  37,  whose 
gearing  was  well  oiled, 
Pig>  .$7.  and  whose  handle  was 

replaced  by  a  grooved 
wheel,  round  which  was  a  cord  supporting  E  : — 


APPLIED    MECHANICS.  89 


R. 

Heaistanc 
Overcoi 

100  r 

200 
300 
400 
500 
600 
700 
800 

E. 
e  just                                                   Effort  jus 
ne.                                                 Overcome  ] 
bs          8-5 

t  able  to 
lesistance. 

Ibs. 

H 
» 
M 

12-8 

17-0 

21-4 

25-6 

29-9 

34-2 

38-5 

We  found  that  E  fell  forty  times  as  rapidily  as  R  rose,  and 
you  may  have  imagined  that  the  mechanical  advantage  was 
forty,  or  that  a  weight,  E,  could  lift  a  weight,  R,  forty  times  as 
great  as  itself.  This  would  be  true  if  there  were  no  friction  ; 
but  we  see  that  in  practice  it  is  not  the  case.  Plot  the  above 
values  of  E  and  R  on  squared  paper,  and  you  will  find  that,  if 
the  weight  R  is  increased  1  lb.,  E  must  be  increased  '0429  Ib. ; 
and  also  that  when  R  is  0,  an  effort  E  of  4-21  Ibs.  is  needed  to 
cause  a  slow  motion  of  the  crane ;  so  that  the  law  is 

B  =  4-21  +  -0429  R. 

Namely,  multiply  the  resistance  R  in  pounds  by  the  fraction 
•0429,  and  add  4'21  :  the  answer  is  the  effort  required  to  lift 
R.  When  you  have  worked  out  this  rule,  employ  it  in  finding 
how  much  effort,  E,  is  required  to  lift  a  ton  with  such  a  crane. 
— Answer,  100-3  Ibs. 

The  word  power  is  generally,  but  very  unscientifically,  used 
as  the  name  of  the  force  which  I  have  called  E,  the  effort. 
Power  may,  of  course,  be  used  in  many  senses  by  newspapei 
writers,  but  when  used  by  the  engineer  it  is  a  technical  term, 
meaning  the  rate  of  doing  work.  If  a  weight  of  1,000  Ibs. 
falls  100  feet  in  two  minutes,  it  does  1,000  x  100  or  100,000 
foot-pounds  of  work  in  two  minutes,  or  50,000  foot-pounds  of 
work  in  one  minute.  Now,  33,000  foot-pounds  of  work  done 
in  one  minute  is  called  a  horse-power,  and  hence  our  falling 
weight  gives  out  50,000-^-33,000  or  1-5  horse-power.  Ten 
horse-power  means  ten  times  33,000  foot-pounds  of  work  done 
in  one  minute.  The  idea,  then,  of  power  is  an  idea  of  work 
done  in  a  certain  time. 

72.  Economical  Efficiency. — Take  any  pair  of  numbers  from 
the  above  table,  say  E  =  8 '5  Ibs.,  when  R  =  100  Ibs.  Let  us 
suppose  that  E  is  moving  at  the  rate  of  forty  feet  per  second, 
then  we  know  that  R  is  rising  at  the  rate  of  one  foot  per 


90  APPLIED  MECHANICS. 

second.  E  is  giving  out  the  power  8 -5  x  40,  or  340  foot-pounds 
per  second;  R  is  receiving  100  foot-pounds  per  second.  The 
ratio  of  the  power  usefully  employed  to  the  power  given  to  the 
machine  is  called  the  efficiency  of  the  machine,  so  that  our 
crane  has  an  efficiency  100-^-340,  or  '294.  Sometimes  the 
efficiency  is  put  in  the  form  of  a  fraction ;  sometimes  we  say 
that  it  is  2 9 '4  per  cent.,  meaning  that  the  machine  employs 
usefully  29 -4  per  cent,  of  the  energy  given  to  it. 

Now  take  another  pair  of  numbers,  say  E  =  38 '5,  R  ==  800, 
and  let  E  fall  forty  feet  in  one  second,  as  before.  We 
now  get  as  our  answer  -519 — that  is,  more  than  half,  or 
51 '9  per  cent.,  of  the  power  given  to  the  crane  is  usefully 
employed.  We  see,  then,  that  as  the  power  given  to  the  crane 
is  greater,  the  efficiency  is  also  greater.  This  arises  from  the 
fact  that  the  friction  of  the  unloaded  crane  is  always  entering 
into  the  calculation ;  and  if  we  take  the  case  where  no  resist- 
ance, R,  is  being  overcome,  and  E  must  be  4-21  Ibs.,  we  shall 
find  an  efficiency,  0,  because  work  is  being  given  to  the  crane, 
and  none  is  coming  out  usefully*  You  will  always  find  that  the 
power  usefully  given  out  is  a  certain  fixed  fraction  of  the  total 
power  given  to  the  machine,  minus  the  power  required  to  drive 
the  crane  at  the  given  speed  when  it  is  unloaded.  Choose  some 
speed,  say  that  E  falls  forty  feet  per  second ;  find  the  total 
power  or  40  E  ;  find  the  usefully  employed  power  1  x  R  for 
every  case  of  the  above  table.  Plot  your  answers  on  squared 
paper,  and  you  will  find  this  rule  :  if  PO  is  the  power  required 
to  drive  the  crane  at  the  same  speed  when  unloaded,  if  u  is 
the  useful  power,  and  T  is  the  total  power  supplied : — 

T  =  l-761  U  +  PO 

73.  In  some  machines  there  have  been  attempts  to  ap- 
portion the  loss  of  power  to  various  parts.  As  a  rule,  these 
are  very  speculative.  Roughly,  we  may  say  thac  the  frictional 
loss  in  a  steam-engine  may  be  divided  in  the  following  propor- 
tions : — Crank-shaft  bearings  and  eccentric  sleeves,  1 ;  valve, 
if  unbalanced,  '6 ;  valve,  if  balanced,  *05 ;  piston  and  rod,  *4 ; 
crosshead  and  slides,  *2  ;  crank  pin,  '14 ;  total  loss  because  of 
uir  pump,  0'3  to  0'5.  It  is  fairly  obvious  that  on  account  of 
the  great  weight  of  the  fly-wheel  and  other  parts  much  of  the 
energy  loss  in  a  steam-engine  is  the  same  whether  the  engine 
is  giving  out  much  or  little  power,  and  in  many  cases  for 
purposes  of  calculation  this  assumption  may  be  made. 

*  See  Appendix. 


APPLIED    MECHANICS. 


91 


74.  The  work  of  this  and  the  succeeding  two  pages  is  more 
suggestive  than  any  other  work  in  this  book.  From  the  follow- 
ing results  of  experiment  with  a  gas-engine,  show,  by  plotting 
on  squared  paper  and  correcting  for  errors  of  observation,  that 
if  i  is  the  indicated  horse-power,  B  the  brake  horse-power,  o  the 
cubic  feet  of  gas  per  hour,  including  what  is  used  for  ignition,  then 

G  =  20-3  i  +  8,  G  =  20-4  B  +  45,  B  =  1—1-8. 


I 

B 

G 

Q/I 

G/B 

Efficiency. 

13-4 

11-6 

280 

20-9 

24-1 

•166 

10-2 

8-4 

216 

21-2 

25-7 

•155 

7'3 

5-4 

156 

21-4 

28-9 

•123 

4-6 

2-9 

104 

22-6 

37-9 

•112 

1-8 

0 

45 

25-0 

— 

0 

When  we  plot  the  values  of  G  and  i  and  of  G  and  B  on 
squared  paper,  we  find  points  lying  very  nearly  in  straight 
lines.  Assuming  that  they  ought  to  lie  in  straight  lines,  we 
find  the  above  laws  satisfied. 

Let  the  student  fill  in  the  columns  showing  G/I  and  G/B. 
Also  from  the  brake  horse-power,  and  knowing  that  one  cubic 
foot  of  gas  per  hour  means  an  actual  supply  of  energy  of  a 
quarter  horse-power,  let  him  fill  in  the  column  of  efficiencies. 
[The  calorific  power  of  one  cubic  foot  of  average  coal  gas  may 
be  taken  to  be  from  500,000  to  540,000  foot-pounds;  of 
Dowson  gas,  it  is  about  124,000  foot-pounds.] 

Again,  taking  the  following  experimental  results  from  an 
oil-engine  (one  pound  of  oil  being  taken  to  give  out  11,700 
centigrade  heat  units  in  burning),  i  being  the  indicated,  B  the 
brake  horse-power,  and  o  the  pounds  of  oil  used  per  hour. 


I 

B 

0 

O/I 

O/B 

Efficiency. 

7-41 

6-77 

6-4 

•86 

•95 

•128 

8-33 

6-88 

6-8 

•82 

•99 

•122 

4-71 

3-62 

5-0 

1-06 

1-38 

•088 

0-89 

0 

3-1 

3-48 

— 

0 

92 


APPLIED    MECHANICS. 


Let  the  student  show  that  o  =  0-505  I  +  2-62,  o=  0*52 
B  -f  31,  B  =  0-98  i  —  -89.  Let  him  also  fill  in  the  columns 
o/i  and  O/B.  Prove  that  1  Ib.  of  oil  per  hour  means  8  '27  horse- 
power actually  supplied,  and  fill  in  the  column  of  efficiency. 

75.  Accept  the  following  measurements.  A  steam-engine 
employed  in  driving  a  dynamo  machine  delivering  electric 
energy  to  customers,  each  load  being  kept  steady  for  four 
hours,  each  measurement  being  the  average  of  the  results 
obtained  during  the  four  hours.  I  is  indicated  horse-power  ; 
B  the  brake  horse-power  measured  by  a  transmission  dynamo- 
meter ;  the  electrical  horse-power  E  is  obtained  by  multiplying 
amperes  and  volts  to  get  the  power  in  watts  and  dividing  by 
746  ;  c  is  the  coal  used  per  hour  ;  and  w  is  the  weight  of  steam 
used  by  the  engine  per  hour.  The  governor  acted  upon  the 
throttle-valve,  and  not  upon  the  cut-off. 


I 

B 

Amp&res. 

Volts. 

B 

w 

c 

190 

163 

1,050 

100 

143 

4805 

730 

142 

115 

730 

100 

96 

3770 

544 

108 

86 

506 

100 

69 

3080 

387 

65 

43 

219 

100 

29 

2155 

218 

19 

0 

— 

— 

— 

1220 

— 

First  plot  the  values  of  I  and  w,  I  and  B,  E  and  B,  I  and  c 
on  squared  paper.  It  will  be  found  that  there  is  approxi- 
mately a  linear  law  in  every  case.  See  if  you  get  some  such 
laws  as — 

w  =  800  +  21 1 

B  =  -95  i  -  18 

B  =  -93  B  -  10 

c  =  4-2  i  -  62 


Now  produce  a  few  more  columns  of  numbers  and  study  them. 
Give  w  -f-  i  and  c  -r-  i.  Give  w  -j-  B  and  c  -f-  B.  Give  w  -f-  E 
and  c  -f  E.  Also  give  w  -r  c. 

76.  Students  may  compare  the  above  results  with  the 
following  average  measurements  made  at  an  electric  supply 
station  in  1891 :— 


APPLIED    MECHANICS. 


93 


i 

E 

w 

c 

Average  for  7  hours 
1  1  a.m  to  6  p.m. 

80-3 

57-1 

3,268 

552 

Average  for  6  hours. 
6  p.m.  to  midnight. 

227-7 

163-2 

7,122 

742 

Average  for  11  hours. 
Midnight  to  11  a.m. 

37 

23-64 

2,143 

232 

24  hours. 
11  a.m,  to  11  a.m. 

97-3 

68-3 

3,718 

453 

Here  it  will  be  found  that  although  the  load  was  constantly 
varying  even  when  the  averages  for  the  24  hours  are  taken  with 
the  others,  we  have  linear  laws  between  i,  E  and  w,  w=1150  + 
26-25  i,  and  E  =  *72  i— 2.  But  c  does  not  follow  a  linear  law 
with  the  others.  The  reason  lies  in  the  fact  that  a  spare  boiler 
was  used  during  part  of  the  time,  and  there  is  consequently  a 
greater  consumption  of  fuel  than  if  one  or  two  boilers  had  been 
used  the  whole  time.  Since  we  have  considered  fuel  consumption 
in  the  above  exercises,  it  may  not  be  out  of  place  to  introduce 
here  some  figures  from  the  testing  of  a  water- tube  boiler : — 


Steam  per  hour  from 
and  at  100°  C.  per  Ib. 
of  coal. 

Coal  per  sq.  ft  of 
grate  per  hour. 

Water  evaporated  per 
sq.  ft.  of  total  boiler 
heating     surface     per 
hour.    This  is  not  re- 

10. 

duced  to  100°  0. 

13-40 

7'74 

1-24 

103 

12-48 

18-6 

3-20 

233 

12-00 

29-8 

4-70 

357 

10-29 

66-8 

8'50 

686 

If  w  =  steam  per  hour  per  square  foot  of  grate,  /  =  fuel 
per  hour  per  square  foot  of  grate.     Plotting  w  and /on  squared 
paper,  we  find  a  fair  approach  to  a  linear  law, 
w  =  45  + 


7 


w         45  -i-  Q-7ft 
or    j  =  —  —  »  • o. 


94 


APPLIED    MECHANICS. 


77.  If  o  is  the  total  cost  per  hour  when  the  useful  horse-power 
p  is  being  sent  out  "by  an  hydraulic  or  electric  or  other  supply  com- 
pany (c  includes  interest  and  depreciation  on  first  cost,  rent,  taxes, 
repairs,  wages,  stores,  coal,  water,  office  expenses  and  management), 
and  if  it  is  found  that  there  is  a  simple  law  like  c  =  a  p  +  b,  where 
a  and  b  are  constants,  prove  that  the  average  cost  per  hour  is 
calculated  from  the  average  power  in  exactly  the  same  way  as  the 
real  cost  c  in  any  hour  is  from  the  p  during  that  hour.  For  if  t  is 
time  in  hours,  then  the  cost  during  the  year  is,  if  T  is  the  number 
of  hours  in  a  year, 


p  p  pr 

I   c  .dt  =   I  O  P  +  V)  dt  =  a  \  p 

Jo  Jo  Jo 


Hence  the  average  cost  per  hour  is  - 


Now 


-  fp  .dt  is 

TJo 


+  b. 


the  average  power,  call  it  PTO,  and  we  see  that 


the  average  cost  is  a  pm  +  b.     The  average  power  delivered  in  a 
day  or  year,  divided  by  the  maximum  power,  is  called  the  daily  or 
yearly  load  factor.     If  /  is  the  load  factor  and  PJ  is  the  maximum 
power,  then  average  cost  per  hour  =  A  ?>l  f  -\-  b. 
78.  An  electrical  company  has  arranged  for  a  maximum  output  of 
1,000  horse-power.     It  is  found  that  the  total  cost  per  hour  in  pence  c  is 
c  =  0-8  P  +  350.     If  p  pence  is  charged  for  every  horse-power  hour  sent 
out,  what  is  the  yearly  profit  when  the  average  power  sent  out  day  and 
night  is  pm  ? 

Ans. — Subtract  from  p  Tm  the  average  cost  per  hour  to  the  company 
which  is  0*8  rm  +  350,  and  the  profit  per  hour  is  (p  -  0'8)  rm  -  350. 
The  profit  per  annum  in  pounds  is  therefore  this  multiplied  by  36'5. 
Exercise. — What  charge  per  horse-power  hour  will  give  just  no  profit  ? 

Ans. — x  =  0-8  +  350/pm. 
Thus  if  pm  has  the  following  values,  we  have  the  values  of  x 


Pm      

1,000 

500 

200 

100 

X         

1-15 

1-50 

2-55 

4-30 

79.  We  have  seen  that  if  a  force  E  lb.,  acting  through  e  feet, 
overcomes  a  force  B  lb.,  acting  through  r  feet,  instead  of  the  non- 

frictional  law  B=  —  ....  (1),  we  find  experimentally  some  such 
law  as 


It  is  evident  that  the  constant  A  is  a  frictional  resistance  from  the 
mere  weight  of  the  parts  of  the  machine  ;  a  is  always  found  to  be 


APPLIED    MECHANICS.  95 

greater  than  -.     The  work  done  by  B  in  the  distance  e  feet  is 

K  e  or  ea  R  -f-  A  0,  and  the  useful  work  is  r  R,  so  that  the 

r 

Efficiency  x 


ea  R  -f-  A  e       i   •    *_ 

"*"  a-R. 

It  will  be  observed  that  the  larger  R  is,  the  smaller  is  —  ,  or 

the  more  insignificant  is  the  term  due  to  weight  of  parts  of 
machine,  and  the  more  nearly  does  the  denominator  approach 
unity.  llowever  great  R  may  become,  the  efficiency  cannot 

exceed   -  .     ;  and  as  a  is  always  greater  than  -,  the  efficiency  is 

C      d  & 

always  less  than  unity,  as  was  to  be  expected. 

Denote  a  by  ,  •    ,  where  k  is  always  less  than  1.     Then  the 

/C      6 

efficiency  can  never  be  greater  than  k.  If  e  and  r  are 
feet  per  second,  then  ~  and  ^  represent  the  horse-powers.  Let 

us  call  ,^    the  total  horse-power  T,  and  _^-  the  useful  horse-power 
ooO  ooU 

u,  and  we  shall  call  T  —  u  the  lost  or  wasted  horse-power  L.  Then  as 
ze  ae  nr  A  <?  1  ,  A  0  ,  .  A.k  e 


80.  It  is  interesting  to  know,  from  such  examples  as  we  have 
studied  in  Arts.  73-76,  that  if  T  is  the  indicated  horse-power  of 
a  steam-engine,  and  u  is  the  brake  horse-power,  or  what  is 
given  out  usefully  by  the  crank  shaft,  and  if  this  shaft  drives  a 
dynamo  machine,  and  E  is  the  electrical  horse-power  given  out  ;  or 
if  the  shaft  drives  a  pump,  and  E  is  the  effective  horse-power  of 
the  pump,  we  always  find  (probably  with  only  approximate 
accuracy)  a  linear  law  connecting  T  and  u,  and  u  and  E,  and 
therefore  also  connecting  T  and  E.  ^  Furthermore,  in  a  steam- 
engine  which  does  not  vary  its  period  of  cut-off  —  that  is,  the 
regulation  is  by  throttling  or  the  boiler-pressure  changes  —  if  w  is 
the  poundage  of  steam  per  hour,  there  is  a  linear  law  connecting 
any  two  of  the  quantities  w,  T,  u,  and  E. 

If  we  write  (1)  as 

u  =  £  T  —  a,  or  T  =  r  u  4-  r  a, 

it  is  curious  that  in  so  many  of  our  machines  in  which  the 
transmission  of  power  is  by  mere  shafting  k  should  be  so  much  less 
than  unity  aa  it  is. 

This  occurs  because  a  meve  torque  is  very  seldom  applied  to  a 
shaft  when  power  is  being  given  or  taken.  t  When  the  power  is 
supplied  or  taken  off  by  a  belt,  the  power  is  proportional  to  the 
difference  of  pulls  in  the  belt,  whereas  the  loads  on  the  bearings 
and  the  friction  are  nearly  constant.  Such  kinds  of  driving  tend 
rather  to  increase  a  than  to  diminish  k.  All  spur  or  bevel  gearings 


96  APPLIED    MECHANICS. 

tend  to  diminish  k,  and  only  by  their  mere  weight  do  they  tend  to 
increase  a.  The  poor  efficiency  of  ordinary  lines  of  shafting  is  a 
matter  to  which  few  people  seem  to  have  paid  any  attention  ;  and 
this  is,  I  think,  altogether  due  to  the  absence  of  dynamometer 
couplings  or  other  means  of  forcing  upon  the  attention  the  actual 
amounts  of  power  which  are  being  transmitted  at  each  place. 
Electrical  engineers  have  very  accurate  methods  of  measuring  the 
power  given  out  by  their  dynamos,  and  hence  they  are  concerned 
as  to  the  actual  mechanical  power  supplied  to  them.  It  was 
almost  altogether  due  to  this  that  direct  driving  of  dynamos  from 
engines  took  the  place  of  indirect  methods;  and,  indeed,  I  may 
go  further,  and  say  that  it  was  due  to  this  that  the  great  improve- 
ments have  taken  place  in  steam-engine  manufacture  and  working 
during  the  last  fifteen  years.  If  even  the  practical  engineer  is 
beginning,  therefore,  to  think  of  economy,  it  is  greatly  due  to  the 
fact  that  electricity  is  paid  for  at  so  much  per  unit  of  energy; 
although,  no  doubt,  much  is  also  due  to  the  fact  that  economy  of 
coal  is  very  important  in  ocean-going  steamships. 

81.  In  any  machine  in  which  a  small  effort,  E,  overcomes  a  great 
resistance,  K,  we  usually  find  that  the  frictional  loss  of  energy  in 
the  machine  is  almost  altogether  dependent  upon  R.  Thus  in  the 
inclined  plane  (Fig.  25),  with  E  parallel  to  the  plane,  the  friction 
is  independent  of  E.  When  in  the  inclined  plane  of  small  inclina- 
tion E  is  horizontal,  the  friction  is  almost  altogether  dependent 
on  the  load  to  be  lifted  ;  and  it  is  so  in  a  screw-jack  and  in  the 
differential  pulley-block.  It  is  not  so  much  the  case  in  cranes, 
unless  such  gearing  as  worm-gearing  is  employed. 

Let  us,  then,  suppose  that  the  loss  of  energy  in  the  machine  is 
altogether  due  to  B.  Then  in  the  direct  use  of  the  machine  —  that 
is,  E  overcoming  R  —  if  the  efficiency  is  only  50  per  cent.,  the  loss 
of  energy  is  equal  to  R  r  (if  B.  is  lifted  r  feet),  K  r  being  also  the 
useful  energy.  Consequently,  if  we  try  by  increasing  R  to  R1  to 
make  the  machine  reverse,  the  loss  is  always  RV  ;  or  the  whole 
energy  would  be  wasted  if  the  machine  could  be  supposed  to  move. 
If,  then,  the  direct  efficiency  is  equal  to  or  less  than  50  per  cent., 
the  machine  cannot  reverse  or  overhaul. 

In  any  machine  there  is  some  loss  due  to  E,  and  also  to  the 
weight  of  the  parts  of  the  machine,  as  well  as  to  R,  and  conse- 
quently the  direct  efficiency  must  always  be  less  than  50  per  cent. 
if  the  machine  is  not  to  overhaul.  On  the  assumption  that  the 
lost  energy,  L,  is 


where  A  is  a  constant,  due  to  the  weight  of  parts  of  the  machine, 
and  m  and  n  are  fractions,  we  have,  in  direct  working, 


K  e  —  R  r  =  L, 


1  —  n 


Hence  the  direct  efficiency  r  is  -.  —  ^    ~  n'  ....  (1), 

\H  ~T  *•)  B**  ~T  A 


APPLIED    MECHANICS.  97 

1  ** 

m  ,   i         .    r-     Again,  in  reversed  working 

nlr  —  E  e  =  L, 

RV  —  E  e  =  m  RV  -f  «  E  e  -J-  A, 
_•¥{!*»  »)-.-* 


The  reversed  efficiency y  =  21  =  O- 
R1/-  (l 

1  his  is  0  or  negative  when 


»T 

As  n1  may  be  made  very  great,  our  condition  of  non-reversibility 
must  be  m  =  1  or  >  1.  Hence  the  direct  efficiency  must  be  equrJ 

to  or  less  than  g  +~  V.    We  therefore  see  that  there  is  no  general 

rule  such  as  many  people  seem  to  believe  in — the  result  of  mis- 
leading mathematics.  We  can  hardly  call  it  a  general  rule  to  say 
that,  if  m  is  equal  to  1,  R  cannot  overcome  E,  because  m  =  I  means 
that  all  it's  energy  would  be  wasted. 


CHAPTER    VI. 


MACHINES.       SPECIAL    CASES. 

82.  Blocks  and  Tackle.  —  It  is  very  good  to  have  a  general 
law  telling  us  about  machines  in  which  there  is  no  friction. 
That  law  vou  now  know.  The  mechan- 
ical work  given  to  a  machine  is  equal  to 
the  work  given  out  by  it,  unless  it  is 
stored  up  in  the  machine  itself  by  the 
coiling  of  a  spring  or  in  some  other  way. 
But,  besides  knowing  the  law  itself,  it  is 
well  to  know  what  it  leads  to  in  certain 
special  cases.  Take,  for  instance,  a 
pulley-block,  Fig.  38.  It  is  evident 
here  that  if  we  have  three  pulleys  in 
the  block  B,  if  the  effort  P  acts  through 
six  inches,  w  will  only  rise  one  inch, 
and  therefore  P  will  balance  six  times 
its  weight  at  w  if  there  is  no  friction. 
The  mechanical  advantage  is  therefore 
six. 

This  is  a  case  in  which,  it  is  less  dif- 
ficult than  usual  to  trace  the  loss  of 
energy  due  to  friction.  If,  when  the  pull 
of  a  cord  is  PO  on  the  overcoming  side  of  a 
pulley,  the  tension  on  the  other  side  is 
given  by  PI  =.  a  PO  —  b,  where  a  is  lf«s 
than  1,  and  b  does  not  depend  much  on  the 
weight  of  a  sheave,  hut  rather  on  the  vis- 
cosity  of  the  rope,  we  may  take  it  that 

P2  =  a  P,  —  b,  and  so  on.  I  shall  leave  it  to  students  to  work 
out  an  algebraic  expression  connecting  w,  the  sum  of  p,,  P2,  PS, 
etc.  ,  and  PO.  w  will  include  the  weight  of  the  lower  block.  But 
it  is  good  to  take  a  numerical  example.  For  instance,  lei 
PJ  =  0'9  PO  —  3,  the  forces  being  in  pounds.  Then 

p2  =  0-81      PO  -    2-7      -3  =  0-81      PO  -    5-7, 
pg  =  0-729    PO  -    5-13    -3  =  0-729    PO  -    8-13, 
P4  =    -6561  PO  -     7-317  -  3  =    -6561  PO  -  10-32, 


Fig.  88. 


Hence 


-5905  PO  -    9-288  -  3  =    -5905  PO  -  12-29, 
-6315  PO  —  11-06    -3=    -5315  PO'-  H-06; 

=  4'2171  PO  -  53-50,  or  PO  =  0-227  w  +  12  7. 


APPLIED    MECHANICS. 


99 


•1  (\1     TV 

The  efficiency    *  ~  .237  w  +  12.? 

Tf 

Hence,  however  great  w  may  he,  the  efficiency  is  less  than  70 '5 
per  cent.     When  w  =  53  '6  the  efficiency  is  only  half  this. 


l+^r 


effect 


Fig.  89. 


83.  This  is  a  fairly  good  example  of  the  cumulative 
of  friction.     If  we  give 

power  PO  to  a  machine 

whose   efficiency  is  e0, 

if  this   machine   gives 

power  to  another  whose 

efficiency  is  el  and  so 

on,    the    power    given 

out   by   the  last   of  a 

series   of    machines   is 

PO  x  eQ  x  e^  x  e2  x  etc. 

If,  as   in  transmitting 

power  for  a  great  dis- 

tance by  means  of  ropes, 

all   the  contrivances  are  the   same,   we   have  the  compound 

interest  law  of  diminution  of  power.  (See  Exercise,  page  232.) 

Students  may  find  it  interesting  to  study  a  machine  or  method 

of  transmission  of  power  in  a  manner  allied  to  what  follows 

in  Art.  91  of  this  chapter. 

84.  Inclined    Plane.  —  Again,   take    the  inclined    plane, 
Fig.  39  ;  w  is  a  weight  which  may  roll  down  the  plane  without 
friction,  let  us  suppose;  P  is  the  pull  in  a  cord  which  just 
prevents  w  from  falling.     The  cord  is  parallel  to  the  plane. 
Evidently  when  w  rises  from  level  A  to  level  B  the  cord  is 
pulled  the  distance  A  B  ;  that  is,  w  multiplied  by  the  height  of 
the  plane  is  equal  to  P  multiplied  by  the  length  of  the  plane. 
Thus,  if  w  is  1,000  Ibs.,  and  the  length  of  the  plane  10  feet  for 
a  rise  of  2  feet,  then  ten  times  P  is  equal  to  2,000,  or  p  is 
200  Ibs. 

Barrels  and  boilers  are  often  raised  along  an  inclined  plane, 
ropes  or  chains  held  fast  at  the  top  of  the  plane  passing  round 
the  cylindric  object  and  back  towards  the  top  of  the  plane 
where  force  is  applied  to  them.  In  this  case  p  is  evidently 
only  half  what  is  stated  above.  As  the  weight  rolls  on  the 
double  rope  or  chain  there  is  not  much  friction.  In  Art.  90 
we  shall  consider  the  effect  of  friction  in  the  inclined  plane. 

85.  The  Screw.  —  Again,  suppose  there  is  no  friction  in  the 


100 


APPLIED   MECHANICS. 


screw  A  B,  Fig.  40  ;  if  it  rises  it  lifts  a  weight  say  of  3,000  Ibs. 
Now,  if  the  screw  make  one  turn  it  rises  by  a  distance  equal 
to  its  pitch;  that  is,  the  distance  (measured  parallel  to  the  axis) 

between  two  threads. 
Say  that  the  pitch 
is  '02  foot,  then 
when  the  screw 
makes  one  turn  it 
does  work  on  the 
weight  3,000  x  '02, 
or  60  foot-pounds. 
But  to  do  this,  p 
must  fall  througli  a 
distance  equal  to 
the  circumference 
of  the  pulley  A, 
about  which  I  sup- 
pose the  cord  to  be 
wound.  Suppose 


Fig.  40. 


the  circumference 
of  the  pulley  to  be 
6  feet,  then  p  multiplied  by  6  must  be  60,  or  p  is  10  Ibs.  The 
rule,  then,  for  a  screw  is  this — effort  multiplied  by  circum- 
ference of  the  pulley  equals  resistance  multiplied  by  pitch  of 
screw.  It  is  not  usual  to  have  a  pulley  and  a  cord  working 
a  screw;  it  is  more  usual  to  have  a  handle,  and  to  push  or 
pull  at  right  angles  to  the  handle.  Instead  of  the  circum- 
ference of  the  pulley,  we  should  take,  then,  the  circumference 
of  the  circle  described  by  the  point  where  the  effort  is  applied 
to  the  handle. 

Example. — A  steam-engine  gives  to  a  propeller  shaft  in 
one  revolution  60,000  foot-pounds  of  work ;  the  pitch  of  the 
screw  is  1 2  feet.  What  is  the  resistance  to  the  motion  of  the 
vessel?  Answer:  The  resistance  in  pounds  multiplied  by  12 
gives  the  work  done  in  overcoming  this  resistance,  and  this 
work  must  be  equal  to  60,000  foot-pounds ;  hence  the  resistance 
to  the  motion  of  the  vessel  is  5,000  Ibs.  [It  would  be  more 
correct  to  say  that  this  is  the  work  done  per  foot-travel  of  tho 
vessel,  assuming  no  slip  of  the  screw.] 

Screws  are  used  for  many  purposes.  When  used,  as  in  bolts 
to  fasten  things,  the  threads  are  triangular  in  section.  The 
Whitworth  thread- is  shown  in  Fig.  41.  A  B  is  the  pitch ;  G  H 


APPLIED   MECHANICS. 


101 


is  -96  of  the  pitch,  the  angle  H  B  j  being  55° ;  the  corners  are 
rounded,  so  that  I  E  is  '64  of  the  pitch.  The  Sellers  thread 
used  in  America  has  an  angle  of  60°.  Fig.  42  shows  a  square 
thread.  There  is  less  friction  and  less  wear  with  this  form  of 
screw,  and  it  is  used  when  accuracy  of  motion  is  important. 
As  there  is  only  half  the  amount  of  material  resisting 
shearing,  the  square  thread  has  only  half  the  strength 

of  a  triangular  thread.     The  Buttress  thread 

of   Fig.  43  has   strength   and   accuracy  as  to 

motion;  it  is  used  when 

the  important   motion   is 

in  one  direction  only. 
A  student  has  plenty 

of  opportunity  of  examin- 
ing examples  of  the  use  of 

screws.     To  fasten  things 

together   we    have   many 

kinds  of  bolts  and  many 

forms  of  heads  and  nuts, 

and  many  ways  of  locking 

nuts.  Other  fastenings 
such  as  pins  and 
and  cottars,  come  \ 

before  his  eyes  every  day      Fig.  42.  Fig.  43. 

in  the  workshops,  and  he 
must  become  familiar  with  them  and  their  usual  proportions 
both  in  the  shops  and  the  drawing-office.  He  must  make 
a  very  careful  examination  of  such  a  tool  as  a  good  screw- 
cutting  lathe,  and  especially  of  the  mechanism  of  the  slide 
rest.  The  calculation  of  the  proper  change  wheels  for  the 
cutting  of  a  particular  screw  is  a  very  simple  matter  ;  but  all 
so  simple  as  it  is,  he  must  work  out  some  examples. 

He  must  note  the  shapes  of  threads  of  screws  for  wood,  for 
sand  or  mud,  and  for  water;  the  use  of  screw  piles  as  the 
supports  of  structures,  the  forms  of  screw-propeller  for  steam- 
ships and  wind-mills,  and  screws  used  as  fans  for  blowing  air. 

86.  Wheel  and  Axle. — If  A  and  B,  Fig.  44,  are  two  pulleys  or 
drums  on  the  same  axis  and  having  cords  round  them,  a  small 
weight,  p,  hung  from  A,  will  balance  a  larger  weight,  w,  hung 
from  B.  For,  suppose  that  one  complete  "turn  is  given  to  the 
axis,  P  falls  a  distance  equal  to  the  circumference  of  A  whilst 
w  is  rising  a  distance  equal  to  the  circumference  of  B.  Hence 


Fig.  41. 


102 


APPLIED    MECHANICS. 


p  x  circumference  of  A  =  w  x  circumference  of  B, 
or,  what  really  conies  to  the  same  thing, 

p  x  diameter  of  A  =  w  x  diameter  of  B  .  .  .  .  (1), 

or  p  x  radius  of  A  =  w  x  radius  of  B. 

In  practice  A  is  often  a  handle  and  B  a  sort  of  barrel  on 
which  a  chain  may  be  wound.    Or  the  axle  may  be  compound, 
consisting  of  two  parts,  the  diameters  D  and 
jg^^  d  being  nearly  equal,  the  rope  being  coiled 

round  them  in  opposite  directions  so  as  to 
form  a  loop,  upon  which  hangs  a  pulley.  In 
this  case  (1)  becomes 


Fig.  44 


. 
p  x  diameter  or  A  =  w  x    —  K— 

A 

Or,  again,  A  may  not  be  on  the  same  shaft 
as  the  barrel  B,  but  may  gear  with  it  through 
spur  and  bevel-gearing.     Thus,  in    a  hand 
crane,  the  handle  may  turn  many  times  for 
one  turn  of  the  barrel.     Also  there  may  be 
9  snatch-block,  so  that  when  two  feet  of  chain 
are  coiled   on  the   barrel,  the   weight   rises 
only  one  foot.     There  is  no  mystery  about 
tooth  gearing,  and  anyone  who  has  looked  at 
a  crane  knows  how  by  merely  measuring  the 
length  of  the  handle,  or,  rather,  the  circum- 
ference of  the  circle  described  by  the  handle, 
and  the  amount  of  chain  coiled   on  in  one 
revolution  of  the  barrel  (this  is  larger  than  the 
circumference  of  the  barrel  itself),  and  counting  the  number  of 
teeth  of  the  driving  and  driven  wheels,  what  is  the  velocity 
ratio,  and,  therefore,  the  hypothetical  mechanical  advantage. 
I  write  for  men  who  go  about  among  machinery,  and  the  most 
illiterate  workman  knows  well  how  speeds  of  shafts  depend 
upon  numbers  of  teeth.     Also,  all  my  readers  have  seen  com- 
binations of  barrels,  snatch-blocks,  blocks  and  tackle  and  crab- 
winches.     A  worm  and  worm  wheel,  or  other  kinds  of  scr^w 
gearing  may  also  be  imagined. 

Example.  —  The  handle  of  a  crab  or  crane  is  18  inches  long; 
20  inches  of  chain  are  wound  on  in  one  revolution  of  the 
barrel.  The  barrel  is  driven  from  the  handle  by  a  train  of 
wheels  —  driver  15  teeth,  follower  64  teeth;  driver  16  teeth, 
follower  60  teeth  (the  value  of  this  train  is  said  to  be 


APPLIED    MECHANICS. 


103 


m&kes  16  revolutions  for 


^j  -  ™,  orifi  )>  so 

one  of  the   barrel;    the  chain,  lifts  the  weight  through  the 
agency  of  a  block  and  tackle,  with  3  sheaves  below. 

When  the  barrel  turns  once,  20  inches  of  chain  are  coiled 

20 
on,  and  therefore  the  weight  rises  -~  —  «  inches  or  3J  inches. 

The  handle  turns  16  times,  and  the  hand 
moves  through  16  x  36  TT  inches.  Hence 
the  velocity  ratio  of  hand  to  weight  is 

-   =   543.     Often  we  have  a  means 

65 

provided  of   disengaging   the   spur   wheels, 

driver  16,  follower   60,  and  so  we  can  de- 

i  /» 

crease   the   velocity   ratio   to    543  x  —  -  or 

60 

about  145. 

87.  A  differential  pulley-block  is  shown 
in  Fig.  45.  When  the  chain  E  is  pulled, 
it  turns  the  two  pulleys,  or  rather  one 
pulley  with  two  grooves,  B  and  c.  Now  c 
is  a  little  smaller  than  B,  so  that,  although 
at  D  the  chain  is  lifted,  it  is  lowered  at  F. 
If  the  circumference  of  B  is  2  feet  and  that 
of  c  is  1-95  feet,  then,  when  E  is  pulled  2 
feet,  D  is  lifted  2  feet,  but  p  is  lowered 
1-95  feet,  so  that  there  is  0'05  foot  of  chain 
less  than  before  in  the  parts  D  and  F,  and 
the  pulley  G  rises  the  half  of  this,  or  '025 
foot.  If  R  is  2,000  Ibs.,  then  2,000  x  -025, 
or  50,  must  be  equal  to  the  pull  E  multiplied 
by  2,  hence  E  is  25  Ibs.,  or  an  effort  of 
25  Ibs.  is  able  to  overcome  a  resistance  of 
2,000  Ibs.  The  general  rule,  then,  for  the 
differential  pulley-block  is,  effort  E  multi- 
plied by  circumference  of  larger  groove  B  is 
equal  to  resistance  R  multiplied  by  half  the 
difference  between  the  circumference  of  the  two 
grooves  B  and  c.  You  will  find  that  this  rule  comes  to  the 
same  thing  —  effort  multiplied  by  diameter  of  B  is  equal  to 
resistance  multiplied  by  half  the  difference  between  the  diameters 


104 


APPLIED   MECHANICS. 


of  B  and  c.     The  grooves  are  furnished  with  ridges  to  catch 
the  links  of  the  chain,  so  that  there  shall  be  no  slipping. 

It  is  only  when  we  experimentally  measure  the  effort  B 
which  will  slowly  overcome  the*  resistance  R  in  this  pulley-block 
as  actually  made,  that  we  see  how  great  is  the  f  rictional  waste  of 
energy.  It  is  in  consequence  of  this  that,  however  great  the  re- 
sistance R  may  be,  it  will  not  fall,  even  when  there  is  no  force 
exerted  at  E.  This  property  of  not  "overhauling"  makes  the 
differential  pulley-block  the  very  useful  implement  which  we 
know  it  to  be  in  a  machine  shop.  It  is  the  characteristic  of 
any  machine  which  has  a  very  great  velocity  ratio  that  if  its 


Fig.  46 


Fig  48. 


efficiency  is  less  than  half,  it  will  not  overhaul  (see  Art.  81), 
and  lifting  machines  which  do  not  overhaul  are  often  very 
convenient. 

88.  Equilibrium  in  one  Position. — In  all  the  machines 
which  we  have  hitherto  considered,  we  could  give  motion  without 
altering  the  balance  of  the  forces,  but  there  are  many  machines 
in  which  the  mechanical  advantage  alters  when  motion  is  given. 
In  such  cases  you  will  employ  your  general  principle,  but  you 
must  make  your  calculation  from  a  very  small  motion  indeed. 
For  instance,  in  the  inclined  plane,  if  the  cord  which  prevents 
the  weight  from  falling  is  not  parallel  to  the  plane — say  that 
it  is  Hke  M,  Fig.  46 — you  will  find  that  the  necessary  pull 
depends  on  the  angle  the  cord  makes  with  the  plane.  Now, 
suppose  that  the  cord  pulls  the  carriage  from  b  to  c,  evidently 
the  angle  of  the  cord  alters.  The  question  is,,  what  is  p,  that 
it  may  support  w  in  the  position  shown  in  the  figure  ?  We 


APPLIED    MECHANICS.  105 

know  that  it  will  be  different  after  a  little  motion,  but  what  is 
it  now  1  Imagine  such  a  very  small  motion  from  b  to  c  to 
occur  that  the  angle  of  the  cord  does  not  alter  perceptibly,  and 
now  make  a  magnified  drawing,  Fig.  47.  P  has  not  fallen  so 
much  as  the  distance  b  c,  it  has  only  fallen  the  distance  b  a  (c  a 
is  perpendicular  to  b  a).  In  the  meantime  the  weight  w  has 
been  lifted  the  distance  k  c.  Hence, 

w  x  k  c  ought  to  be  equal  to  P  x  b  a. 

Thus,  if  you  measure  k  c  and  b  a  on  your  magnified  drawing  to 
any  scale  you  will  find  the  relation  between  p  and  w.  Another 
way  of  finding  the  same  relationship  is  this.  We  know  that 
the  weight  of  w  acting  downwards,  the  pull  in  the  cord,  and  a 
force  acting  at  right  angles  to  the  plane,  are  the  three  forces 
which  keep  w  where  it  is.  Draw  a  triangle  whose  three  sides 
are  parallel  to  the  directions  of  these  three  forces.  Fig.  48, 
with  arrow-heads  circuital;  then  x  and  y  are  in  the  propor- 
tion of  w  and  p.  Here  we  have  used  the  principle  called 
"  the  triangle  of  forces  "  to  find  P. 

EXERCISES. 

1.  Find  the  force  parallel  to  the  plane  required  to  draw  a  weight  of 
2  cwt.  up  a  smooth  inclined  plane.     Height  of  plane,  3  ;  length,  5. 

Ans.,  1-2  cwts. 

2.  In  a  screw-jack  the  pitch  of  the  screw  is  f  inch  ;  radius  of  circle 
described  by  hand,  19  inches  ;  find  the  velocity  ratio.     It  is  found  that  a 
force  A  at  the  handle  of  30  Ibs.  will  overcome  a  weight  of  2,300  Ibs.,  and 
one  of  10  Ibs.  will  overcome  a  weight  of  500  Ibs.  ;  what  law  connects  A 
and  w  ?    When  w  is  3,000  Ibs.  ,  what  is  A  ?     What  is  the  efficiency  ? 

An*.,  318-47  ;  A  =  ^  +  4j  ;  37g  Ibs.  ;  25  per  cent. 

3.  The  handle  of  a  lifting-jack  measures  24  inches  in  length;   the 
pitch  of  the  screw  is  f  inch  ;  what  force  applied  at  the  end  of  the  handle 
would  be  required  to  raise  a  load  of  22  cwt.,  the  effect  of  friction  being 
neglected?  Ans.,  6-125  Ibs. 

4.  Pitch  of  screw-propeller,  18  feet;  slip,  10  per  cent.;  speed  of  ship, 
15  knots;  find  the  revolutions  per  minute.     What  is  the  thrust  if  the 
actual  horse-power  spent  by  the  propeller  is  2,000,  and  the  waste  by  surface 
friction  is  30  per  cent  ?    What  is  the  torque  in  the  shaft  ? 

Ans.,  93-83,  13'57  tons;  50  ton-feet. 

5.  The  British  Association  rules  for  the  pitch  and  diameter  and  index 
number  of  screw-threads  for  instrument  work  are 


Taking  the  ^alues  of  0,  1,  2,  3,  ....  12  for  s,  calculate  p  and  d,  and  keep 
for  reference  in  a  table.  Try  to  what  extent  the  rule  p  =  '08  d  -\-  '04  for 
the  pitch  and  outside  diameter  of  triangular  screws  agrees  with  the  well- 
known  Whitworth  table. 


106  APPLIED    MECHANICS. 

6.  What  must  be  the  difference  in  the  diameters  of  a  compound  wheel 
and  axle  so  that  the  velocity  of  E  may  he  eighty  times  that  of  R,  the 
length  of  the  handle  heing  2  feet?  Am.,  1-2  inches. 

7.  The  weight  on  a  crane  is  carried  hy  a  snatch-block ;  the  chain  goes 
to  a  barrel  on  which  17  inches  are  wrapped  in  one  revolution.     The 
barrel  is  driven  by  wheels,  which  give  it  1  revolution  for  15  revolutions 
of  the  handle,  and  the  hand  describes  a  circle  of  21  inches  radius  ;  what  is 
the  velocity  ratio  ?  Ans.,  232-8. 

8.  In  a  single-purchase  crab  the  pinion  has  12  teeth  and  the  wheel 
has  78  teeth,  the  diameter  of  the  barrel  (or  rather  of  the  chain  on  the 
barrel)  being  7  inches,  and  the  length  of  the  lever-handle  14  inches.     It 
is  found  that  the  application  of  a  force  of  15  Ibs.  at  the  end  of  the  handle 
suffices  to  raise  a  weight  of  280  Ibs. ;  find  the  efficiency  of  the  machine. 

Ans.,  0-718. 

9.  In  the  differential  pulley,  if  the  weight  is  to  be  raised  at  the  rate 
of  5  feet  per  minute,  and  the  diameters  of  the  pulleys  of  the  compound 
sheave  are  7  and  8  inches,  at  what  rate  must  the  chain  be  hauled  ? 

Ans. ,  80  feet  per  minute. 

10.  In  a  differential  pulley  in  which  the  velocity  ratio  of  fall  to  lift  is 
30,  a  pull  of  7  Ibs.  will  just  raise  a  load  of  24  Ibs.,  and  a  pull  of  25  Ibs.  a 
load  of  240  Ibs.     Find  the  pull  required  to  lift  600  Ibs.,  and  the  efficiency 
of  the  machine  when  such  a  weight  is  being  raised. 

Ans.,  55  Ibs.  ;  0'36. 

89.  "When  one  body  touches  another,  and  there  is  equilibrium,  if 
there  is  no  friction,  this  means  that  there  is  no  tangential  force  at 
the  surface.  The  force  with  which  either  body  acts  on  the  other 
has  no  tangential  component.  If,  however,  there  is  friction,  and 
we  assume  that  the  friction  is  ju  R  where  R  is  the  force  normal  to 
the  rubbing  surface,  it  is  evident  that  the  total  force  at  the  place 
makes  an  angle  <f>  with  the  normal  if  tan.  <£>  =  /j..  If  the  total 
force  makes  an  angle  less  than  <J>,  its  tangential  component  is  less 
than  the  friction,  and  there  can  be  no 
motion.  If,  then,  two  bodies  A  o  B  and 
»oc  touch  at  o,  and  N  o  N1  is  their  com- 
mon normal  at  o,  so  long  as  the  direction 
of  the  force  acting  between  the  bodies  lies 
inside  the  cone  Q  o  p,  whose  axis  is  o  N, 
the  angle  PON  or  Q  o  N  being  </>,  there  will 
be  no  sliding  motion  or  rubbing  at  o. 
Thus,  when  the  block  lies  en  the  inclined 
plane  (Fig.  50),  the  total  force  trans- 
mitted between  block  and  plane  is  w,  the 
vertical  weight  of  the  block.  So  long 
as  the  angle  is  less  than  <£,  there  is  no 
Fig.  49.  sliding.  But  the  angle  is  the  angle  of 

the  plane ;  hence  sliding  is  about  to  begin 

when  we  are  increasing  the  angle  of  the  plane,  and  it  has  become 
as  much  as  </>. 

If  the  body  D  o  c  is  at  rest,  and  A  o  B  moves,  touching  the  first, 
the  motion  is  one  of  sliding,  rolling,  or  spinning,  or  combinations 
6i'  these  three.     If  there  is  no  sliding,  so  that  o  is  momentarily  at 
tho  instantaneous  motion  of  A  o  B  must  be  an  angular  velocity 


APPLIED   MECHANICS. 


107 


7  about  some  axis  through  o.     If  this  axis  makes  an  angle  6  with 
the  tangent  plane,  there  is  an  angular  velocity  of  rolling  y  cos.  0 


Fig.  50. 


about  an  axis  in  the  tangent  plane,  and  an  angular  velocity  of 
spinning  7  sin.  9  about  the  common  normal  to  the  two  planes. 

If  in  Fig.  49  the  surfaces  are  cylindric  and  the  rolling  angular 
velocity  is  u,  and  if  r  is  the  radius  of  curvature  of  A  o  B  at  o,  and 
if  R  is  the  radius  of  curvature  at  o  of  the  fixed  surface  DOC,  and  if 
v  is  the  linear  velocity  of  the  point  of  contact,  it  is  easy  to  show 
that 


„-.(!+!)* 

\r       R/ 


W 


The  centres  of  curvature  are  supposed  to  be  on  different  sides  of 
the  tangent  plane. 

If  in  Fig.  50  there  is  friction  between  w  and  the  plane,  and 
motion  up  the  plane  is  steadily  taking  place,  draw  R,  making  an 
angle  N  o  R  =  <J>,  with  o  N  the  normal.  Let  P 
show  the  direction  of  the  pulling  force,  and  w 
of  the  weight.  Knowing  only  w  to  calculate  P 
and  R,  we  have  simply  to  use  the  triangle  of 
forces.  We  are  given  one  side  and  the  angles 
to  find  the  other  sides.  Thus  draw  AB  repre- 
senting w  to  scale ;  and  draw  B  c  an  I  AC 
parallel  to  the  two  unknown  forces,  with  the 
arrow-heads  going  circuitally.  Measure  B  c 
and  c  A,  and  the  forces  P  and  R  are  known. 

Analytically. — Resolve  all  the  forces  horizon- 
tally and  vertically,  and  we  have  (see  Art.  31),  if 
the  angle  p  o  D  is  a,  as  P  makes  an  angle  o  +  0 
with  the  horizontal,  and  R  makes  an  angle  R  o 
complement  of  (<J>  +  0), 

p  cos.  (a  -f  0)  =  R  sin.  (<J>  +  0) *(1) 

p  sin.  (a  +/)  +  B  cos.   <J>  +  0}  =  W  .  .  .  .  (2). 

AS  we  really  do  not  wish  to  know  R,  use  its  value  from  (1)  in  (2), 


Fig.  51. 
which  is  the 


108  APPLIED   MECHANICS. 

and  we  find  P  sin.  (a  +  0)  +  cos.  (p  +  6}  **£'{**%  =  *• 

w  sin.  (<f>  +  8) 
P  ***  oos.  ($  +  8)  cos.  (a  +  0)  +  sin.  (<J>  +  0)  sin.  (a  +  0)  ~~ 


w  sin,  ((ft  +  0)  sin.  (0  +  4>) 


If  p  is  the  force  required  to  just  prevent  the  body  sliding  down 
the  plane,  it  is  necessary  to  draw  the  angle  NOR  upwards,  instead 
of  downwards,  and  to  take  —  $  for  </>  in  the  formula.  Then 

P=w8in-;*--*i t*). 

cos.  (a  +  <p) 

Example  1. — If  a  =  o.  p  =  w  — —^- for  upward  motion ; 

cos.  <f> 

p  =  w  — — for  downward  motion,      p  is  0  for  downward 

COS.  (f) 

motion  when  0  =  </>.  Under  these  circumstances  the  body  will 
just  be  about  sliding  down  under  the  action  of  its  own  weight. 

When  0  ==  <{>,  it  takes  a  force  P  =  2  w  sin.  <£  to  drag  the  body 
up  the  plane. 

Notice  that  when  the  body  is  pulled  up  the  plane,  if  0  is  a 
small  angle,  expanding  sin.  (<J>  +  0),  and  dividing,  we  find  p  =  w 
(tan.  <f>  cos.  0  +  sin.  0). 

Taking  cos.  0  =  1,  we  have,  since  tan.  <f>  =  /t,  p  =  /tw  +  w 
sin.  0. 

Now  fj.vf  is  the  force  that  must  be  exerted  if  we  have  no 
inclination  0,  and  w  sin.  0  is  the  force  that  must  be  exerted  if  we 
have  inclination  but  no  friction.  Hence  the  rule  usually  employed 
in  calculating  the  pulling  force  on  a  vehicle  (Art.  47) — namely, 

the  total  pull  when  the  inclination  is  1  in  n  f  or  sin.  6  =  -  j — is 

equal  to  the  pull  necessary  on  a  level  road  plus  -  th  of  the  weight. 

It  is  only  true  when  0  is  small. 

Example  2. — If  a  =  —  0,  so  that  P  acts  horizontally,  p  =  w 

sin.  (0  —  $) 


=  w.  tan.  (0  —  <J>)  for  downward  motion. 

Example  3.  —  If  <6  =  o,  so  that  there  is  no  friction,  p  =  w  —  :  —  . 

cos.  a 
Example  4.  —  Find  a  so  that  p  for  upward  motion  may  be  a 

minimum.     That  is,  what  value  of  a  will  cause  Sm'  ;  -  ^  to  be 

cos.  (a  —  <p) 

a  minimum  ?  That  is,  what  value  of  a  will  make  cos.  (a  -  <f>) 
a  maximum  ?  Evidently  it  is  a  maximum  when  a  —  <f>  =  0 
or  a  =  <f>,  and  then  p  =  w  sin.  (<£  +  0).  This  important  result 
seems  to  be  completely  ignored  by  men  who  deal  practically  with 
traction  problems. 

Example  5.  —  If  there  is  no  force  P,  what  is  the  acceleration 


APPLIED   MECHANICS.  109 

down  the  plane?  "We  must  first  find  what  value  of  P  acting 
parallel  to  the  plane  would  just  "be  overcome.  This  is  given  in 

Example  1  as  w  — i — ""      .     Now,  this  force  acts  upon  the  mass 
cos.  <j> 

— ,  and  acceleration  is  force  •*•  mass;  so  that  the  acceleration  is 
</ 

g  — '_± — ~  TV.  Of  course,  when  <j>  =  0,  this  becomes  the  well- 
known  g  sin.  0. 

Example  6. — When  the  nut  of  a  square-threaded  screw  is 
turned,   the   surface    of    the   thread   is    like  an  inclined   plane, 

the  tangent  of  whose  inclination,   0,  is        .  ,   if  d  is  the  mean 

diameter  of  the  thread  (or  the  diameter  of  the  pitch  cylinder  of  the 
screw).  The  nut  presses  upon  the  inclined  plane  exactly  as  in 
Example  2,  and  hence  p  =  w  tan.  (0  -J-  </>)  ....  (1)  if  P  over- 
comes w,  the  total  weight  to  he  lifted  when  the  nut  is  turned ;  or 
p  =  w  tan.  (8  —  </>)....  (2)  if  w  overcomes  P.  If  the  length 
of  the  handle  to  which  the  real  force  A  is  applied  he  I,  then 

f  A  =  £  P  rf  or  A  =  £  ^ ;  so  that  (1)  and  (2)  "become 
A  =  i|wtan.  (0  +  </>) (1), 

A  =  i  *  w  tan.  (0  -  </>)  .  .  .  .  (2). 

(1)  and  (2)  are  equal,  of  course,  if  <f>  is  0,  and  then 

,  d 
A  =  I  -,  w  tan.  6. 

In  this  case,  as  there  is  no  friction,  the  efficiency  is  1,  and  the 

mechanical  advantage  is  —  —     2  ?       ....  (3),      The  mechanical 
A        d  tan.  e 

2  I 
advantage  from  (1)  when  there  is  friction  is  ^tan  (0  +  0) ^' 

Hence  the  efficiency  when  there  is  friction  is  e  =     n/g  ,  ^ (4)« 

When  w  overcomes  A  the  efficiency  is  similarly  e  —  — — ^ — -^  . .  (5). 

Both  (4)  and  (5)  "become  1  when  there  is  no  friction. 

It  is  an  easy  exercise  in  the  calculus  to  show  that  (4)  is  a 

maximum  when  6  =  j  —  ^  <f>  .  .  .  .  (6),  and  that  (5)  is  a  maximum 

when  0=|+i</, (7). 

If,  then,  a  screw  is  to  work  as  much  in  one  way  as  the  other,  it 
seems  reasonable  to  use  6  =  J-  or  45°  as  the  angle  of  its  mean 

spiral. 

Note  that  when  d  is  as  small  as  0,  w  cannot  overcome  A,  and 
the  screw  will  not  overhaul  or  reverse.     In  this  case,  if  A  over- 


110  APPLIED    MECHANICS. 

comes  w,  e  = '-¥. .     When  a  lubricant  is  used  we  cannot  be 

Tan.  /  <p 

sure  that  the  here  assumed  law  of  friction  holds  ;  that  is,  ju  is  not 
a  constant.  When  a  lubricant  is  not  used  it  is  safe  to  say  that  in 
no  actual  screw-jack  does  one  find  6  approaching  <f>  in  value. 

Note  that  when  there  is  no  friction  the  mechanical  advantage 
.    w  2 1  pitch          . ,    . 

13  —  =  — »  where  tan.  6  =  - — r  ;  so  that 

A        d  tan.  0'  ird 

W  _       2lird        =   27rl 
A        d  x  pitch        pitch' 

or  circumference  described  by  the  end  of  the  handle  -j-  pitch,  the 
rule  given  already. 

I  have  not  considered  the  very  considerable  loss  of  efficiency 
due  to  friction  because  the  load  which  is  lifted  is  being  kept  from 
turning.  Even  the  carefully-constructed  and  well-lubricated  square- 
threaded  screw-jack  in  my  laboratory  has  an  efficiency  of  only 
about  '25  even  at  the  highest  loads.  The  ordinary  jack  used  in 
workshops  has  a  very  much  smaller  efficiency  than  this.  When 
the  screw  has  a  triangular  thread  we  may  assume  that  with  the 
same  kinds  of  rubbing  surface  the 
co-efficient  of  friction  has  greatly 
increased  ;  in  reality  it  is  the  normal 
pressure  on  the  thread  which  has 
increased. 

90.  When  &  putting  force  r,  making 
an  angle  ft  with  the  normal,  is  applied 
to  move  a  block  which  is  being  pressed 
against  a  surface,  the  tangential 
component  which  overcomes  friction 
Fig.  52.  is  F  sin.  ft.  The  normal  component 

F  cos.  ft  diminishes  R,  so  that  the  fric- 
tion is  jw  (R  —  r  cos.  £).     Hence,  when  sliding  occurs, 
F  sin.  ft  =  fjL  (R  -  F  cos.  0), 

/U.  R  .  ,-. 

~~  sin.  ft  +  p  cos.  ft 

For  any  given  value  of  /*  it  is  easy  to  find  the  value  of  ft  which 
will  make  F  a  minimum.  In  fact,  the  denominator  of  (1)  is  a 
maximum  when  cos.  ft  =  ^  sin.  ft,  or  p.  tan.  ft  =  I,  or  ft  is  the 
complement  of  0,  the  angle  of  repose. 

But  if  F  is  a  pushing  force,  we  have  F  sin.  ft  =  p  (R  -f  F  cos.  ft) ; 

so  that  F  =  -: — ^—^ .    In  this  case  F  becomes  infinity 

sin.  (3  —  M  cos.  ft 

if  /3  is  less  than  the  complement  of  <J>.  For  an  angle  (3  which 
is  less  than  90°  the  pushing  F  must  be  greater  than  the  pulling.  F. 
When  one  piece  of  machinery  drives  another  at  a  sliding  contact 
this  great  distinction  between  pushing  and  pulling  must  be 
remembered. 

EXEECISES. 

1.  A  weight  of  5  cwt.  resting  on  a  horizontal  plane  requires  a 
horizontal  force  of  100  Ibs.  to  move  it  against  friction.  What,  in  that 
case,  is  the  value  of  the  co-efficient  of  friction  ? 


MECHANICS. 


Ill 


2\  A  weight  of  50  Ibs.  is  supported  by  friction  alone  on  an  inclined 

plane  ;  what  is  the  force  of  friction  ?  Angle  of  plane,  sin      •§-.     Ans.,  20  Ibs. 

3.  A  body  placed  on  a  horizontal  plane  requires  a  horizontal  force  equal 
to  one-half  its  own  weight  to  overcome  the  friction.  If  the  plane  be  gradually 
tilted,  at  what  angle  will  the  body  begin  to  slide?  Ans.,  26°  34'. 

4.  Find   the   force   parallel  to  the   plane  required  to  draw  a  weight 
of  40  Ibs.  lip  a  rough  inclined  plane  if  fj,  =  ^,  the  inclination  of  the  plane 
being  such  that  a  force  of  12  Ibs.  acting  at  an  angle  of  15°  to  the  plane 
would  support  the  weight  if  the  plane  were  smooth.         Ans.,  24'35  Ibs. 

5.  On  a  rough  inclined  plane  it  is  found  that  a  body  is  just  supported 
on  it  by  a  horizontal  force  equal  to  three-quarters  the  weight  of  the  body. 

Find  the  co-efficient  of  friction.     Angle  of  plane,  sin  ~~  f.          Ans.,  rs¥. 

6.  Two  unequal  weights,  Wj  and  w2,  of  the  same  material  on  a  rough 
inclined  plane  are  connected  by  a  string  which  passes  over  a  fixed  pulley 
in  plane.     Find  inclination  of  plane  when  the  system  is  in  equilibrium. 

Ans.,  tan  "' 


Wj  —  W2 

7.  Two  rough  bodies,  wx  and  w2,  rest  on  an  inclined  plane  and  are 
connected  by  a  string,  the  inclination  of  string  to  the  horizontal  being  the 
same  as  the  plane.     If  the  co-  efficients  of  friction  are  MI  and  p2  respectively, 
find  the    greatest    inclination  of    the    plane  when  the    system  is   in 
equilibrium.  '   Ang    tan  -1  A*i  wt  +  /*a  wa 

\fl  —  wa 

8.  A  plane  is  inclined  to  the  horizontal  at  24°;  there  is  a  load  of 
1,200  Ibs.  on  it,  the  co-efficient  of  friction  being  '18.     Determine  (1)  the 
force  required  to  just  draw  the  load  up  the  plane  ;  (2)  the  force  required 
to  just  prevent  it  slipping  down  ;  (3)  the  force  required  to  support  it  if 
there  be  no  friction,  the  direction  of  the  line  of  action  of  the  force  being 
in  each  case  23°  with  the  plane,  and  above  it.     Determine  the  correspond- 
ing results  if  the  direction  of  the  force  be  parallel  to  the  plane,  and  again 
if  the  force  be  horizontal.     Obtain  also  the  least  force,  in  magnitude  and 
direction,  which  is  necessary  in  each  of  the  three  cases  first  referred  to  ; 
that  is,  for  raising,  lowering,  and  supporting  without  friction.     There  are 
twelve  results  in  all. 

Ans., 


Effort  E  in  pounds. 

E  -Making  23° 
with  plane. 

K  parallel  to 
plane. 

E 

horizontal. 

K  a  mini- 
mum. 

Upward  motion. 

692 

685 

816 

074 

Downward  motion. 

342 

291 

295 

286 

No  friction. 

530 

488 

534 

488 

112 


APPLIED    MECHANICS. 


9.  A  load  of  1,200   Ibs.  has  to  be  pulled  along  a  horizontal  plane. 
Determine  in  magnitude  and  direction  the  least  force  necessary  to  do  this, 
the  co-efficient  of  friction  being  taken  0-4.  Ans.,  446  ;  21°  48'. 

10.  Determine  the  efficiency  in  the  case  of  a  screw  2^  inches  diameter, 
in  which  there  are  four  threads  to  the  inch,  taking  the  co- efficient  of 
friction  -04.  Ans.,  0'442. 

11.  Taking  the  mean  diameter  of  the  threads  of  a  1  inch  bolt  to  be 
•92  inch,  number  of  threads  per  inch  8,  the  co-efficient  of  friction  reduced 
to  an  equivalent  square  thread  '17,  find  the  turning  couple  required  to 
overcome  an  axial  force  of  2£  tons.  Ans.,  -20  ;  46*1  pound-feet. 


JOINTS   WITH   FRICTION. 

91.  When  we  know  the  resultant  of  the  loads  or  forces  on  a  pin 
ar  journal,  and  the  law  of  friction,  we  can  calculate  the  resultant 
force  on  the  eye  or  step.  Thus,  if  no  re- 
presents the  resultant  load  w  upon  the  journal 
s  P,  and  there  is  also  a  twisting  couple  turning 
the  journal  in  the  direction  of  the  arrow  T, 
the  journal  rides  up  in  the  step  till  P  is  the 
point  of  contact,  and  until  o  P  Q  =  <£  is  the 
angle  of  repose,  P  Q  being  a  vertical  force  equal 
to  w,  and  this  is  the  resultant  of  the  forces 
with  which  the  step  acts  on  the  journal;  ju,  the 
coefficient  of  friction,  is.  tan.  o  P  Q. 

It  is  usually  stated  in  books  that  the  resul- 
tant force  between  an  eye  and  pin  must  act 
through  some  point  such  as  P  in  the  surface  of  contact.     To  show 
that  this  is  a  mistake,  imagine  A  B  and  B  c  two  pieces  with  a  pin 
joint  at  B.    These  letters  indicate  the  centres  of  pins  at  A,  B,  and 


Fig.  53. 


0.  If  we  neglect  the  weights  of  the  pieces,  imagine  frictionless  pins 
at  A  and  c,  then  any  force  communicated  from  A  to  c  through  the 
frictional  joint  B  is  in  the  straight  line  A  c ; 
for  the  resultant  force  of  the  pin  A  upon  the 
piece  A  B  acts  through  the  centre  A,  and  the 
resultant  force  of  the  piece  B  c  on  the  pin  c 
acts  through  the  centre  c.  Imagine  the  pin 
at  B  to  be  rigidly  fixed  to  B  c.  The  force 
between  A  B  and  c  B  at  the  joint  B  must 
be  in  the  direction  A  c ;  and  we  can  have 
so  much  friction  at  B  that  A  c  does  not 
Fig,  56.  necessarily  act  through  any  point  of  the 


APPLIED    MECHANICS.  113 

surface  of  contact.  We  cannot,  however,  imagine  this  happening  if 
the  pin  and  eye  at  B  meet  only  at  one  point  or  line  at  right  angles  to 
the  paper ;  and  it  is  therefore  important  that  pins  and  eyes  should  be 
&f  slack  fit,  and  we  shall  in  future  imagine  them  to  he  so.  We  avoid 
unnecessary  complication  if  we  imagine  that  when  two  pieces  act 
through  a  pin,  the  pin  is  rigidly  attached  to  one  of  them.  Let  c  n 


Fig.  58. 

and  E  P  be  two  frictional  pins  acting  on  the  piece  o  H.  Let  A  and 
B  be  their  centres.  Let  the  coefficients  of  friction  be  a  and  b,  or 
the  angles  of  repose  <f>  and  01.  Let  the  radii  of  the  pins  be  A  and 
B.  Describe  about  A  a  circle  with  the  radius  A  sin.  <f>,  or 

a  b 

A    /-.    ,     a.  and  about  B  a  circle  with  the  radius   B    ,-,       ^~. 
v  1  +  a2'  V  1  +  b3 

These  may  be  called  the  friction  circles  of  the  pins. 

If  there  were  no  friction,  the  equal  and  opposite  forces  at  A  and 
B  acting  on  the  pin  would  be  in  the  direction  A  B.  They  are  really 
in  the  direction  of  a  common  tangent  to  the  two  circles.  There 
are  four  positions  of  this  common  tangent  depending  upon  the 
direction  of  relative  sliding  of  pin  and  piece  at  each  place. 

There  are  certain  cases  in  which  we  see  the  direction  without 
trouble.  Thus :  if  A  (Fig.  57)  is  the  centre  of  the  crosshead  of  a 


Fig.  57. 

steam-engine,  and  c  of  the  crank  shaft.  If  the  resultant  force  due 
to  the  piston-rod,  etc.,  is  F  at  A,  with  no  friction,  we  know  that  F 
sec.  BAG  is  the  pushing  force  in  the  connecting-rod  A  B,  and  the 
moment  of  this  about  c  is  F  sec.  BAG  X  CD  cos.  B  A  c  =  F  x  c  D.  If,  then, 
we  had  no  friction  at  the  slide,  but  only  in  the  pins  A  and  B,  as  we 
see  that  both  frictions  diminish  the  turning  moment,  the  line  of 
resistance  touches  the  two  friction  circles  drawn  at  A  and  B  in  the 
manner  shown  in  the  figure,  and  the  turning  moment  on  the  crank 
shaft  is  F  x  c  B.  Happily,  in  steam-engines  that  are  well  attended  to 
the  friction  circles  are  so  small  that  we  may  practically  neglect 
them,  and  consider  the  force  between  a  rod  and  pin  to  be  truly 
through  the  centre  of  the  pin.  Should  we,  however,  as  in  the  case 
of  the  hinged  joint  of  an  arch,  fear  friction  at  the  joint,  it  is 
worth  while  remembering  that  whatever  be  the  loads  on  a  piece, 


114 


the  necessary  equilibrating  forces  at  the  pins  are  to  be  fo\md 
tangential  to  the  friction  circles  there.  Thus,  if  A  and  B  (Fig.  58) 
are  the  centres  of  the  hinges  of  the  arch  A  c  B,  draw  the  friction 
circles  at  A  and  B.  It  is  not  enough  to  find  the  line  of  resistance 

ACT?,  but  also  other  pos- 
sible ones  not  passing 
through  the  centres  A  and  B, 
but  touching  the  friction 
circles  there.  If  a  is  the 
inclination  of  the  centre  line 
of  the  arch  at  A,  and  r  is  the 
radius  of  the  friction  circle 
there,  there  will  be  no  great 

Pig.  58.  error  in  adopting  this  rule ; — 

The  line  of  resistance  may 

start  from  any  point  between  A7  and  A"  in  the  vertical  through  A, 
if  A'A  =  A  A*  =  r/sin.  a. 

The  smaller,  therefore,  the  radius  of  the  cylindric  surface  or 
hinge  of  a  hinged  arch  the  better.  When  an  arch  ring  abuts  at  a 
cylindric  surface  of  large  radius,  I  cannot  see  much  distinction 
between  it  and  an  arch  fixed  at  the  ends,  except  that  there  may  be 
no  tension  in  the  joint,  or  rather  that  the  line  of  resistance  may 
not  pass  outside  the  joint. 

92.  Body  turning  about  an  Axis. — In  Fig.  59  we  have  a  body 
which  can  move  about  an  axis.     It  is  acted  on  by  a  number  of 

cords  exerting  forces 
which  just  balance 
one  another.  Now, 
if  you  make  this  ex- 
periment you  will 
find  that  you  must 
keep  your  finger  on 
the  body,  because  it 
is  in  such  a  state 
that  a  very  small 
motion  either  way 
causes  the  forces  to 
no  longer  balance. 
:  Suppose,  however, 
you  were  to  let  the 
Fig.  69.  cord  A  be  wound  on 

a  pulley  whose  radius 

is  equal  to  the  distance  A  o  j  the  cord  B  on  a  pulley  whose 
radius  is  equal  to  B  o,  and  so  on,  you  would  have  the  arrange- 
ment shown  in  Fig.  60,  which  differs  from  Fig.  59  in  that  a 
small  motion  has  no  effect  on  the  balance.  Now  what  is  the 


APPLIED   MECHANICS. 


115 


Fig.  60. 


Condition  of  balance  in  this  case?    Suppose  one  complete  turn 

given   to   the   axis,  every  cord   shortens   or   lengthens   by  a 

distance  equal  to  the 

circumference  of  the 

pulley  on  which  it 

is  wound.  Let  A  and 

B  lengthen,  and  let 

c  shorten,  then  we 

know  that  the  work 

done    by   A   and   B 

must  be  equal  to  the 

work  done  against  c. 

Hence, 

Pull  in  A  x  cir- 
cumference of 

A'S         pulley, 

together  with 

pull    in   B  x 

circumference 

of  B'S  pulley,  must  be  equal  to  pull  in  c  x  circumference 

of  c's  pulley. 

We  might,  however,  use  the  diameters  or  radii  of  the  pulleys, 
and  so  we  see  that  in  Fig.  60  there  is  balance  if 

Pull  in  A  x  A  o,  together  with  pull  in  B  x  B  o,  equals  pull  in 

C  X  CO. 

The  pull  in  A  x  A  o  is  really  the  tendency  of  A  to  turn  the 
body  about  the  axis,  and  in  books  on  mechanics  it  is  called  the 
moment  of  the  force  in  A  about  the  axis  o.  The  law  is  then, 
if  a  number  of  forces  try  to  turn  a  body  and  are  just  able  to 
balance  one  another,  the  sum  of  the  moments  of  the  forces 
tending  to  turn  the  body  against  the  hands  of  a  watch  must  be 
equal  to  the  sum  of  the  moments  of  the  forces  tending  to  turn 
the  body  with  the  hands  of  a  watch.  We  sometimes  say : — 
"  The  algebraic  sum  of  the  moments  of  all  the  forces  is  zero." 
That  is,  we  regard  one  kind  of  moment  as  positive  and  the 
other  as  negative.  Another  way  of  putting  the  proof  is  this  : — 
If  a  much  magnified  drawing  be  made  showing  a  very  small 
motion  through  the  angle  0,  it  will  be  seen  that  the  work  done 
by  c  being,  either  c  x  motion  of  point  of  application  resolved 
in  direction  of  c,  or  motion  of  point  x  resolved  part  of  c 
in  direction  of  motion;  in  either  case  this  work  is  equal  to 
0  X  0  c  x  0,  that  is,  moment  of  c  x  0.  Hence,  total  work  = 


116  APPLIED   MECHANICS. 

algebraic  sum  of  moments  x  6,  and  if  total  work  is  0,  then  the 
algebraic  sum  of  the  moments  is  0.  When  work  is  done  upon 
a  body  in  turning  it,  observe  that  the  work  in  foot-pounds  is 
equal  to  the  moment  in  pound-feet  multiplied  by  the  angle  in 
radians  turned  through.  If  a  constant  moment  M  acts  upon  a 
body  rotating  at  a  radians  per  second,  Ma  is  the  work  done 
per  second.  If  n  is  the  number  of  revolutions  per  minute, 
2  TT  n  is  the  angle  per  minute,  and  hence  M  2  IT  n  +  33,000  is 
the  horse-power. 

[No  doubt  the  student  has  already  become  quite  familiar 
with  the  idea  expressed  by  "work=  force  x  distance."  It 
has  just  been  shown  how,  by  the  very  simplest  transformation, 
this  expression  becomes,  in  the  case  where  a  force  is  producing 
turning  about  an  axis,  "  work  done  =  moment  x  angle  turned 
through  in  radians."  Although  not  required  for  the  study  of 
the  portion  of  the  subject  now  under  consideration,  yet  we 
may  remind  the  student  that  at  a  later  stage  it  will  be  an 
advantage  to  him  if  he  has  accustomed  himself  to  calculating 
work  done  when  the  turning  moment  (called  the  torque)  and 
angle  turned  through  are  given.] 

93.  The  Lever. — Thus,  for  example,  a  lever  is  a  body  such 
as  I  have  spoken  about,  capable  of  turning  about  an  axis.  You 
will  find  that  our  general  rule  of  work  and  this  rule  of 
moments  will  give  the  same  result.  If  two  forces  act  on  a 
lever,  they  will  balance  when  their  moments  about  the  axis  are 
equal;  that  is,  when  p,  multiplied  by  the  shortest  distance 
from  the  fulcrum  or  axis  to  the  line  in  which  p  acts,  is  equal 
to  w  multiplied  by  the  distance  of  the  fulcrum  from  the  line 
in  which  w  acts. 

If  a  number  of  forces  balance  when  acting  on  a  lever,  tfie 
sum  of  the  moments  tending  to  turn  the  lever  against  the  hands 
of  a  watch  must  be  equal  to  the  sum  of  the  moments  tending  to 
turn  the  lever  with  the  hands  of  a  watch. 

It  must  be  remembered  that  if  the  body  acted  upon  has  its 
centre  of  gravity  somewhere  else  than  in  its  axis,  then  we  must 
consider  that  the  weight  of  the  body  is  a  force  acting  vertically 
through  its  centre  of  gravity. 

Example. — The  safety  valve,  Fig.  61,  must  open  when  the 
pressure  on  the  valve  is  just  100  Ibs.  per  square  inch.  The 
mean  area  of  the  valve  A,  on  which  we  assume  that  the  pres- 
sure acts,  is  3  square  inches;  CD  is  2  inches,  E  is  50  Ibs.,  the 
weight  of  the  lever  is  6  Ibs.,  and  its  centre  of  gravity  is  6  inches 


APPLIED   MECHANICS. 


117 


from  D  ;  where  must  '&  be  placed  1  All  distances  are  measured 
horizontally.  Here,  the  upward  force  is  100  x  3,  or  300  Ibs., 
and  its  moment  about  D  is  300  x  2,  or  600.  The  moment  of 
the  weight  of  the  lever  is  6  x  6,  or  36.  The  moment  of  the 
weight  E  is  50  x  the  required  distance  from  D.  Hence,  600 — 
36,  or  564  divided  by  50,  is  the  answer;  11-28  inches  from 
D.  Thus  we  find  where  the  mark  100  ought  to  be  placed. 

Let  the  student  repeat  this  for  pressures  of  90,  80,  and 
70  Ibs.  per  square  inch,  stating  in  each  case  the  distance  of  the 
weight  from  D.  What  are  the  distances  apart  of  the  marks  ? 

This  is  not  the  place  to  consider  the  various  forms  of  safety 


0 


O 


Fig.  81. 

valve  employed  by  engineers.  The  force  which  fluid  at  rest 
can  exert  against  a  valve  is  not  the  same  as  when  the  valve  is 
open  and  the  fluid  is  moving ;  but  it  is  very  important  that  a 
valve  should  stay  open,  allowing  the  fluid  to  escape.  This  is 
effected  partly  by  using  a  peculiar  shape  of  valve,  but  more 
usually  by  letting  E  come  closer  to  the  fulcrum  when  the  valve 
opens.  In  the  above  figure  the  valve  seat  is  conical ;  in 
practice,  a  flat  seat  is  now  much  more  commonly  used,  the 
breadth  of  the  seat  being  very  small. 

Exercise. — A  weighbridge  consists  of  three  levers  whose 
mechanical  advantages  help  each  other ;  I  mean,  the  short 
arm  of  each  supports  the  long  arm  of  the  next.  Suppose  that 
the  weights  of  all  parts  are  arranged  so  as  just  to  be  balanced 
when  no  load  is  on  the  bridge,  and  that  the  mechanical 
advantages  of  the  three  levers  are  8,  10  and  12,  what  load 
will  be  balanced  by  a  weight  of  15  Ibs.  ?  Answer — 14,400  Ibs. 
Suppose  that  it  is  the  first  of  these  levers  that  is  alterable 


118  APPLIED    MECHANICS. 

(that  is,  there  is  a  sliding  weight),  what  is  its  mechanical 
advantage  altered  to  when  the  load  is  16,000  Ibs.  1  Answer — 
It  was  8 ;  it  now  becomes  increased  in  the  proportion  of  1 6,000 
to  14,400,  so  that  it  becomes  8-88S9. 

Show  that  the  graduation  of  the  lever  with  the  sliding 
weight  is  in  equal  divisions  for  equal  alterations  of  the  load. 
Do  this  by  finding  the  position  of  the  sliding  weight  for 
various  loads. 

Friction  is  greatly  got  rid  of  in  weighing-machines  by  using 
steel  knife  edges  as  fulcrums  of  levers  resting  on  hard  steel  or 
agate  plates.  Students  must  examine  actual  specimens.  The 
common  chemical  balance  must  be  examined.  There  is  one 
thing  about  it  which  may  trouble  the  student.  "Why  are 
short-armed  balances  now  preferred  to  the  older  forms'?"  The 
short-armed  balance  has  less  moment  of  inertia,  and  this 
causes  it  to  be  quicker  in  its  motions,  so  that  time  is  saved. 
(See  Art.  453.)  But  there  is  much  more  to  be  said  about  it 
than  this.  Indeed,  in  this  as  in  all  other  cases,  to  thoroughly 
understand  one  machine  requires  a  knowledge  of  the  whole  of 
applied  mechanics  and  applied  physics.  I  am  not  now  dis- 
cussing any  one  machine  exhaustively.  I  was  strongly  tempted 
to  take  up  the  thorough  consideration  of  one  machine  and  call 
this  the  study  of  applied  mechanics ;  and  if  I  had  a  student 
with  a  particular  interest  in  one  machine  this  would  be  the 
very  best  way  to  put  before  him  the  study  of  applied  mechanics. 
The  method  I  have  adopted  in  this  book  is  to  illustrate  eacli 
principle  by  means  of  a  machine  in  which  that  principle 
happens  to  appear  most  important.  The  defect  of  the  method 
arises  from  its  causing  a  student  to  think  that  he  knows  all 
about  a  machine  when  he  only  knows  the  most  important 
principle  of  applied  mechanics  which  is  illustrated  by  it.  The 
cure  for  this  academic  training  defect  comes  when  a  student  is 
compelled  to  take  a  special  interest  in  some  one  machine,  and 
it  is  then,  in  practical  work,  that  he  really  is  learning  applied 
mechanics.  We  can  only  partially  correct  the  defect  by  our 
numerical  exercise  and  laboratory  work  and  experience  in  the 
workshop. 

Students  looking  at  weighing-machines  ought  particularly 
to  notice  that  when  objects  to  be  weighed  are  placed  not  on 
swinging  scale-pans,  but  on  fairly  firm  platforms,  the  construc- 
tion of  the  balance  must  be  such  as  to  make  the  measurement  to 
be  independent  of  the  position  of  the  object  upon  the  platform. 


APPLIED    MECHANICS.  119 

EXERCISES. 

1.  A  uniform  straight  bar,  2  feet  long,  weighs  5  Ibs. ;  it  is  used  as  a 
lover,  and  an  8  Ibs.  weight  is  suspended  at  one  end ;  find  the  position  of 
the  fulcrum  where  there  is  equilibrium.  Ans.,  4'5  inches. 

2.  A  lever  safety-valve  has  the  following  dimensions  : — Mean  diameter 
of  valve,  3  inches ;  weight  of  valve,  8  Ibs. ;  distance  of  centre  of  valve 
from  fulcrum,  2£  inches ;  weight  of  lever,  16  Ibs. ;  distance  of  its  centre  of 
gravity  from  fulcrum,  13  inches.     Find  where  a  weight  of  35  Ibs.  must 
be  hung  from  the  lever,  so  that  the  steam  may  blow  off  at  85  Ibs.  per 
square  inch.  Ans.,  36-4  inches  from  fulcrum. 

3.  A  B  c  D  is  a  rectangle,  and  A  c  a  diagonal.     In  A  c  take  a  point  o, 
such  that  A  o  is  a^third  of  A  c.     Forces  of  30,  10,  and  5  act  from  A  to  B,  c 
to  B,  and  D  to  c  respectively.     If  A  B  =  3  inches,  and  B  c  =  4  inches, 
write  down  the  moment  about  o  of  each  force,  with  its  proper  sign,  and 
find  their  algebraic  sum.  Ans., -40;   +20;   +— ;   -  — . 

3  3 

4.  The  diameter  of  the  safety-valve  of  a  steam-boiler  is  4  inches ; 
the  weight  on  the  lever  is  90  Ibs.,  and  distance  from  centre  of  the  valve 
to  the  fulcrum  is  2'5  inches ;  what  must  be  the  distance  of  the  point  of 
suspension  of  the  weight  from  the  fulcrum  in  order  that  the  valve  will 
just  lift  when  the  pressure  of  steam  in  the. boiler  is  80  Ibs.  per  square 
inch  ?    Length  of  lever,  30  inches ;  weight,  12  Ibs. ;  distance  of  centre  of 
gravity  from  fulcrum,  14  inches.  Am.,  26'059  inches. 

5.  A  ball  weighing  4  Ibs.  is  fixed  to  one  end  of  a  bar  hanging  vertical, 
the  other  end  of  which  can  turn  about  a  fixed  axis.     If  the  ball  be  pulled 
out  by  a  force  which  acts  horizontally  through  the  centre  of  the  ball, 
what  will  be  the  amount  of  this  force  necessary  to  keep  the  bar  at  rest  in 
a  position  such  that  the  bar  makes  30°  with  the  vertical  ?    If  the  length 
of  the  bar  be  30  inches,  what  would  be  the  force  if  applied  at  a  point  on 
the  bar  10  inches  from  the  centre  of  the  ball  ?  4    t  _        __ 


120 


CHAPTER    VII. 

ELEMENTARY   ANALYTICAL   AND    GRAPHICAL    METHODS. 

94.  IN  Fig.  62  we  have  another  example  of  the  fact  that  when 
there  is  a  displacement  of  a  point,  o,  in  the  direction,  o  T,  and 
a  force,  p,  acts  in  the  direction  o  A,  the  angle  A  o  Q  being  called 
a,  the  work  done  is  the  displacement  multiplied  by  the  com- 
ponent of  P  in  the  direction  o  T.  This  component  is  p  cos.  a,  or 
it  may  be  defined  as  the  projection  of  o  A,  the  line  which 
represents  the  force,  upon  o  T. 

The  student  must  note  what  we  mean  by  the  projection  of 

a  line.  The  projection  of 
A 

-T 

I 


Fig.  62. 


0  A  upon  the  line  o  T  is  o  Q. 
If    the   arrow-head    is   not 
put  upon  o  Q,  the  order  of 
the  letters  will  indicate  the 
sense  of  the  action.     Thus 

1  shall  use  Q  o  to  mean  a 
sense   opposite   to  o   Q.       Now   note   that   if  we  have  lines, 
o  A,  A  B,  B  c,  c  D,  and  D  E  (Fig.  63),  what  I  call  the  sum 
of  their  projections  upon  any  line  is  really  the  projection   of 
o  E.     If  I  have  a  closed  polygon,  the  sum  of  the  projections 
of  its  sides  upon  any  line  is  zero,  if  the  sense  of  every  side  is 
circuital  round   the   polygon  with  the  sense  of  all  the  rest. 
If   the  sides  are  parallel  to,  and  proportional  to  forces,  this 
means  that  the  sum  of  all  the  components  of  all  the  forces  in 
any  direction  whatsoever  is  zero. 

Now  suppose  we  have  a  body  acted  upon  by  any  number  of 

forces  in  all  sorts  of  directions 
(we  can  only  illustrate  this 
by  strings  and  weights) ;  and 
if  the  weight  of  every  portion 
of  the  body  itself  be  con- 
sidered also  acting;  and  if 
the  body  is  in  equilibrium, 
the  principle  of  work  tells 


Fig.  68. 

a    small    translationaJ 


us  that  if  the  body  receives 
displacement,    «,    in    any    direction 


APPLIED   MECHANICS.  12l 

whatsoever  (translation  means  that  every  point  moves  parallel 
to  the  motion  of  every  other  point,  and  through  the  same 
distance),  then  the  total  work  done  by  all  the  forces  is  zero. 
But  the  work  done  by  any  force  is  its  component  in  the  direc- 
tion of  the  motion  multiplied  by  the  displacement ;  hence,  we 
see  that  the  sum  of  the  components  in  any  direction  whatsoever 
must  be  zero.  This  is  a  condition  which  must  hold  when  any 
body  is  in  equilibrium.  But  it  is  evident  that  we  have  the 
same  rule  when  we  say  that  the  forces  are  proportional  to  the 
sides  of  a  closed  polygon,  the  senses  of  the  forces  being  cir- 
cuital round  the  polygon.  If  the  forces  are  not  in  one  plane, 
the  polygon  will  be  what  is  called  a  gauche  polygon,  but  if  the 
amounts  and  directions  of  the  forces  are  given  in  plan  and 
elevation,  the  polygon  may  be  drawn  in  plan  and  elevation. 
To  give  the  analytical  rule  more  simply : — Take  three  lines 
mutually  at  right  angles  to  one  another,  o  x,  o  Y,  o  z ;  project 
all  the  forces  upon  o  x,  they  balance  (that  is,  the  algebraic  sum  of 
their  projections  is  zero) ;  project  all  the  forces  upon  o  Y,  they 
balance  ;  project  all  the  forces  upon  o  z,  they  balance.  These 
three  analytical  conditions  are  the  same  as  the  graphical  rule, 
"  the  force  polygon  is  closed." 

Perhaps,  however,  we  had  better  confine  our  attention  to 
forces  all  in  one  plane.  The  analytical  rule  is  :— The  algebraic 
sum  of  all  the  horizontal  components  is  zero,  or  the  horizontal 
components  balance;  the  algebraic  sum  of  all  the  vertical 
components  is  zero,  or  the  vertical  components  balance. 
These  two  conditions  are  the  same  as  "  the  force  polygon  is 
closed." 

EXERCISE. 

Given  the  [following  forces,  find  their  equilibrant — 
Force  o  x  of  50  Ibs.,  force  o  A  of  20  Ibs.,  the  angle,  x  o  A, 
being  33°;  force  M  o  of  56  Ibs.,  the  angle  XOM  being  150°; 
force  o  G  of  100  Ibs.,  the  angle  x  o  G  (all  angles  are  measured 
counter-clockwise)  being  217° ;  force  H  o  of  70  Ibs.,  the 
angle  x  o  H  being  315°.  Now  let  a  student  draw  these  on 
paper.  I  have  assumed  them  all  to  be  in  lines  through  a 
point  o,  but  this  was  for  ease  in  setting  the  question. 
The  forces  need  not  act  through  one  point.  Let  him, 
by  drawing,  or  by  using  a  table  of  sines  and  cosines,  find 
the  components  of  all  in  the  direction  o  x  and  o  Y,  if 
o  Y  is  perpendicular  to  0  x  (Art.  31).  I  make  these  projec- 
tions to  be — 


122 


APPLIED   MECHANICS. 


In  the  direction  o  x. 


50  cos.   0  =  60 
20  cos.  33°  =  16-773 

-f    56  cos.  30°  =  48-497 

-  100  cos.  37°=  -  79-863 

-  70  cos.  45°  =  -  49-497 

Algebraic  sum  =  2  x  =  —  14-09 


In  the  direction  OY 


50   sin.  0   =   0 
20  sin.  33°  =  10-893 
—    56  sin.  30°  =  —  28 
-  100  sin.  37°  =  -  60-181 
-f-     70  sin.  45°  =  49'497 

Algebraic  sum  =  2  Y  =  —  27 '791 


The  Greek  letter  2  is  used  to  mean  "  the  sum  of  all  such 
terms  as."  x  means  any  p  cos  a,  and  Y  any  p  sin  a.  We  have 
found  2x  and  SY ;  call  them  x  and  Y  respectively.  Draw  o  c 
=  -  14-09,  o  B=  -  27-79,  and  complete  the  rectangle  o  B  R  c ; 
then  o  R  is  the  resultant  and  R  o  the  equilibrant.  It  is 

evident  that  o  R2  =  \/x2  +  Y2=31'16.      Also  tan  R  o  C=Y/X= 
1-971  ;  therefore  Roc=63°  7',  and  x  o  R=243°  7'. 

The  student  must  get  accustomed  to  symbols  which  so 
greatly  shorten  our  use  of  language. 
Our  rule  is: — If  P15  P2,  etc.,  are 
forces  in  a  plane,  and  if  they  make 
angles  alf  a2,  etc.,  with  any  line,  say 
a  horizontal  line,  then  Pl  cos.  a^  = 
Xj,  P2  cos.  a2  =  x2,  etc.,  are  the 
horizontal  components  of  the  forces. 
The  directions  of  arrow-heads  must 
be  noted,  and  the  algebraic  sum  of 
the  horizontal  components  is  de- 
noted by  X  =  2P  cos.  a,  or  2x.  In 
the  same  way  Y=  £p  sin.  a,  or  SY, 
denotes  the  algebraic  sum  of  the 
vertical  components  Y1=P1  sin.  a]} 
Y2=P2  sin.  a2,  etc.  Then  the  re- 
sultant, R,  of  all  the  forces  is  the 
resultant  of  x  horizontally  and  Y  vertically,  and  R2  =  x2  -f  Y2, 
and  if  R  makes  the  angle  a,  with  the  horizontal,  then  tan.  a  == 
Y  -r-  x.  We  have  only  found  the  amount,  clinure  and  sense 
of  R.  We  do  not  yet  know  where  it  acts.  (The  word  clinure 
was,  I  believe,  invented  by  the  late  Prof.  Thomson ;  the  word 
direction  implies  more  than  what  we  try  to  express  by  clinure.) 


Fig.  63A. 


APPLIED   MECHANICS.  123 

Similarly,  if  the  forces  are  not  in  the  same  plane,  and  if  each  p 
makes  the  angles  o,  j8,  and  7,  with  three  standard  lines  at  right 
angles  to  one  another  (usually  called  the  axes  of  x,  y,  and  z),  then 
if  p  cos.  o  is  called  x  and  SP  cos.  o  =  x,  if  p  cos.  0_is  called  T  and 
2i«  cos.  £  =  Y,  and  if  P  cos.  7  =  z  and  SP  cos.  7  =  z,  then  R2  =  x2 
-f  x2  +  Y2  and  the  cosines  of  the  angles  which  R  makes  with 
the  axes  are  x/B,  Y/B,  and  z/B. 


95.  Any  physical  quantity  which  is  directional  is  called  a 
Vector  (such  as  a  displacement,  a  velocity,  an  acceleration,  a 
force,  a  stress,  the  flow  of  a  fluid,  etc.),  and  may  be  represented 
by  a  straight  line.    The  length  of  the  line  represents  the  quantity 
to  some  scale  of  measurement ;    the  line's  clinure  represents 
the  clinure  of  the  vector,  and  the  barb  of  an  arrow  represents 
its  sense.     It  is  easy  by  actually  drawing  lines  and  measuring 
their  lengths  to  solve  problems  which  would  otherwise  require 
a  good  deal  of  mathematical  knowledge.     This  sort  of  graphi- 
cal   calculation    having    proved   useful,   it   has   attracted  the 
attention  of  men  who  have  leisure  enough  to  make  an  elaborate 
study  of  its  methods.     It  has,  unfortunately,  become  a  com- 
plicated weapon  with  which  these  men  can  attack  all  sorts  of 
problems  which  are  much  more  easily  solved  in  other  ways. 

We  shall  only  use  graphical  methods  where  they  happen  to 
be  the  best  methods.  Now  a  force  is  a  vector ;  it  has  magni- 
tude, clinure,  and  sense,  but  it  is  more  than  a  vector ;  it  has 
a  fourth  quality  not  possessed  by  ordinary  vectors — namely, 
actual  position  in  space.  Settling  any  one  point  in  its  line  of 
application  settles  its  position  when  its  clinure  is  known.  But 
if  we  are  told  that  forces  all  act  at  a  point,  they  are  added 
exactly  as  all  mere  vector  quantities  are  added.  When,  then, 
I  speak  of  finding  the  resultant  or  equilibrant  of  forces  at 
a  point,  I  may  be  said  to  speak  of  any  vectors. 

96.  Forces  acting  at  a  Point— The  line  A  B  (Fig.  64) 
represents  a  force  in  clinure  by  its  own  clinure, 

in  amount  by  its  length  to  any  scale  we  please, 
and  in  sense  by  its  arrow-head,  which  shows 
that  the  action  of  the  force  is  from  A  to  B. 

It  would  not  be  correct  to  call  this  the  force 
B  A,  because  this  is  opposed  to  the  sense  of  the 
arrow-head.  The  forces  A  o,  o  B,  o  c,  o  D 
(Fig.  65),  all  act  upon  .a  small  body,  o,  or  their 
lines  of  action  when  produced  all  pass  through  pig.  &t 


APPLIED  MECHANICS. 


Fig.  66. 


a  point,  0,  in  a  large  rigid  body.  The  amount  of  each  force 
is  shown  by  the  length  of  the  line,  representing  it  to  some 
scale.  Now  to  add  these  forces  together  in  the  most  perfect 
manner — that  is,  to  find  a  force  called  their  resultant,  which 
shall  be  exactly  equivalent  in  its  effects 
to  all  the  above  forces  acting  together 
— we  draw  a  polygon  (Fig.  66).  Each 
side  of  this  polygon  is  parallel  to,  and 
proportional  to  a  force  in  Fig.  65  ; 
thus,  the  side  A  corresponds  to  the  force 
A  o,  and  the  arrow-heads  agree,  and 
lastly  the  action  indicated  by  the 
arrow-heads  is  circuital.  Fig.  66  is 
always  called  the  force  polygon. 
When  it  is  unclosed,  as  it  is  in  the 
present  case,  we  know  that  the  forces 
A  o,  etc.  (Fig.  65),  are  not  in  equili- 
brium. To  keep  A  o,  o  B,  etc.,  in  equilibrium,  a  new  force, 
called  the  equilibrant,  must  be  introduced  corresponding  to 
the  side  E  (shown  dotted),  which  will  close  the  polygon,  its 
arrow-head  being  circuital  with  the  others.  Now  if  we 
want  the  resultant  of  A  o,  o  B,  o  c,  o  D,  it  evidently  acts 
through  o  and  corresponds  to  E,  Fig.  66,  but  with  arrow- 
head reversed.  The  resultant  of  a  number  of  forces  is  equal 
and  opposite  to  their  equilibrant. 

Prove  now  the  following  statements  by  actual  drawing  : — 
1st.  The  resultant  of  any  number  of  forces 
does  not  depend  on  the  order  in  which  they 
are  drawn  as  sides  of  the  polygon. 

2nd.  Any  lines  or  forces  whatever  which 
form  a  closed  polygon  in  any  given  order  will 
form  a  closed  polygon  if  drawn  in  any  other 
order. 

3rd.  In   adding  forces  we  may  first  find 
the  resultant  of  some  of  the  forces,  and  then 
add  together  this  resultant  and  all  the  rest  of  the  forces.     The 
answer  will  always  be  the  same,  however  we  may  group  the 
forces  before  adding  them. 

When  the  forces  are  not  all  in  one  plane  the  polygon  must 
be  drawn  by  descriptive  geometry,  and  to  draw  it,  and  so  find 
the  resultant  or  equilibrant  as  the  closing  side,  is  an  excellent 
graphical  exercise. 


Fig.  66. 


APPLIED    MECHANICS.  125 

In  working  exercises  we  recollect  the  fact  that  the  resultant 
of  all  the  forces  due  to  gravity  is  called  the  weight  of  the  body 
and  acts  through  its  centre  of  gravity  (seepage  136).  This  one 
force  replaces  the  millions  which  are  due  to  gravity.  When 
we  observe  that  we  have  only  three  forces  acting  upon  a  body 
at  rest,  we  know  that  they  must  be  in  a  plane  and  act  through 
a  point  unless  they  are  parallel.  In  the  ordinary  books  on 
mechanics  there  are  many  problems  which  are  easily  solved 
if  we  remember  this  fact.  Four  forces  in  equilibrium,  if  not 
all  in  one  plane,  must  meet  in  one  point.  When  a  body  touches 
a  smooth  surface  we  know  that  the  force  which  there  acts  upon 
the  body  is  normal  to  the  smooth  surface.  When  a  body 
touches  a  rough  surface  we  know  that  the  force  which  there 
acts  upon  the  body  makes  an  angle  with  the  normal,  and  the 
limiting  value  of  this  angle  when  sliding  is  about  to  occur  is 
what  is  called  the  angle  of  repose,  or  the  angle  whose  tangent 
is  p,  the  co-efficient  of  friction  there.  It  is  astonishing  what 
a  number  of  exercises  are  easily  worked  out  if  one  will  only 
recollect  these  few  general  principles. 

The  student  will  at  this  place  work  again  the  exercises  of 
page  111.  It  ought  to  be  getting  clear  to  him  that  the  most 
difficult  analytical  work  has  really  nothing  in  it  more  com- 
plicated than  these  exercises  have.  The  exercises  are  usually 
called  easy,  certainly  students  get  easily  into  the  way  of 
working  them,  and  I  am  always  sorry  to  notice  too  great 
an  ease  of  this  kind.  It  often  indicates  shallowness  of 
comprehension.* 

EXERCISES. 

1.  Forces  o  A  =  30  Ibs.,  o  B  =  50  Ibs.,  co  =  15  Ibs.,  DO  =  80  Ibs., 
o  E  =  150  Ibs. ;  the  angles  are  B  o  A  =  45°,  c  o  A  =  90°,  D  o  A  =  135°, 
E  o  A  =  270°.     Find  the  resultant  analytically  and  graphically. 

Ana.,  223  Ibs.  at  an  angle  of  303"  with  o  A. 

2.  Sheer  legs  eacn  50  feet  long,  30  feet  apart  on  horizontal  ground, 
meet  at  point  c,  which  is  45  feet  vertically  above  the  ground  ;  stay  from 
c  is  inclined  at  40°  to  the  horizontal ;  load  of  10  tons  hanging  from  a 
Find  the  force  in  each  leg  and  in  stay.  dns.,  7 '8  tons ;  6 '4  tons. 

97.  In  many  engineering  problems,  when  forces  A,  B,  c,  D, 
etc.,  are  given,  it  is  sometimes  important  to  be  able  to 
show  graphically  the  resultant  of  A  and  B,  the  resultant  of  A,  B, 
and  c,  the  resultant  of  A,  B,  c,  D,  and  so  on.  Thus  (Fig. 
67)  A,  B,  c,  D,  etc.,  are  given  forces. 

Dra.w  the  unclosed  force  polygon  (Fig.  68).  Join  the  point 
0  with  each  corner  of  the  force  polygon.  From  the  point  B' 
•  See  Appendix. 


126 


APPLIED   MECHANICS. 


where  A  and  B  meet  (Fig.  67)  draw  a  line  B'  c'  parallel  to  the 
line  o  B  c  (Fig.  68)  (o  B  c  is  the  line  from  o  to  the  corner 
where  B  and  c  meet) ;  from  c'  draw  c'  D'  parallel  to  o  c  D, 
and  so  on.  Then  B'  c'  represents  the  position  and  direction, 


and  o  B  c  represents  to  scale  the  resultant  of  the  given  forces 
A  and  B.  Similarly  D'E'  represents  the  position  and  direction, 
and  ODE  represents  to  scale  the  resultant  of  the  given  forces 
A,  B,  c,  and  D.  Note  the  arrow-heads  of  the  resultants  we 
have  found.  The  line  A  B'  c'  D'  E'  F'  (Fig.  67)  is  usually  called 
a  line  of  resistance. 

98.  We  have  in  Art.  96  confined  our  attention  to  the  forces 
acting  upon  a  small  body,  or  forces  which  all  pass  through  one 
point  if  they  act  on  a  large  body.  But  in 
Fig.  67  and  in  our  description  we  assumed  a 
large  body,  and  our  forces  were  any  forces 
whatsoever.  We  gave  it  a  small  motion  of 
translation,  and  obtained  an  important  result 
from  the  consideration  that,  on  the  whole,  no 
work  was  done  upon  the  body.  Now,  let  us 
assume  that  any  point  o  in  the  body  is  fixed, 
or,  rather,  that  an  axis  o  is  fixed,  the  axis 
being  at  right  angles  to  the  plane  in  which 
all  the  forces  act ;  about  this  axis  we  assume 
that  the  body  may  rotate.  Consider  the  work 
done  by  all  the  forces  during  any  small 
rotation  6 ;  it  is  zero.  But  the  work  done 
by  any  force  is,  as  we  have  already  seen 
(Art.  92)  the  moment  of  the  force  multiplied 
by  0.  Hence  the  sum  of  all  the  moments  of  all  the  forces  about 
0  is  zero  if  there  is  equilibrium.  But  any  small  plane  motion 


Fig.  68. 


APPLIED    MECHANICS.  127 

of  the  body  whatsoever  we  know  to  be  resolvable  into  a 
motion  of  translation  and  a  rotation  about  some  axis  o  at 
right  angles  to  the  plane.  Hence  the  law  of  work  tells  us 
that  if  any  system  of  forces  is  in  equilibrium  their  compo- 
nents in  any  direction  balance  one  another  and  theii 
moments  about  any  axis  balance  one  another. 

In  the  numerical  exercise  Art.  94  let  the  respective  forces 
(only  given  in  amount,  clinure,  and  sense  as  yet)  be  at  the 
following  distances  from  a  certain  point,  which  I  shall  call  s. 
The  sign  +  means  that  a  force  tends  to  turn  the  body  against 
the  hands  of  a  watch,  o  x  is  at  the  distance  -f-  5  feet,  o  A 
is  at  —  2  feet,  MO  is  at  —  7  feet,  o  G  is  at  +  3  feet,  n  o 
is  at  —  4  feet.  Let  the  student  now  draw  these  forces  in  their 
proper  positions  relatively  to  s.  The  sum  of  their  moments 
is  50  x  5  -  20  x  2  -  56  x  7  +  100  x  3  —  70  x  4,  or 
—  162  pound-feet.  This  is  the  moment  of  their  resultant 
which  is  31-158  Ibs. ;  so  that  its  distance  from  s  is  —  5-2  feet, 

99.  If  the  cliiiure  and  sense  of  a  force  p  be  given,  it  is  also  neces- 
sary to  give  some  point  through  which  its  direction  passes.  Thus, 
let  all  the  forces  be  in  one  plane ;  let  p  make  an  angle  a  with  the 
horizontal ;  let  the  co-ordinates  of  the  given  point  be  x  and  y 
referred  to  the  axes  of  as  and  y. 

If  p  cos.  a  is  x  and  P  sin.  a  is  Y,  then  as  the  moment  of  p 
about  any  asis  is  equal  to  the  sum  of  the  moments  of  x  and  Y, 
taking  moments  about  the  origin,  the  moment  of  p  is  Y  x  -  x  y. 

If  R  is  the  resultant,  its  components  are  2  x  and  2  Y  ;  call  them 
x  and  Y.  Also,  if  x  and  y  are  the  co-ordinates  of  a  point  in  R, 
WY  —  y  x  =  2  (Y  a;  —  x  y). 

For  equilibrium  we  must  have  2  x  =  0  .  .  (1),  2  Y  =  0  .  .  (%\ 
2(YaJ-xy)  =  0..(3). 

Notice  that  we  may  have  (1)  and  (2)  true  without  (3)  being 
true.  In  this  case  the  system  of  forces  reduces  to  a  mere  couple 
whose  moment  about  any  point  is  2  (Y  x  —  x  y} ;  such  a  system  is 
called  a  torque.  If  we  choose  any  point  in  the  plane,  we  can 
replace  any  system  of  forces  by  a  single  force  through  this  pointj, 
together  with  a  couple  whose  moment  is  the  sum  of  the  moments 
of  the  system  of  forces  about  this  point.  This  is  often  an 
exceedingly  important  fact  to  remember.  (See  Art.  100.)  The 
student  ought  to  work  many  numerical  exercises  graphically  and 
analytically. 

Example  1.— A  beam,  ABODE,  is  supported  at  A  and  E  by 
forces  x  and  y.  The  load  at  B  is  3  tons,  and  A  B  =  4  feet ;  load  at 
c  is  2  tons,  and  A  c  is  7  feet ;  load  at  D  is  2|  tons,  and  A  D  is  9  feet ; 
AEIS  12  feet. 

Here  z  +  y  =  3  +  2  +  2f  =  7  J  tons.  Taking  moments  about 
A,  the  moments  with  the  hands  of  a  watch  are 

3x4  +  2x7  +  2£  x  9  =  48£  ton-feet. 


12.8  APPLIED    MECHANICS. 

The  moment  against  the  hands  of  a  watch  is  y  x  12,  and  CD 
has  no  moment,  because  it  acts  at  A.  Hence  our  second  equation 
is  12  y  =  48|,  y  =  4 '04 17  tons,  and  therefore  x  is  3'4583  tons. 

Example  2. — We  neglected  the  weight  of  the  beam  itself  in 
Example  1.  If  its  centre  of  gravity  is  4  feet  from  A,  and  if  the 
weight  of  the  beam  is  half  a  ton,  and  if  x'  and  y'  are  the  additional 
supporting  forces, 

a/  -f  y'  =  \,          \  x  4  =  /  x  12. 
Hence,  y'  =  $  ton,          x'  =  1  ton. 

Example  3. — In  Fig.  69  o  A  is  part  of  a  beam.  Considering 
only  the  portion  of  beam  to  the 
right  of  the  section  at  o,  let  the 
loads  downward,  PI}  P2,  and  P3, 
and  the  supporting  force  upward, 
Q,  be  given.  Let  the  perpendicular 
distances  from  o  be  alt  «2,  «3,  and 


q.     Find  a  force  through  o,  and 

pi  '    ~T      a  couple    to  Balance   the    given 

Q        forces.     ^Call  the  force  s.     If  it 
acts  downward  at  o,  its  amount 
Fig-  69.  must  be  s  =  Q  —  PJ  —  *P2  —  ps. 

If  the  couple  is  called  M  and  it 
acts  tending  to  turn  the  body  o  A  round  o,  with  the  hands  of  a  watch 

M  =  Q  q  -  P!  «!  —  P2  »a  -  pa  %• 

When  we  come  to  consider  beams  we  shall  call  s  the  shearing  force 
and  M  the  bending  moment  at  the  section  o. 

The  student  will  at  this  point  work  the  Exercises  1,  2,  3,  of  pages 
134-5,  as  well  as  the  following : — 

1.  A  uniform  beam,  20  feet  long  and  supported  at  its  ends,  has  weights 
of  1,3,  2,  and  4  cwts.  placed  at  distances  of  2,  8,  12,  and  15  feet  respect- 
ively from  one  end.     Taking  the  weight  of  the  beam  to  be  5  cwts.,  find 
the  reactions  at  each  of  the  supports.  Ans.,  7  cwts. ;  8  cwts. 

2.  Draw  any  line  o  x,  and  lines  o  p,  o  Q,  o  R,  o  s,  o  T,  making  angles  of 
28°,  62°,  118°,  220°,  and  305°  respectively  with  o  x.     Consider  that  forces 
act  along  these  lines,  their  amounts  being  25,  34,  14, 42,  and  18  Ibs.    Find 
the  amount  and  direction  of  the  force  which  balances  these.     In  doing 
this  first  determine  the  components  of  each  force  in  the  directions  o  x  and 
o  Y,  and  arrange  these  in  columns  as  shown  in  Art.  94. 

Ans.,  15-67  Ibs. ;  232°-2  with  o  x. 

3.  A  trap-door  of  uniform  thickness,  5  feet  long  and  3  feet  wide,  and 
weighing  5  cwts.,  is  held  open  at  angle  of  35°  with  the  horizontal  by 
means  of  a  chain.     One  end  of  the  chain  is  fixed  to  a  hook  placed  4  feet 
vertically  over  the  middle  point  of  the  edge  on  which  the  hinges  are,  the 
other  end  being  fixed  to  the  middle  point  of  the  opposite  edge.     Deter- 
mine the  force  in  the  chain  and  the  force  at  each  hinge. 

Ans.,  2-65  cwts. ;  2'5  cwts. 

4.  A  uniform  beam,  weighing  2  cwts.,  is  suspended  by  means  of  two 
chains  fastened  one  at  each  end  of  the  beam.    When  the  beam  is  at  rest  it 
is  found  that  the  chains  make  angles  of  100°  and  115°  with  the  beam; 
find  the  tensions  in  the  chains.  Ans.,  1  cwt. ;  Tl  cwt. 


APPLIED   MECHANICS.  129 

5.  AB  is  a  horizontal  uniform  bar  1^  feet  long,  and  F  a  point  in  it  10 
inches  from  A.  Suppose  that  A  B  is  a  lever  that  turns  on  a  fulcrum  under 
F,  and  carries  a  weight  of  40  Ibs.  at  B  ;  weight  of  lever,  4  Ihs.  If  it  is  kept 
horizontal  by  a  fixed  pin  above  the  rod,  7  inches  from  F  and  3  inches  from 
A,  find  the  pressure  on  the  fulcrum  and  on  the  fixed  pin 

An*.,  89-14  Ibs.  ;  45-14  Ibs. 

100.  When  the  forces  are  not  in  one  plane,  let  the  force  P  make 
angles  a,  /3,  and  7  with  the  three  axes  of  co-ordinates.  Let  a  point 
in  the  direction  of  P  be  a?,  y,  z.  If  P  cos.  a,  or  x,  P  cos.  #,  or  Y,  and 
p  cos.  «y,  or  z  be  the  three  components  of  P,  we  can  use  x,  Y,  and  z 
instead  of  p  for  all  purposes.  Positive  directions  are  the  direc- 
tions of  increasing  x,  y  or  z.  Thus  the  moment  of  p  about  any 
axis  is  equal  to  the  sum  of  the  moments  of  x,  Y,  and  z  about  the 
axis.  Attention  must  be  paid  to  the  sense  of  each  force,  and 
whether  it  tends  to  turn  the  body  against  or  with  the  hands  of  a 
clock.  The  student  ought  to  spend  time  in  fixing  clearly  in  his 
mind  the  truth  of  the  following  statements  :  —  The  moment  of  p 
about  the  axis  of  x  is  z  y  —  Y  z  ;  the  moment  of  p  about  the  axis 
of  y  is  x  z  —  z  x  ;  the  moment  of  p  about  the  axis  of  z  is  Y  x  —  x  y. 
Hence  we  see  that  if  R  is  the  resultant,  its  components  are  2  x, 
2  Y,  2  z,  and  the  sum  of  their  squares  is  R2,  which  is  therefore 
easily  calculated  ;  also  each  of  them,  divided  by  R,  is  a  direction 
cosine  of  R.  Again,  if  x,  yy  and  z  be  the  co-ordinates  of  a  point 
in  H,  then 

y2z-z2Y=2(zy  -  Y  z), 

2  2  x  -  a?  2  z  =  2  (x  2  -  z  a?), 


The  value  of  R  and  its  clinure  (the  angles  which  it  makes  with  (he 
axes)  having  already  been  found,  these  equations  enable  the  posi- 
tion of  a  point  in  the  resultant  to  be  found  ;  so  that  R  is  completely 
determined. 

For  equilibrium  we  must  have  2x  =  o,  2Y  =  o,  2z  =  o, 
2  (z  y  —  Y  z)  =  o,  2  (x  z  —  z  a?)  =  o,  2  (Y  x  —  x  y)  =  o. 

Given  a  set  of  forces,  it  is  evident  that  we  can  always  sum  them 
into  a  resultant  force  acting  at  any  point  we  please  to  choose, 
together  with  a  couple  about  some  axis.  If  we  are  not  given  the 
point,  it  is  always  possible  to  reduce  any  system  of  forces  to  a 
resultant,  and  a  couple  whose  axis  is  the  resultant  force. 

101.  The  Link  Polygon.  —  We  shall  now  consider  graphical 
methods  of  dealing  with  forces  which  do  not  necessarily  act 
through  one  point.  Take,  for  example,  the  forces  of  1,  2,  3,  4 
(Fig.  70).  Draw  the  unclosed  force  polygon  1',  2',  3',  4'  (Fig. 
71),  with  its  sides  parallel  to  and  proportional  to  the  forces,  and 
the  arrow-heads  circuital.  Now  the  dotted  line  b  a,  with  its 
arrow  non-circuital  with  the  rest,  is  parallel  to  and  proportional 
to  the  resultant  of  all  the  given  forces.  But  this  does  not  tell 
us  where  the  resultant  force  is  situated,  although  it  tells  us  its 
direction  and  amount. 


130 


APPLIED    MECHANICS. 


From  any  point,  o  (Fig.  71),  draw  a  line  to  the  junction  of 
1'  and  2'  (it  is  easier  to  say  draw  the  line  o  1'  2'),  o  2'  3',  etc., 
to  all  the  angles  of  the  force  polygon.  Now  construct  a  new 
unclosed  polygon,  with  its  corners  on  1,  2,  3,  4  (Fig.  70),  and 
its  sides  parallel  to  o  1'  2',  o  2'  3'  etc.  (Fig.  71),  its  last  side 
being  parallel  to  oa,  and  its  first  parallel  to  06.  We  have 
now  found  the  point,  5  (Fig.  70),  where  the  first  and  last 
sides  of  the  link  polygon  meet  The  resultant  of  the  forces 


Fig.  70. 

1,  2,  3,  4,  passes  through  this  point,  5,  and  corresponds  to 
the  closing  side,  ba,  in  direction  and  magnitude.  The  new 
polygon  is  called  the  link  polygon  of  the  forces  relative  to 
the  pole,  0.  The  position  of  A,  the  point  at  which  we  start  to 
draw  the  link  polygon,  may  be  chosen  anywhere  on  1,  and  hence 
there  may  be  any  number  we  please  of  link  polygons  for  a 
given  position  of  the  pole,  o.  Again,  there  are  any  number 
we  please  of  link  polygons  corresponding  to  any  other  posi- 
tion of  o,  and  we  can  choose  o  where  we  please.  Any  student 
who  studies  this  in  the  light  of  what  he  did  in  Art.  96,  will 
see  that  the  link  polygon  really  consists  of  a  system  of  links 
which  would  be  in  equilibrium  under  the  given  set  of  forces 
and  the  force  we  have  found ;  and  since  the  mere  links  only 
introduce  forces  which  are  of  themselves  in  balance,  being 
equal  and  opposite  in  each  link,  the  system  of  forces  acting  at 
the  joints  must  balance. 

Suppose  we  find  that  when  we  are  given  the  forces  1,  2,  3, 
and  4  (Fig.  72),  and  we  draw  the  force  polygon  (Fig.  73),  and 
any  link  polygon  (Fig.  72),  that  these  are  both  closed,  let  us  prove 
that  the  forces  are  in  equilibrium. 

A  system  of  forces  acting  on  a  rigid  body  is  not  affected  by 
introducing  any  number  of  forces  which  separately  balance  one 
another.  Now  let  a  force  represented  by  the  length  of  the  line  o 
1'  2'  act  at  the  point  A  in  the  direction  B  A,  its  sense  being  shown 
by  the  arrow-head  near  A,  and  let  an  equal  force  act  at  B  in  the 


APPLIED   MECHANICS.  131 

direction  AB,  its  sense  being  opposite  to  that  of  the  force  at  A. 
These  two  forces  are  in  equilibrium  with  one  another,  and  they 
cannot  therefore  affect  the  original  system  of  forces  in  any  way. 


Fig.  72.  Fig.  73. 

Similarly,  the  forces  shown  by  the  arrow-heads  in  B  c,  c  D,  D  A  are 
introduced,  every  pair  balancing  one  another. 

Now  we  see  that  the  three  forces  at  the  point  A  are  in  equi- 
librium with  one  another,  because  they  are  parallel  to  and  pro- 
portional in  amount  to  the  sides  of  the  triangle  omn  (Fig.  73), 
and  corresponding  arrow-heads  would  run  right  round  the  triangle. 
Similarly,  there  is  equilibrium  at  every  other  corner  of  the  link 
polygon  A  B  c  D  ;  hence  all  the  forces  are  in  equilibrium,  and  hence 
the  forces,  1,  2,  3,  4,  taken  by  themselves,  must  be  in  equilibrium. 

The  theorems  which  we  wish  students  to  prove  by  construction 
can  be  proved  to  be  generally  true,  reasoning  from  the  fact  that  a 
number  of  forces  acting  at  a  point  can  only  have  one  resultant. 

102.  We  see,  then,  that  the  force  polygon  alone  is  sufficient 
to  find  the  resultant  of  any  number  of  forces  if  the  forces  meet  at 
a  point,  but  we  need  also  the  link  polygon  if  the  forces  do  not 
meet  at  a  point. 

The  link  polygon  really  shows  that  the  sum  of  the  turning 
moments  of  the  forces  1,  2,  3,  4  (Fig.  70)  about  any  point  is  equal 
to  the  moment  of  the  resultant  about  the  same  point.  The  force 
polygon  pays  no  regard  to  turning  moments  of  forces ;  it  merely 
tells  us  about  the  resultant  of  the  forces,  supposing  that  they  ail 
pass)  d  through  the  same  point. 

103.  You  ought  by  actual  drawing  to  illustrate  the  truth 
of  the  following  four  ways  of  putting  one  statement.     If  you  use 
coloured  inks  your  drawings  will  be  more  instructive. 

1st.  The  resultant  of  any  number  of  forces  is  independent 
of  the  order  in  which  we  draw  them  in  the  force  polygon,  and 
draw  between  them  the  sides  of  the  link  polygon. 

2nd.  In  adding  forces  we  may  first  find*  the  resultant  of 
some  of  the  forces,  and  then  add  together  this  pesultant  and  all 
the  other  forces.  The  answer  will  always  be  the  same,  however 
we  may  group  the  forces  before  adding,  them. 


132  APPLIED   MECHANICS. 

3rd.  If  the  force  polygon  of  a  number  of  forces  is  closed, 
and  if  we  can  draw  a  closed  link  polygon,  then  all  the  link 
polygons  we  may  draw  will  also  be  closed. 

4th.  If  any  other  pole  be  taken  in  Fig.  71,  and  another  link 
polygon  be  drawn  and  a  new  point  5  (Fig.  70)  is  found,  both  of 
the  points  so  found  lie  in" a  straight  line  parallel  to  b  a  of  Fig.  71. 

You  will  also  find,  and  it  is  easy  to  prove,  that  the  locus  of  the 
point  in  which  any  two  sides  of  the  link  polygon  meet  is  parallel 
to  the  line  which  closes  the  corresponding  portion  of  the  force 
]>olygon.  Again,  if  b  is  taken  as  pole  instead  of  o,  the  last  side  of 
the  link  polygon  is  found  to  be  in  the  direction  of  the  resultant  of 
the  forces  1,  2,  3,  4;  and,  generally,  any  side  of  the  link  polygon 
is  in  the  direction  of  the  resultant  of  the  corresponding  number  of 
the  given  forces.  Thus,  if  b  is  taken  as  pole,  4,  5  becomes  the 
resultant  of  the  forces  1,  2,  3,  4,  and  34  becomes  the  resultant  of 
the  forces  1,  2,  3.  It  is  evident  from  this  that  the  direction  of  the 
resultant  of  any  two  forces,  or  of  any  number  of  forces,  which  meet 
at  a  point  passes  through  their  point  of  intersection. 

A  system  of  forces  may  not  reduce  to  a  resultant  force,  but  be 
equivalent  to  a  couple.  When  this  is  the  case  the  force  polygon  is 
closed,  and  the  first  and  last  sides  of  any  link  polygon  that  may  be 
drawn  are  parallel  to  one  another.  You  may  also  find  it  worth 
your  while  to  prove  by  construction  this  statement :  if  two  link 
polygons  are  drawn  for  two  positions  of  the  pole  o,  the  correspond- 
ing sides  of  the  two  polygons  meet  in  points  which  lie  in  a 
straight  line  parallel  to  the  line  joining  the  two  positions  of  the 
pole  o. 

If  you  have  been  able  to  make  a  few  drawings  such  as  I  have 
been  speaking  about,  and  so  take  an  interest  in  this  easy  and 
instructive  method  of  working  mechanical  exercises,  you  ought  to 
work  by  means  of  it  some  such  exercises  as  the  following : — 

104.  1.  Exercises. — In  Exercise  2  of  page  128,  draw  the  force 
polygon,  taking  the  forces  in  the  order  OP,  OR,  o  Q,  o T,  o  s,  and 
observe  that  the  resultant  and  equilibrant  are  the  same  as  beiore.  Obtain 
also  the  resultant  of  o  p,  o  R,  o  T,  and  the  resultant  of  o  a,  o  s,  and  show 
that  these  have  a  resultant  equal  to  the  resultant  of  the  five  forces. 

2.  Draw  a  line  H  K  47  inches  long.  On  H  K  take  points  A,  B,  c,  distant 
from  H  0-4  inch,  1-5  inch,  and  3-3  inches  respectively.  Now  draw  A  p,  B  Q, 
c  R  inclined  at  angles  of  72°,  57°,  and  37°  with  H  K.  Suppose  that  a  force 
of  214  Ibs.  acts  from  K  to  H,  one  of  576  Ibs.  from  R  to  c,  one  of  132  Ibs.  from 
Q  to  B,  and  one  of  237  Ibs.  from  P  to  A.  (1)  Draw  a  force  polygon  to  deter- 
mine the  amount,  clinure,  and  sense  of  the  resultant.  (2)  Take  any  pole 
o  and  draw  a  link  polygon  to  determine  the  lateral  position  of  the  result- 
ant. In  giving  your  answer  say  what  are  the  perpendicular  distances  of 
H  and  K  from  the  resultant.  (3)  Draw  a  new  force  polygon,  taking  the 
forces  in  a  different  order,  and  observe  that  the  resultant  is  the  same  as 
before  as  regards  amount,  clinure,  and  sense.  With  respect  to  a  new 
pole  o,  now  draw  a  second  link  polygon,  and  observe  that  the  lateral 
position  of  the  resultant  agrees  with  that  obtained  before.  (4)  Take  the 
first  force  polygon  and  choose  a  new  pole,  and  with  respect  to  it  draw  a 


APPLIED    MECHANICS.  133 

new  link  polygon.  Observe  that  this  gives  rise  to  the  same  resultant. 
That  is,  the  closing  sides  of  each  link  polygon  meet  at  points,  all  of  which 
should  lie  on  the  same  straight  line.  (5)  Calculate  the  above  answers 
according  to  the  rules  of  Art.  99. 

Ans.,  (1)  1,066  Ibs. ;  39 '7°  with  HK;  sense  same  as  force  RC;  (2) 
1-312  inches;  1'7  inches. 

105.  Example. — Given  a  set  of  forces,  find  two  forces,  one 
given  in  position,  and  clinure  the  other  to  pass  through  a  given 
point  P,  such  that  these  two  forces  will  balance  the  given  set. 
The  ingenious  student  will  find  out  how  to  do  this  himself; 
other  students  will  benefit  by  the  following  instructions. 

Draw  the  force  polygon  with  one  corner  unknown ;  choose 
a  pole  and  begin  drawing  a  link  polygon  from  the  point  P  ;  the 
side  closing  the  link  polygon  enables  the  missing  corner  of  the 
force  polygon  to  be  found.  This  problem  is  one  of  the  very 
commonest  to  come  before  the  engineering  student.  Thus,  let 
any  structure  (a  roof  principal  or  a  railway  girder,  for  example) 
have  given  loads.  Let  it  be  supported  at  a  hinge  joint  at  P, 
and  upon  rollers  at  Q,  to  allow  for  expansion.  The  direction 
of  the  supporting  force  at  Q  is  known,  and  one  point,  P,  in  tho 
other  supporting  force  is  known. 

106.  A  student  who  sees  the  essential  ideas  underlying  our 
methods  of   working  will  have  pleasure  in  working  curious 
problems,    such  as  the  following  exercise : — Given   a  set  of 
forces  and  given  three  points,  A,  B,  and  c.     Draw  a  link  polygon 
with  three  of  its  sides  passing  respectively  through  A,  B  and  c. 

Hint. — You  must  first  find  the  resultant  of  the  given  forces,  and 
observe  where  its  line  of  action  meets  A  B  in  I ;  i  B  is  the  line 
of  action  of  the  balancing  force  through  B.  Three  forces  now 
act  at  I.  Their  lines  of  action  are  known,  and  the  magnitude 
of  one  force ;  hence  the  amounts  and  senses  of  the  other  two  can 
be  found. 

107.  Exercise.—  Draw  a  rectangle  ABCD,  A  B  =  5  inches,  B  c  =  1-8 
inch.  From  A,  along  A  B,  measure  off  lengths,  A  E,  A  P,  A  o=  1'75,  2*8,  5*7 
inches  respectively.  From  D  along  D  c  measure  off  D  H,  D  K,  D  L  =  1'4,  2-4, 
2-85  inches  respectively.  Suppose  that  forces  of  the  following  amounts  act 
on  a  rigid  body— viz. ,  1,460  Ibs.  from  A  to  H,  1,085  Ibs.  from  E  to  K,  808  Ibs. 
from  F  to  L.  These  are  balanced  by  two  others,  one  of  which  acts  through 
x,  a  point  in  D  A  distance  0'6  inch  from  D,  and  the  other  has  for  its  line  of 
action  Y  Y,  the  line  passing  through  c  and  o.  Find  the  magnitudes  of  the 
balancing  forces  and  the  angle  between  them.  (The  rectangle  is  intro- 
duced merely  as  a  convenient  way  of  settling,  without  a  figure,  the  given 
forces.)  Scale  i  inch  to  100  Ibs.  Ans.,  2,550  IBs. ;  1,140  Ibs. ;  63°. 

108.  Example. — Given  a  set  of  forces,  A.,  B,  c,  D ;  given  also 
three  lines,  x,  Y,  z,  as  the  positions  of  three  forces  which  are  to 
balance  our  given  set.  Find  these  three.  The  method  here 


134 


APPLIED    MECHANICS. 


is  not  so  obvious  as  the  method  of  the  last  example.  Draw 
A',  B',  c',  D',  known  sides  of  the  force  polygon.  Draw,  also, 
after  D',  x'  parallel  to  x  unlimited  in  length.  In  it  choose  o 
the  pole  and  join  to  all  known  corners  of  the  force  polygon. 
Note  that  o  x'  is  itself  two  radiating  lines  from  o,  because  there 
are  two  corners  of  the  force  polygon  in  it.  Now  let  the  point 
where  D  and  x  meet  be  called  a  side  of  the  link  polygon ;  the 
intercept  on  x  till  it  meets  Y  is  another  side,  and  it  will  now 
be  found  that  we  have  sufficient  information  for  the  completion 
of  both  force  and  link  polygons.  As  an  example,  Fig.  74 
represents  a  ladder  whose  centre  of  gravity  is  at  G,  and  weight 
300  Ibs.  A  string  fastened  to  it  at  c  in  the  direction  c  o  keeps 
it  in  equilibrium,  its  end  A  resting  on  the  smooth  wall  o  A,  and 
its  end  B  on  the  smooth  floor  o  B.  Find  the  pull  in  the  string 
and  the  reactions  at  A  and  B. 

The  forces  acting  on  the  ladder  are  shown  by  the  arrows. 
Draw  Y  x  (Fig.  75)  vertically  to  represent  the  weight  of  the 
ladder.  Draw  w  Y  horizontally  of  unknown  length.  Draw 


Fig.  74. 


Fig.  75. 


x  o  parallel  to  0  C,  and  take  o  anywhere  in  this  line.  Use  o  as 
pole  of  the  force  polygon.  Join  o  Y.  Now  the  link  polygon 
is  M  N  P  M,  and  drawing  o  w  parallel  to  P  M,  and  w  z  parallel  to 
B  M,  we  find  that  Y  x  z  w  is  the  force  polygon.  The  lengths  of 
x  z,  z  w,  w  Y  represent  the  forces  at  c,  at  B,  and  at  A. 

The  student  will  find  it  very  instructive  now  to  introduce 
friction  into  this  problem,  and  thus  create  two  new  problems. 
(1)  If  the  pull  in  the  cord  is  just  sufficient  to  prevent  the 
ladder  from  moving.  Here  the  angle  OEM  ought  to  be 
90°  —  0j  where  tan.  fa  is  the  coefficient  of  friction  at  B,  and 
GAP  ought  to  be  90°  +  <j>2  if  tan.  <p2  is  the  coefficient  of  friction 
at  A.  (2)  If  the  pull  in  the  cord  is  just  sufficient  to  produce 
motion.  In  this  case,  o  B  M  =  90°  +  ^  o  A  P  =  90°  —  ^2. 


APPLIED  MECHANICS.  135 

Students  ought  at  this  stage  to  work  a  great  number  of 
exercises. 

EXERCISES. 

1.  InFig.  74,AB  =  20feet,  OB  =  10feet,BC  =  2feet.     The  weight  of  the 
ladder  is  300  Ibs.     Find  the  pull  in  c  o  and  the  reactions  at  B  and  A.     If  p  = 
0-1,  find  the  pull  in  the  rope ;  first,  if  B  is  made  to  approach  o ;  second,  if  B  is 
about  to  recede  from  o.     Ans.,  99-2  Ibs. ;  319  Ibs. ;  98  Ibs. ;  145  Ibs. ;  58*8  Ibs. 

2.  A  horizontal  beam  A  P  has  loads,  at  B  of  1,000  Ibs.,  at  c  of  250  Ibs.,  at 
D  of  1,200  Ibs.,  at  E  of  600  Ibs.     If  A  B  =  5  feet,  A  c  =  9  feet,  A  D=15  feet, 
A  n  =  l  8  feet,   AF  =  24  feet,  find  the  supporting  forces  analytically   and 
graphically.  Ans.j  1,548  Ibs.;  1,502  Ibs.  nearly. 

3.  Draw  a  rectangle  A  B  c  D,  A  B  =  5'2  inches,  and  B  c  =  2-2  inches.     On 
A  K  take  points  E,  F,  o,  H,  K,  L  distant  from  A,  0'9,  1'55,  2'7,  3-25,  4*25, 
and  4-75  inches  respectively.     On  BC  take  BM  =  1-25  inch.     On  CD  take 
N,  o,  P,  Q,  distant  from  c,  1-5,  2-85,  3*7,  and  4  inches  respectively.     On 
D  A  take  D  R  =  1  inch.     Now,  three  forces  act  on  a  rigid  body— one  of 
0-615  ton  from  F  to  o,  one  of  0'536  ton  from  H  to  P,  one  of  0'423  ton  from 
L  to  N.     These  three  forces  are  balanced  by  three  others,  whose  lines  of 
action  are,  x  x  drawn  through  11  and  o,  Y  Y  drawn  through  E  and  Q,  and 
z  z  drawn  through  M  and  K.     Find  the  magnitudes  of  these  balancing 
forces.     Scale  &  inch  to  O'l  ton.     Ans.t  0-930  ton ;  0-375  ton ;  0-815  ton. 

109.  The  distance  x  of  the  centre  of  mass  (usually  called  cen- 
tre of  gravity,  but  only  few  bodies  have  true  centres  of  gravity) 
of  a  body  or  system  of  bodies  from  a  plane  is  obtained  by 
multiplying  each  little  portion  of  mass  m  by  the  distance  x 
of  its  centre  from  the  plane,  adding  together,  and  dividing  by 
the  whole  mass.  The  symbol  for  this  is  x  .  M  =  %  m .  x  where 
M  stands  for  %  m.  Practically  the  engineer  often  finds  the 
centre  of  gravity  by  an  experimental  method. 

The  distance  x  of  the  centre  of  an  area  (usually  called  the 
centre  of  gravity  of  the  area)  from  a  plane  (or  more  usually  of 
a  plane  area  from  a  line  in  its  plane)  is  obtained  by  multiplying 
each  little  portion  of  the  area  a  by  the  distance  x  of  its  centre 
from  the  plane  (or  line),  adding  together,  and  dividing  by  the 
whole  area.  The  symbolic  way  of  representing  this  is  x .  A  = 
3  ax,  where  A  stands  for  the  whole  area. 

If  the  centres  of  the  masses  mv  m2,  w3,  etc.,  are  at  the  dis- 
tances ajj,  a?2,  a?3,  etc.,  from  any  plane,  let  the  sum  of  the  masses 
m1  +  m^  +  etc.  be  called  M,  and  let  x  =  (m^  xl  +  m?  x^  +  etc.)  -r-  M, 
or,  as  we  prefer  to  write  it,  x  =  2  (ma;)  -f-  M.  Similarly,  taking 
distances  from  two  other  planes  at  right  angles  to  the  first,  let 
y  =  2  (m  y]  -r-  M,  z  =  S  (m  z)  +  M.  Let  a?,  y,  z  be  the  co-ordinatea 
of  a  point.  If  WT,  w2,  %,  etc.,  be  the  distances  of  the  masses  from 
any  other  plane,  at  a  distance  a  from  the  origin,  perpendicular  to 
the  direction  (J,  m,  »),  it  is  easy  to  see  that 


136  APPLIED   MECHANICS. 

and  if  u  is  the  distance  of  the  point  already  found  from  the  nexv 
plane,  u  =  Ix  -\-my-\-nz  —  a;  and  using  the  above  values  for 
a?,  y,  2,  and  re-arranging  terms,  we  find  that  u  =  S  (m  u)  -~  M. 
We  see,  then,  that  the  above-mentioned  point  will  be  in  the  same 
position  whatever  be  the  planes  of  reference.  We  call  it  the  centre 
of  mass  or  centre  of  inertia;  and  when  we  are  dealing  with  a 
system  so  small  that  the  forces  of  attraction  upon  it  due  to  gravity 
may  be  regarded  as  parallel  to  one  another,  we  may  call  it  the 
centre  of  gravity.  In  the  same  way,  if  the  centres  of  the  areas 
av  a2,  etc.,  all  in  the  same  plane,  are  at  the  distances  xv  x2,  etc., 
from  a  line  in  the  plane,  and  if  A  is  «j  +  a2  +  etc.,  the  whole  area, 
and  if  x  is  the  distance  of  the  common  centre  of  area  (often  called 
the  centre  of  gravity  of  the  area)  from  the  line,  then 

etc.)  -r-  A. 


110.  We  can  use  a  graphical  statics  method  of  finding  the 
centre  of  mass,  or  area  G.     Thus,  let  there  be  masses  or  areas, 
Wj,  w2,  etc.,  whose  centres  are  at  the  points  1,  2,  etc.  (Fig.  76). 
Draw  parallel  forces  11,  22,  etc.,  in  any  direction  proportional 
to  ml5  m2,  etc.,  and  find  the  resultant  by  the  above  method. 
Suppose   M   N   to  be  the  direction  of   the   resultant.       Now 
repeat   the   process,  taking   the  parallel   forces   of   the  same 
magnitudes   as   before,  but   in  a  different  direction,  and   let 

RIP  be  the  direction  of  the  resultant.  Evidently 
M,  where  these  lines  meet,  is  the  centre  of 
gravity  of  the  masses  or  areas.  This  method 
may  often  prove  useful,  for  areas  especially. 
Thus,  to  find  the  centre  of  gravity  of  any 
given  area,  divide  it  into  any  suitable  number 
of  parts,  so  that  the  centre  of  gravity  and  area 
of  each  part  may  be  found  easily.  If  we 
divide  the  area  by  parallel  lines,  these  lines 
Fig.  76.  maJ  be  drawn  equidistant,  and  the  area  of 

each  part  is  approximately  given  by  the  length 
of  the  line  which  separates  it  from  either  of  its  neighbours. 
A  repetition  of  the  process  has  been  employed  to  determine 
the  moment  of  inertia  of  the  area  about  any  given  line. 

111.  I  do  not  advise  students  to  adopt  this  link  polygon 
method  of  finding  centres  of  gravity  or  of  calculating  moment 
of  inertia.     A  practical  engineer  will  always  apply  the  ordinary 
formula  to  find  the  centre  of  gravity  of  an  area.     Thus,  if 
you  want  the  centre  of  gravity  of  the  figure  M  N  o  P  (Fig.   77), 
draw  two  parallel  lines,  G  H,  K  o,  touching  the  figure  at  two 
opposite  sides.     Draw  a  line,  K  G,  at  right  angles  to  G  IT,  and 


APPLIED    MECHANICS. 


137 


divide  it  into  any  number  of  parts,  each  equal  to  d.  Draw 
the  lines  A  B,  c  D,  etc.,  parallel  to  G  H,  so  that  they  are  at  the 
distance  d  apart,  the  distance  from  A  B  to  G  H,  or  from  Y  z  to 
K  o  being  |  d.  It  is  evident  that  if 
N  x  P  is  a  line  parallel  to  G  H  through 
the  centre  of  gravity,  then  approxi- 
mately 

.  .    AB  +  3cD  +  5EF  +  etc. 
a  x  =  \  d  . 


-7- 


\ 


r/- 


Pig.  77. 


AB+CD  +  EF  +  etC. 

We  have  thus  obtained  one  line 
through  the  centre  of  gravity,  and  in 
a  similar  way  we  may  obtain  another 
such  line,  and  their  point  of  intersec- 
tion is  the  centre  of  gravity  required. 

Sometimes  we  choose  as  our  line 
of  reference,  a  line,  G  H,  which  cuts 
through  the  area ;  in  this  case  distances  on  one  side  of  the  line 
are  to  be  considered  negative ;  and  if  G  H  happens  to  pass 
through  the  centre  of  gravity,  of  course  the  sum  %  ax  is  Q. 

We  often  cut  the  shape  of  an  irregular  area  from  sheet 
zinc  and  balance  it  in  two  positions  on  a  straight  edge  to  find 
the  centre  of  gravity  of  the  area. 

Exercises.— I.  Masses  whose  centres  are  in  a  straight  line  at  A,  B,  c,  D 
are  4  Ibs.,  8  Ibs.,  7  Ibs.,  6  Ibs. ;  where  is  the  common  centre  of  mass  if 
A  B  =  0-5  feet,  A  c  =  2  feet,  and  A  D  =  2£  feet  ?  Ans.,  A  G  =  1-32  ft. 

2.  A  disc  8  inches  diameter,  2  inches  thick,  with  a  hole  4  inches 

« 5".. .... 


2.85' 


3. 71" 


Pig.  78. 


<- 5.64". 


<- 8'    -> 

diameter ;  centre  of  hole,  1  inch  from  centre  of  disc ;  where  is  the  centre 
of  mass  ?  Ans. ,  $  inch  from  centre. 

3.  ABC  is  an  equilateral  triangle,  each  side  being  3  inches  long; 
particles  whose  masses  are  1,  2,  3  are  placed  at  A,  B^  c  respectively ;  find 
their  centre  of  gravity  by  construction,  and  note  its  distance  from  A. 

Ant..  2-18  inches. 

F* 


138  APPLIED   MECHANICS. 

1.  ABC  D  is  a  square  lamina  of  uniform  density;  E,  F  are  the  middle 
points  of  A  B  and  B  c.  If  the  corner  of  the  square  is  turned  down  along 
the  line  E  F,  so  that  B  comes  on  to  the  diagonal  A  c,  find  the  centre  olf 
gravity  of  the  lamina  under  the  new  circumstances. 

Ans.,  ^g  of  the  diagonal  from  the  centre. 

6.  A  circular  disc,  8  inches  in  diameter,  has  a  hole  2  inches  in 
diameter  punched  out  of  it,  the  centre  of  the  hole  being  3  inches  from  the 
circumference  of  the  disc.  Find  the  centre  of  gravity  of  the  remaining 
portion.  Ans.,  0*0667  inch  from  centre  of  larger  circle. 

6.  A  thin  plate  of  metal  is  in  the  shape  of  a  square  and  equilateral 
triangle,  having  one  side  common;  the  side  of  the  square  is  12  inches 
long.    Find  the  centre  of  gravity  of  the  plate. 

Ans.,  2'86  inches  from  centre  of  square. 

7.  Find  the  centres  of  area  of  the  areas  in  Fig.  78. 

112.  To  find  the  moment  of  inertia,  i,  of  a  great  number  of  littlo 
masses  about  an  axis,  multiply  each  mass  by  the  square  of  its 
distance  from  the  axis,  and  add  up.     If  the  whole  mass  is  M,  we 
often  write  M  K2  ==  i ;    and  when  i  and  M  are  known  we  can 
calculate  k,  which  we  call  the  radius  of  gyration  of  the  mass  about 
the  axis.     To  find  what  is  called  the  moment  of  inertia,  i,  of  a  great 
number  of  little  areas  about  a  line  in  their  plane,  multiply  each  by 
the  square  of  its  distance  from  the  line,  and  add  up.     If  A  is  the 
whole  area,  and  A  k2  =  i,  we  call  k  the  radius  of  gyration  of  the 
area  about  the  line. 

113.  Just  as  we  found  centres  of  gravity,  so  we  may  obtain  the 
moment  of  inertia,  i,  of  any  area  about  any  line;  or,  as  is  often 
the  case,  suppose  we  wish  to  find  the  moment  of  inertia  of  M  N  o  p 
(Fig.  77)  about  NXP,  a  line  which  passes  through  the  centre  of 
gravity.  Evidently  the  moment  of  inertia  about  G  H  is  approximately 
i  =  ^3(AB_|_9CD_|_25EF  +  etc.)/4. 

Students  ought  to  practise  this  method  first  upon  a  rectangle 
and  a  circle  whose  moments  of  inertia  have  been  worked  out  for 
them  by  the  calculus. 

Now,  it  is  well  known  that  the  moment  of  inertia  of  an  area 
ttbout  any  line  is  equal  to  its  moment  of  inertia  about  a  parallel  line 
through  its  centre  of  gravity,  together  With  the  product  of  the  area 
into  the  square  of  the  distance  between  the  two  lines.  Hence,  the 
moment  of  inertia  about  Nxpisi0  =  i  —  ox2.  rf  (A  B  +  CD  +  EP 
+  etc.). 

It  will  be  found  in  practice  that  this  easy  way  of  carrying  out 
simple  ideas  is  better  than  the  compli- 
cated use  of  the  link  polygon  method 
for  finding  moment  of  inertia.  If  the 
area  may  be  divided  into  rectangles 


D  —  !  whose  sides  are  parallel  to  and  per- 

pendicular to  the  axis,  we  need  not 
subdivide  these  rectangles.  It  must  be 
remembered  that  the  moment  of  inertia 

i of  any  given  area  such  as  a  rectangle 

^  about  any  axis  is  equal  to  the  area 
Fig-  79-  multiplied  by  the  square  of  the  dis- 

tance of  its  centre  of  gravity  from  the 


APPLIED    MECHANICS.  139 

axis,  plus  the  moment  of  inertia  of  the  area  about  a  parallel  line 
through  its  centre  of  gravity.  Thus  the  rectangle  A  B  c  D 
(Fig.  79)  is  known  to  have,  about  N  M  N,  the  moment  of  inertia 

A.  B      B  C^ 

—  so  that  the  moment  of  inertia  of  the  rectangle  about 
It 

ofoo'ia 

A  B  .  B  C3  /B  C2  A 

— ,jy—  +  AB  .  B  C  .  M  O3,  Or  A  B  .  B  C  (  -^  +  M  oM 

It  is  mainly  by  the  use  of  this  rule  that  we  have  found  the 
moments  of  inertia  of  the  various  sections  shown  in  Table  VI.,  and 
all  these  ought  to  be  worked  out  as  exercises  by  students. 

The  student  will  find  it  good  exercise  to  take  a  few  sections  of 
angle-iron,  T-iron,  rails,  and  other  specimens  of  rolled  iron,  and 
find  by  the  above  graphical  method  the  position  of  centre  of 
gravity  of  each  section,  and  the  moment  of  inertia  of  each  area 
about  any  line  through  the  centre  of  gravity.  The  exact  forms 
ought  to  be  taken  from  real  specimens.  If  the  area  is  symmetrical, 
one  line  through  the  centre  of  gravity  can  always  be  found  by 
mere  inspection. 

In  using  this  or  any  other  graphical  method,  it  is  well  to  know 
what  is  the  error  involved  in  having  each  strip  of  width  rf, 
instead  of  being  infinitely  narrower.  We  ought  to  add  to  the  sum 
in  (2)  the  moment  of 'every  strip  about  its  own  central  line — that 
is,  d3  (A  B  +  c  D  +  E  F,  etc.)^  or  A  tf2/12,  if  A  represents  the  total 
area.  It  is  easy  to  see  that  in  a  rectangle  of  depth  D,  if  we  divide 
into  strips  of  breadth  d,  the  fractional  error  is  ^/D3. 

114.  When  the  moments  of  inertia  of  an  area  about  any  three  axes 
through  a  point  are  known,  the  moment  of  inertia  about  any  other 
axis  through  the  same  point  may  be  found ;  because  if  a  distance 
be  measured  from  the  point  along  an  axis  which  is  equal  to  the 
reciprocal  of  the  radius  of  gyration  of  the  area  about  the  axis,  the 
extremities  of  all  such  measured  distances  lie  in  an  ellipse.  The 
principal  axes  of  the  area  are  in  the  directions  of  the  major  and 
minor  axes  of  this  ellipse.  Thus,  if  for  any  area,  M  N  p  B,  (Fig.  80), 

the  least  moment  of  inertia  is  about  an  axis,  o  A,  and  is  — -»  and  if 

O  A 

the  greatest  moment  of  inertia  is  about  OB,  and  is  -^-s,  then 

O  B 

A  B  A7  B'  being  an  ellipse  whose  major  and  minor  axes  are  A  A'  and 
B  B',  the  moment  of  inertia  about  an  axis,  o  c,  is  -^. 

115.  To  know  the  principal  moments  of  inertia  of  an  area  is 
important  for  many  purposes.  It  is  specially  important  in  regard  to 
struts.  A  strut  will  bend  in  such  a  way,  that  the  axis  through  the 
centre  of  gravity  of  a  section,  at  right  angles  to  the  plane  of  bend- 
ing, is  the  axis  about  which  there  is  least  moment  of  inertia  of  the 
section.  The  ellipse  lets  us  see  the  moments  about  all  axes.  To  draw 
it  for  any  particular  section,  if  the  section  is  symmetrical,  as  in  sections 
of  Fig.  80,  we  know  that  the  axis  of  symmetry  and  the  axis  at  right 
angles  to  it  are  the  principal  axes  of  inertia.  II  the  section  is  not 


140 


APPLIED   MECHANICS. 


symmetrical,  as  in  the  case  of  an  angle-iron,  we  find  the  moment 
of  inertia  about  three  axes  as  OP,  o  Q,  o  R  of  Fig.  81,  and  we  set 
A  off  the  distances  o  p,  o  Q,  o  R  to 

represent  the  reciprocals  of  the  radii 
of  gyration.  We  now  have  the 
graphical  problem:  given  three 
points  P  Q,  R  of  an  ellipse,  and  its 
centre  o  to  draw  the  ellipse.  Mr. 
Harrison  thinks  the  following  solu- 
tion better  than  any  other. 

With  centre  o,  radius  o  ci,  de- 
scribe a  circle,  and  through  H, 
where  P  R  cuts  o  Q,  draw  the  chord 
pr  such  that  pn  :  nr  =  p  H  :  HR. 
Draw  the  radius  05  perpendicular 
to  o  Q  ;  through  *  draw  s  M,  *  N 


Pig.  80. 


parallel  respectively  to  p  o,  or;  and/  through  N  and  M  draw  lines  parallel 
to  R  o,  o  p,  intersecting  in  s ;  then  o  s  is  conjugate  to  o  Q. 


Fig.  81. 


For  proof,  suppose  the  ellipse  orthogonally  projected  into  a 
circle,  and  let  the  smaller  figure  be  similar  to  this  projection. 
This  figure  can  be  drawn,  remembering  that  parallel  lines  project 
into  parallel  lines  the  mutual  ratios  of  whose  lengths  remain 


APPLIED    MECHANICS. 


141 


unaltered ;  also  that  conjugate  diameters  project  into  perpendicular 
diameters.  The  solution  given  consists  in  drawing  this  figure 
with  oq  coinciding  with  o  a,  and  then  locating  s. 

To  determine  the  principal  axes,  through  s  draw  a  line  (not 
shown)  perpendicular  to  o  Q,  and  on  it  take  two  points  D,  E, 
opposite  ways  from  s,  such  that  SD  =  SE  =  OQ.  Then  the  axes 
•of  the  ellipse  are  respectively  equal  to  the  sum  and  difference  of 
o  D  and  o  s,  and  the  major  axis  bisects  the  angle  D  o  E.*  • 

MOMENT  OF  INEKTIA  OF  A  RECTANGLE. 

116.  The  moment  of  inertia  of  a  rectangle  about  the  line  o  o 
through  its  centre,  parallel  to  one  side. — Let  A  B 
=  b,  xc  =  d.  Consider  the  strip  of  area  between 
o  P  =  y  and  o  Q  =  y  +  8y.  Its  area  is  b  .  8y, 
and  its  moment  of  inertia  about  o  o  is  b  .  y*-  .  8y ; 
so  that  the  moment  of  inertia  of  the  whole 
rectangle  is 


B 


JH  r*d 

y*.dy  or  b\ 
-**  \--\d 


bd3. 


Fig.  82. 


The  moment  of  inertia  of  the  area  ABC  about 

an  axis  o  at  right  angles  to  the  area  is  equal  to 

the  sum  of  the  moments  of  inertia  ix  and  iy,  about 

o  x  and  o  Y,  axes  at  right  angles  to  one  another 

in  the  area.     For  if  a  is  a  portion  of  area  at  p,  ix 

is  the  sum  of  all  such  terms  as  a .  P  R2,  iy  is  the  sum  of  all  such  terms 

as  a .  p  a2 ;  and  the  sum  of  each  such  pair  of  terms  is  a  term  a .  o  p2. 
117.  Moment  of  inertia  of  a  circle  about  its  centre. — Consider 

the  ring  of  area  between  the 
ladii  r  and  r  -\-  Sr.  Its 
area  is  2,trr .  Sr  more  and 
more  nearly  as  Sr  is  made 
smaller  and  smaller,  and  its 
moment  of  inertia  is  2  trrs  .Sr. 
The  integral  of  this  is 
|irR4  for  a  circle  of  radius 
R.  The  square  of  the  radius 
of  gyration  is  ^R2.  Now, 
in  this  case  ix  =  iy,  each 
being  half  of  ^TrR4 ;  so  that 
the  moment  of  inertia  of  a 
circle  about  its  diameter 
is  jTTR4,  or  ^jTTD4  if  D  is  the 
diameter. 

118.  The  student  ought 

to  be  able  to  prove  the  propositions  referred  to -in  Art.  112  : — 
1.  As  to  mass  or  inertia. — To  prove  that  the  moment  of  inertia 

about  any  axis  is  equal  to  the  moment  of  inertia  about  a  parallel 

axis  through  the  centre  of  gravity  together  with  the  whole  mass 

multiplied  by  the  square  of  the  distance  between  the  two  axes. 

Thus,  let  the  plane  of  the  paper  be  at  right  angles  to  the  axes. 

*  See  Appendix. 


Fig.  88. 


142 


APPLIED   MECHANICS. 


Let  there  be  a  little  mass  m  at  P  in  the  plane  of  the  paper.  Let  o 
be  the  axis  through  the  centre  of  gravity  of  the  whole  mass,  and 
o7  be  the  other  axis.  We  want  the  sum  of  all  such  terms  as 
m .  (o'  r)2. 

Now,  (o'  p)2  =  (o'  o)2  +  o  P2  +  2  .  o  o' .  o  Q,  where  Q  is  the 
foot  of  a  perpendicular  from  p  upon  o  o7,  the  plane  containing 
the  two  axes.  Then,  calling  2  m .  (o'  v)2  by  the  name  i,  calling 
2  m .  o  p2  by  the  name  i0,  the  moment  of  inertia  about  the  axis  o 
through  the  centre  of  gravity  of  the  whole  mass,  then 

i  =  (o'  o)2  2  m  +  I0  +  2  .  o  o' .  2  m  .  o  Q. 

But  2  m  .  o  Q  means  that  each  portion  of  mass  m  is  multiplied  by 
its  distance  from  a  plane  at  right  angles  to  the  paper  through  the 
centre  of  gravity,  and  this  is  zero.  So  that 
the  proposition  is  proved.  Or,  letting  2  m  be 
called  M,  the  whole  mass, 

I  =  I0  +  M.(0'0)2. 

2.  To  prove  the  proposition  about  areas. 
Let  o  o  (Fig.  85)  be  the  axis  through  the 
centre  of  gravity,  and  o'  o'  a  parallel  axis. 
We  want  i,  the  sum  of  all  such  terms  as  a 
(o'  r)2,  and  this  is  the  same  as  2  a  .  o  p2  + 
5  2a  .  o  P  .  o  o'-f  2a  .  o  o'2.  But 
Fig.  84. 


and  this  is  0  from  our  definition  of  centre  of  gravity ;  so  that 
i  =  I0  +  o  o'  2  a. 

EXERCISES. 

1.  A  fly-wheel  has  a  rim  of  rectangular  section,  the  outside  and  inside 
radii  being  8  feet  and  7  feet.     What  is  the  error  in  assuming  the  radius 
of  gyration  to  be  7" 5  feet? 

Ans.  The  true  radius  of  gyration 
is  7  '5 16  feet,  and  hence  the  assumed 
radius  of  gyration  is  '21  per  cent, 
wrong.     It  would  give  a  moment  of    - 
inertia  '4  per  cent,  wrong. 

2.  Find  the  radii  of  gyration  of 
the  sphere  of  Table  II.,  p.  251,  about  a 
line  touching  its  surface ;  the  solid 
cylinder  about  a  line  touching  its    "" 
outside,  parallel  to  the  axis ;  the  rod 
about  a  line  at  right  angles  to  it  at  one 
end.     Ans.,  -5916  rf;  -514  d;  -5773*. 

3.  What  error  is  introduced  in  Table  II.,  by  neglecting  the  size  of  the 
section  of  the  prismatic  rod  ? 

4.  Two  homogeneous  spheres  of  weights  12  and  20  Ibs.,  radii  0-2  and 
0*3  feet,  their  centres  5  feet  apart ;  find  the  distance  of  G,  the  centre  of 
gravity,  from  o,  the  centre  of  the  smaller ;  find  i0,  the  moment  of  inertia 
about  an  axis  through  G  at  right  angles  to  o  G  ;  find  the  moment  of  inertia 
about  o  G  itself ;  find  the  moment  of  inertia  about  any  line  G  A  if  the 
angle  OGAISO.     Ans.,  37'5  inches  ;  27131;  131-3  ;  131-328  +  27000  sin.2  a. 


Fig.  85. 


APPLIED    MECHANICS.  143 

119.  Given  the  moments  of  inertia  of  a  lamina  about  a  pair  of 
axes  at  right  angles  in  the  plane  of  the  lamina,  to  find  the  moment  of 
inertia  about  any  other  axis  through  the  intersection  and  lying  in 
the  plane.  Let  the  moment  of  inertia  about  ox  be  ix  and  the 
moment  of  inertia  about  o  Y  be  iy  ;  o  p  is  any  other  axis,  the 
angle  x  o  P  being  o.  Let  the  distance  of  a  small  portion  of  area 
at  Q  from  the  three  axes  be  called  y\  a/,  and  7  ;  then 

7  =  y'  cos.  o  —  a/  sin.  o, 

and      T2  =  y'2  cos.2  o  -f-  a/2  sin.2  a  —  2  x'  y1  sin.  a  cos.  a. 
Hence,  if  i  is  the  moment  of  inertia  of  the  whole  area  about  the 
new  axis,   I  =  ix  cos.2  a  +  iy  sin.2  a  —  2ixy  sin.  a  cos.  a  ....  (I), 
where  ixy  is  written  to  mean  the  product  of  inertia  about  the  axes 
x  and  y,  or  the  sum  of  all  such  terms  as  "  portion  of  area  x  xy" 

NOW,    let   I  =  -A.,      ix  =    -A,      Iy  =    A-,      ixy  =  JL. 

When  A  is  the  area,  r,  r2,  and  rx  are  the  reciprocals  of  radii  of  gyra- 
tion ;  but  *  has  no  name. 

If  we  take  a  point  p  in  the 
line  o  P,  such  that  o  p  =  r,  and  let 
the  co-ordinates  of  this  point 
relatively  to  the  original  axes  be 
a?  and  y,  then  (1)  becomes 


, 

which    is    the    equation    to    an 

ellipse.      "We  see,  then,  that  if  a 

distance  proportional   to   the   re- 

ciprocal of  the  radius  of  gyration  Fig-  86. 

about  an  axis  be  measured  along 

that  axis  from  o,  the  points  so  found  lie  in  an  ellipse.   When  the  point 

o  is  the  centre  of  the  area  this  ellipse  is  called  the  momenta!  ellipse 

of  the  area,  and  its  principal  axes  are  the  principal  axes  of  inertia. 

If  they  had  been  chosen  as  the  axes  of  reference,  evidently  i^,  the 

product  of  inertia  relatively  to  them,  would  have  been  zero. 


APPENDIX   TO   CHAPTER   VII. 

Mr.  J.  Harrison,  of  the  Royal  College  of  Science,  has  been 
kind  enough  to  prepare  the  following  short  account  of  the 
general  principles  involved  in  the  graphical  study  of  forces 
when  they  do  not  act  in  one  plane,  and  readers  are  referred  to 
"Graphics,"  by  Prof.  R.  H.  Smith,  for  more  detailed  information. 
Forces  in  space  and  framed  structures  in  three  dimensions. 

Problem  1. — To  find  the  resultant  of  a  given  system  of 
forces  in  space,  whose  lines  of  action  all  pass  through  a  point. 

A  force  in  space  is  conveniently  defined  by  two  orthogonal 
projections  of  the  line  which  represents  it.  These  projections 


144 


APPLIED    MECHANICS. 


represent  the  resolved  parts  of  the  force  respectively  parallel 
to  the  planes  of  projection. 

To  compound  the  given  system,  add  the  forces  as  vectors, 
Two  projections  of  the  gauche  polygon  representing  the  vector 
addition  are  required. 

Let  P  Q  R,  P'  Q'  R'  (Fig.  87),  be  the  given  plans  and  eleva- 
tions of  the  lines  of  action  of  the  forces. 

The  plan  p  q  r,  and  the  elevation  p'  q  r'  of  the  vector 
polygon  can  be  at  once  drawn,  since  the  lengths  of  the  sides 
are  supposed  given  as  part  of  the  data.  Then  a  line,  8,  s', 


Pig.  87. 


through  the  given  point  o,  o',  parallel  and  equal  to  the  closing 
side  s,  s  of  the  vector  polygon,  represents  the  required  resultant. 
Problem  2.  —  The  lines  of  action  of  a  concurrent  system  of 
forces  in  equilibrium,  in  space  being  given,  and  the  magnitudes 
of  all  the  forces  except  three,  to  find  the  three  magnitudes. 


APPLIED    MECHANICS.  145 

In  Fig.  87  the  forces  P  P',  Q  Q',  R  B',  are  those  given 
completely ;  the  magnitudes  of  L  L',  M  M',  N  N',  are  required. 

Compound  the  known  forces  into  a  single  force  s  s',  by 
means  of  the  vector  polygon  shown  in  plan  and  elevation. 

To  find  L  L',  observe  that  its  magnitude  must  be  such  that 
the  component  perpendicular  to  the  plane  of  M  N  shall  be  equal 
to  the  component  of  s  s',  perpendicular  to  the  same  plane. 

Draw  a  new  elevation  on  a  plane  perpendicular  to  the 
plane  of  M  N.  Let  s"  or  o"  A"  be  the  new  elevation  of  s.  Draw 
A"  B"  parallel  to  H"  N",  to  meet  L"  in  B"  ;  then  o"  B"  is  the 
magnitude  of  L",  the  elevation  of  L  on  x'  y  .  Project  B"  to  B. 
Then  o  B  is  the  length  of  the  plan  L. 

Draw  I  parallel  and  equal  to  o  B,  and  close  the  vector 
polygon  in  plan  by  drawing  the  lines  mt  n  respectively  parallel 


Fig.  88. 

to  M  and  N.  The  elevation  of  the  polygon  can  now  be  drawn 
The  true  lengths  (not  shown)  of  the  lines  I  I',  m  m',  n  n'  would 
give  the  actual  magnitudes  of  the  three  forces,  L,  M,  N. 

Problem  3. — To  reduce  two  given  forces^  acting  in  directions 
at  right  angles  to  each  other,  to  a  single  force  and  a  couple,  the 
plane  of  the  couple  being  perpendicular  to  the  line  of  action  of 
the  single  force. 

Let  P  and  Q  (Fig.  88)  be  the  giyen  forces,  and  let  A  B  of 
length  a  be  the  line  meeting  both  P  and  Q  at  right  angles. 

At  B  introduce  the  equal  and  opposite  forces  PX  and  P2  as 
shown,  in  a  line  parallel  to  P  and  equal  to  it  in  amount. 


146 


APPLIED   MECHANICS. 


Fig.  89. 


APPLIED   MECHANICS.  147 

Compound  Fl  and  Q  into  Rr  In  the  plane  A  B  RX  draw  A  D, 
making,  with  A  B,  the  angle  DAB  =  QBR1  =  0;  and  drcaw 
B  D  perpendicular  to  A  D.  Introduce  the  forces  R  and  R2,  each 
equal  and  parallel  to  RJ  as  shown. 

Then  the  required  single  force  equivalent  to  p  and  Q  is  R, 
and  the  required  couple  R  x  C  D. 

For  let  the  couple  p  P2  be  represented  by  the  axis  B  N  =  p  x  a\ 
resolve  this  into  B  M,  M  N,  where  M  N  is  perpendicular  to  B  M. 

Then  B  M  =  B  N  x  cos.  Q  =  P  a  cos.  Q  =  R  a  sin.  B  cos.  Q  = 
R  x  c  D. 

And  M  N  =  B  N  x  sin.  B  =  P  a  sin.  Q  =  R  a  sin.2  0  =  R  x  B  c. 

So  R  x  C  D  represents  the  component  couple  B  M,  and  R,  R2  may- 
be taken  as  the  couple  represented  by  M  N.  Now  R2  cancels  RI; 
and  there  remain  R  and  R  x  C  D. 

The  two  outer  lines  at  c  D  may  be  taken  to  represent  the 
couple.  The  convention  adopted  as  to  sign  is,  that  when  the 
arrow-heads  on  the  couple  axis,  and  on  the  force  line  point  the 
same  way,  as  in  the  figure,  the  tendency  of  the  forces  is  to 
produce  a  right-handed  screw  motion,  and  conversely. 

Problem  4. — To  compound  a  given  general  system  of  forces 
in  space. 

Let  the  lines  of  action  of  the  forces  be  supposed  cut  by  any 
plane,  and  at  the  points  of  intersection  resolve  the  forces 
respectively  along  and  perpendicular  to  the  plane.  Compound 
each  of  these  two  sets.  The  system  is  thus  reduced  to  two 
non-intersecting  forces  at  right  angles.  If  desired,  these  two 
forces  may  be  compounded,  as  in  the  last  problem. 

The  given  system  illustrated  in  Fig.  89  consists  of  three 
forces,  p,  Q,  R,  the  projections  of  the  lines  of  action  of  which 
on  three  planes  mutually  perpendicular  are  shown,  as  are  also 
the  three  projections  of  the  vector  polygon,  drawn  as  if  the 
given  system  were  concurrent. 

The  system  may  be  reduced  to  two  forces — one  in  the 
horizontal  plane  x  Y,  the  other  vertical. 

To  find  the  former,  draw  a  link  polygon  in  plan  .with 
respect  to  any  pole  o.  Thus,  s  =  s  is  the  horizontal  force. 

To  find  the  vertical  force,  two  methods  are  available. 

(1)  Compound  the  vertical  components.  In  the  figure,  the 
elevations  of  the  components  on  z  x  are  shown,  and  their 
resultant  v'  is  found  by  means  of  a  link  polygon  drawn  with 
respect  to  any  pole  o,.  In  a  similar  manner  (not  shown),  the 
elevation  on  Y  z  of  the  resultant  vertical  force  could  be  foundj 


1-A8  APPLIED    MECHANICS. 

thus  completely  determining  the  vertical  force  and  giving  the 
plan  H  of  its  line  of  action. 

(2)  Or  thus : — On  z  x,  by  means  of  a  link  polygon  drawn 
to  any  pole  o',  find  s',  the  elevation  of  the  line  of  action  of  the 
component  force  of  the  system  parallel  to  the  plane  z  x.  And 
similarly,  on  Y  z,  find  s",  the  elevation  of  the  component  parallel 
to  Y  z.  Let  these  intersect  the  ground  lines  respectively  in  h' 
and  h",  projections  from  which  give  H,  the  plan  of  the  line  of 
action  of  the  vertical  force  required.  Its  magnitude  is  at  d'. 

A  further  construction  is  shown  for  reducing  the  forces,  as 
in  the  last  problem.  The  line  of  action  of  the  single  force  thus 
obtained  is  called  the  central  axis  of  the  system.  The  couple 
is  equal  to  the  moment  of  system  about  the  central  axis,  and 
may  be  shown  to  be  the  minimum  couple.  There  is  no  other 
axis  about  which  the  forces  have  a  less  moment. 

Note. — If  the  given  system  contain  couples  as  well  as 
forces,  treat  those  independently,  adding  their  axes  as  vectors. 
Resolve  this  couple  parallel  and  perpendicular  to  the  central 
axis,  and  add  to  the  other  part  of  the  system. 

Problem  5. — To  determine  the  stresses  in  a  non-redundant 
framed  structure  of  three  dimensions  under  given  loads. 

The  criterion  for  a  non-redundant  stiff  frame  in  three  dimen- 
sions, with  non-rigid  joints,  is  that  the  number  of  bars  must 
be  six  less  than  treble  the  number  of  joints.  For  this  relation 
is  evidently  true  in  the  simplest  example — viz.,  for  a  frame  of 
six  bars  forming  a  pyramid ;  and  in  building  up  a  frame  which 
shall  be  stiff,  each  new  joint  requires  three  new  bars.  If  any 
portion  consist  of  a  plane  frame  with  redundant  members  (such, 
for  example,  as  a  plane  quadrilateral  with  crossed  diagonals), 
such  redundant  members  are  not  to  be  counted  in  applying  the 
criterion. 

Fig.  90  shows  a  derrick  crane  carried  by  a  braced  triangular 
pier. 

The  stress  diagram  for  the  frame  is  built  up  in  plan  and 
elevation,  applying  Problem  2  in  succession  to  the  several 
joints.  The  order  of  taking  the  joints  is  that  indicated  by 
the  Roman  numerals,  and  the  various  points  on  the  stress 
diagram  are  marked  a,  6,  c,  etc.,  in  order  as  they  are  found. 

If  the  pier  had  been  braced  as  shown  in  Fig.  91,  then  after 
having  drawn  the  stress  diagram  for  the  joints  I  and  II, 
there  would  be  no  new  joint  with  less  than  four  unknowns. 
In  this  case  the  stress  in  the  bar  marked  P  could  be  found  by 


APPLIED    MECHANICS. 


149 


150  APPLIED    MECHANICS. 

resolving  the  forces  at  the  joint  V  perpendicular  to  the  plane 
which  contains  the  other  three  unknowns,  Q,  R,  s.  The  building 
up  of  the  diagram  would  then  proceed  in  the  same  order  as 
before. 

EXERCISES. 

1.  A  vertical  crane  post  is  10  feet  high,  jib  30  feet  long,  stay  24  feet  long, 
meeting  at  a  point  c.     There  are  two  back  stays  making  angles  of  45° 
with  the  horizontal ;  they  are  in  planes  due  north  and  due  west  from  the 
post.     A  weight  of  5  tons  hangs  from  c.     Find  the  forces  in  the  jib  and 
stays — 1st,  when  c  is  south-east  of  the  post ;  2nd,  when  c  is  due  east ; 
3rd,  when  c  is  due  south. 

2.  A  tripod  whose  vertex  is  A,  and  whose  legs  are  A  B,  AC,  A  D,  of  lengths 
8,  8-5,  and  9  feet  respectively,  sustains'a  load  of  2  tons.     The  ends  B,  c,  D 
form  a  triangle  whose  sides  are  B  c  =  7  feet,  CD  =  6  feet,  B  D  =  8  feet,  find 
by  graphical  construction  the  compressive  forces  in  each  leg. 

Ans.,  1-16  ton,  0'55  ton,  0-53  ton. . 

3.  Three  ropes,  each  12  feet  long,  hang  from  the  three  corners  of  a 
horizontal  isosceles  triangle,  A  B  c,  in  which  A  B  and  A  c  are  each  20  feet, 
and  B  c  is  10  feet.     The  ropes  are  joined  at  their  ends  and  support  a  load 
of  1  ton.     Find  the  pull  on  each  rope.     Am.,  0-52  ton,  0-52  ton,  0-92  ton. 


151 


CHAPTER    VIII. 

EXAMPLES    IN   GRAPHICAL   STATICS. 

120.  IN  any  structure,  such  as  the  principal  of  a  roof  and 
many  girders  of  bridges  formed  of  many  different  bars,  if  we 
neglect  the  weights  of,  or  upon,  the  bars,  and  we  assume  that 
the  joints  are  frictionless  hinges,  it  is  easy  to  see  that  the  force 
exerted  by  any  bar  is  in  a  straight  line  between  the  centres  of 
the  hinges  at  its  ends.  For  whatever  may  be  the  many  forces 
between  a  piece  and  pin,  at  the  surface  of  the  pin,  these  forces 
must  all  be  normal  to  the  surface,  since  there  is  no  friction  ; 
they  must  therefore  all  be  directed  in  radial  lines,  and  hence 
their  resultant  must  be  a  radial  line  through  the  centre  of  the 
pin.  In  this  case,  then,  the  joints  of  a  structure  only  being 
loaded  with  known  forces,  it  is  easy  to  calculate  the  pushing 
or  pulling  force  exerted  by  each  piece. 

To  illustrate  this — let  there  be  three  pieces,  whose  centre- 
lines are  A  o,  BO,  and  c  o  (Fig.  92), 
meeting  on  a  pin  whose  centre  is  o.  Sup- 
pose we  know  that  the  piece  c  o  pulls  the 
pin  in  the  direction  o  c  with  a  force  of 
2,000  Ibs.  We  have  then  to  find  the  two 
forces  in  the  given  directions  of  A  o  and 
B  o  to  balance  the  known  force  o  c. 

Draw  the  triangle  (Fig.  93),  whose 
sides,  c,  a,  6,  are  parallel  to  o  c,  A  o,  and 
B  o ;  and  let  c  represent  the  force  o  c 
or  2,000  Ibs.  to  some  scale,  and  let  the 
arrow-head  on  c  represent  the  sense  of  o  c. 
Then  the  lengths  of  b  and  a  represent  the 
other  two  forces  to  scale.  Put  the  circuital 
arrow-heads  on  b  and  a,  and  we  see  that 
o  A  is  a  pulling  force,  and  the  piece  o  A  is  called  a  tie-rod ;  B  o 
is  a  pushing  force,  and  the  piece  B  o  is  called  a  strut.  Note  that 
B  o  is  attached  to  some  other  pin  than  o.  If  we  study  the  equi- 
librium of  the  new  pin  we  must  remember  that  B  o  pushes  it,  and 
we  must  draw  the  new  arrow-head  when  we  study  the  new  pin. 

It  is  very  important  that  the  student  should  illustrate  an 
important  fact  like  this  for  himself  in  the  laboratory.  Fig.  94 
shows  two  real  pieces  A  o  and  B  o  hinged  at  o.  Hang  on  any 


Fig.  93. 


152 


APPLIED    MECHANICS. 


weight  c.  Adjust  the  screw  at  A'  until  the  spring-balance  A'  A 
is  in  a  line  with  o  A,  so  that  it  indicates  the  pull  on  o  A.  The 
push  in  B  o  is  recorded  by  the  spring-balance  B'  B.  When 
things  are  at  rest,  open  the  parts  of  a  two-foot  rule  until  they 


Pig.  94. 

just  fit  the  angle  between  A  o  and  BO,  and  transfer  this  angle 
to  a  sheet  of  paper.  Now  do  the  same  with  the  angle  between 
B  o  and  c  o,  and  test  on  the  paper  if  the  sides  of  a  triangle 
parallel  to  A  o,  B  o,  and  c  o  really  represent,  to  some  scale,  the 
three  forces.  If  they  do  not,  speculate  for  yourself  upon  the 
discrepancy ;  how  much  discrepancy  is  due  to  the  fact  that 
the  forces  as  measured  are  not  truly  the  forces  at  o,  because  of 
the  weights  of  the  parts  ?  How  much  is  due  to  the  fact  that 
the  rule  presupposes  a  pin  with  absolutely  no  friction  ?  If  you 
are  ingenious  and  an  advanced  student,  you  will  try  a  pin  with 
much  friction,  or  perhaps  a  riveted  joint  at  o,  and  so  learn 
more  than  I  can  tell  you. 

EXERCISES. 

1.  In  Fig.  94  the  jib  o  B  makes  an  angle  of  45°  with  the  horizontal ; 
angle  A  o  B,  15° ;  weight  at  c,  5  tons ;  find  the  forces  in  o  A  and  o  B. 

Am.,  13-66  ton,  16'73  ton. 

2.  A  chain  fastened  at  o  (Pig.  94)  goes  round  a  snatch-block  at  c, 
and  then  over  a  pulley  at  o  in  the  direction  o  A  to  the  barrel,  the  total 


APPLIED   MECHANICS.  153 

weight  at  c  being  6  tons ;  find  the  forces  in  j  A  and  o  B.     Dimensions 
eame  as  in  1.  Am.,  10-76  tons;  16'73  tons. 

3.  If  a  wharf  crane,  the  post,  tie-rod  and  jib  measure  15,  20,  and  30 
feet  respectively,  what  would  be  the  stresses  in  each  of  the  three  members 
when  a  load  of  7  tons  is  suspended  over  the  pulley  at  the  jib-head  (1) 
when  the  lifting-chain  passes  from  the  pulley  to  the  drum  parallel  with 
the  jib :  (2)  when  the  drum  is  placed  so  that  the  chain  passes  from  jib-head 
parallel  to  tie-rod  ? 

Ans.,  7  tons  ;  9-3  tons  ;  21  tons  ;  7  tons  ;  2-3  tons  ;  14  tons. 

4.  A  contractor's  portable  hand-crane  has  a  vertical  post  A  B,  to  which 
the  jib  A  c  is  inclined  45°,  and  the  tension -rod  B  c  makes  with  A  B  an  angle 
A  u  c  of  1 20°.     The  back-stay,  from  the  head  of  the  post  B  to  the  extremity 
D  of  the  horizontal  strut  A  D,  is  inclined  at  an  angle  of  45°  to  A  P.     Find 
the  counterbalance  weight  required  at  D  to  balance  a  load  of  10  tons  sus- 
pended from  the  end  c  of  the  jib.     Determine  also  the  nature  and  amount 
of  the  force  in  the  jib  A  c,  and  in  the  rods  B  c  and  B  D.     (The  tension  in 
tho  chain  may  be  neglected.) 

Ans.,  23-66  tons  ;  33-40  tons  ;  27'32  tons  ;  33'46  tons. 

121.  Let  us  now  consider  the  roof-principal  shown  in  Fig.  95, 


Fig.  95. 

Certain  loads  are  given  acting  at  the  joints,  and  we  knoAV 
that  the  structure  is  supported  by  two  forces  or  reactions  at  its 
two  ends.  Our  first  step  is  to  find  these  two  supporting  forces. 
They  must  be  in  equilibrium  with  all  the  external  loads. 

Now,  it  is  well  known  that  we  must  be  given  either  the 
direction  or  some  other  information  about  one  of  these  support- 
ing forces,  else  the  problem  becomes  indeterminate.  It  is 
usual  to  be  told  that  one  or  other  supporting  force  is  vertical. 
This  condition  is  arrived  at  in  practice  by  having  at  one  end  of 
the  structure  a  shoe  with  rollers  resting  on  a  horizontal 
plate  of  iron. 

The  notation  which  we  use  very  materially  simplifies  the 


154 


MECHANICS. 


process  of  calculation.  You  observe  that  every  space  between 
two  forces  in  Fig.  95  is  indicated  by  a  letter.  The  line  which 
separates  the  space  A  from  the  space  B  is  called  A  B,  and  cor- 
responds with  the  line  A  B  in  Fig.  96.  A  point  is  indicated 
by  the  letters  of  the  spaces  which  meet  at  that  point.  Thus, 
A  G  H  is  the  end  of  the  roof -principal. 

Suppose  the  supporting  force  at  the  point  F  G  P  is  known  to 
be  vertical.  We  must  first  find  the  amount  of  the  force  F  G, 
and  the  direction  and  amount  of  the  force  A  G. 

Draw  the  force  polygon,  A  B  c  D  E  F,  Fig.  96.*  We  see 
that  to  close  it  we  need  two  lines  to  join  F  and  A.  Now,  one 


Fig.  90. 

of  these,  F  G,  is  vertical.  Take  o  as  pole.  Join  o  A,  o  B,  o  c, 
o  D,  o  E,  o  F  in  the  usual  way.  Draw  the  link  polygon,  shown 
dotted  in  Fig.  95,  commencing  at  the  only  known  point  of  the 
force  A  G,  namely  A  G  n.  Now,  o  G  (Fig.  96)  is  parallel  to  the 
last  side  of  it,  and  thus  we  find  F  G  and  G  A,  the  supporting 
forces  at  the  end  of  the  principal.  Having  found  the  two 
supporting  forces,  F  G  and  G  A,  we  proceed  as  follows : — We 
have  the  closed  force  polygon  A  B  c  D  E  F  G  A.  The  arrow-heads 
shown  on  this  force  polygon  are  not  to  be  rubbed  out  during 
the  calculation,  and  in  practice  we  mark  them  in  ink.  All 

*  In  an  actual  case  there  would  usually  be  loads  given  at  the  end  joints 
as  well  as  the  others.  In  this  first  example  I  have  only  taken  the  loads  as 
shown. 


APPLIED    MECHANICS.  155 

other  arrow-heads  \vh:ch  we  draw  on  Fig.  96  may  require  to 
be  rubbed  out,  and  ought  only  to  be  marked  in  pencil.  We 
must  begin  our  calculation  at  a  joint  where  only  two  pieces 
meet,  and  where  one  force  which  acts  there  is  given.  Now 
at  the  joint  A  G  H  we  know  the  force  A  G.  In  Fig.  96  draw 
A  H  and  G  H  parallel  to  the  pieces  A  H  and  G  H  of  Fig.  95. 
Put  arrows  on  the  sides  of  the  triangle  G  A  H  circuital  with 
the  arrow  on  G  A.  Now  we  see  by  the  arrows  that  the  piece 
A  H  pushes  the  joint  with  a  force  represented  to  scale  by  th^ 
length  of  the  line  A  H  (Fig.  96).  We  know,  then,  that  A  H 
is  a  strut,  since  it  pushes,  and  we  know  the  total  pushing  force 
in  it.  Similarly,  H  G  is  a  tie,  and  the  total  pulling  force  in  it 
is  represented  by  the  length  of  the  line  H  G  in  Fig.  96. 

We  now  rub  out  the  arrows  which  we  are  supposed  to  have 
drawn  in  pencil  on  the  lines  A  H  and  H  G  (Fig.  96),  and  proceed 
to  the  joint  A  B  I  H.  It  must  be  remembered  that  although  the 
pieces  A  H  and  B  I  are  in  the  same  straight  line,  we  regard 
them  as  two  separate  pieces. 

We  know  the  force  A  B,  we  also  know  that  the  force  with 
which  the  piece  A  H  pushes  the  joint  (we  have  already  found  it 
to  be  a  strut,  therefore  it  pushes  both  joints  at  its  ends)  is 
represented  by  the  length  of  the  line  AH  (Fig.  96).  Draw, 
then,  HI  and  BI  (Fig.  96)  parallel  to  the  pieces  H  I  and  B  I 
(Fig.  95).  We  have  thus  a  polygon  A  B  I  H.  The  force  A  B 
(Fig.  96)  tells  us  how  to  pencil  arrow-heads  circuitally 
round  this  polygon.  When  we  do  this  we  find  that  the  piece 
B  i  pushes  the  joint  with  a  force  represented  by  the  line  B  I 
(Fig.  96),  so  that  B  I  is  a  strut.  Also  I  H  is  a  strut,  in  which 
the  stress  is  represented  by  the  line  I H  (Fig.  96).  We  pro- 
ceed in  this  way  from  joint  to  joint,  always  taking  care  to  rub 
out  our  pencilled  arrow-heads  when  we  proceed  from  one  joint 
to  the  next.  The  lengths  of  the  lines  in  Fig.  96  give  the 
magnitudes  of  the  forces  in  the  pieces  of  the  structure.  It 
is  easy  to  prove  that,  if  no  mistake  is  made,  no  discrepancy 
will  appear  when  the  drawing  is  being  finished. 

If  in  Fig.  96  the  points  K  and  I  were  found  to  coincide, 
this  evidently  means  that  the  piece  K  I  is  unnecessary  in  the 
structure.  If,  again,  we  find  that  we  cannot  close  one  of  our 
little  polygons  in  Fig.  96,  we  ought  to  proceed  to  new  joints, 
and,  possibly,  when  we  again  consider  the  joint  with  which  we 
had  difficulty,  we  shall  be  able  to  close  its  polygon.  If  we 
still  find  difficulty,  it  must  be  caused  by  two  or  more  joints, 


156  APPLIED   MECHANICS. 

and  the  pieces  connecting  these  are  evidently  unnecessary  to 
the  structure.  If  we  find  in  Fig.  96  two  points  with  the  same 
letter,  we  evidently  require  to  add  a  new  piece  to  the  structure, 
which  will  exert  a  force  represented  by  the  distance  between 
these  two  points. 

No  explanation  in  writing  will  enable  the  student  to  master 
this  beautiful  method  of  determining  the  forces  in  structures. 
He  must  select  structures,  apply  loads  to  the  joints,  and  calcu- 
late the  various  forces  for  himself.  When  he  has  made  four 
such  calculations,  he  will  know  nearly  all  that  can  be  said  on 
the  subject.  (Figs.  99  and  101  show  two  examples.) 

122.  Roofs. — It  is  not  my  object  here  to  describe  the  con- 
struction of  a  roof  or  a  bridge.  For  such  information  the 
student  must  examine  real  structures  and  good  drawings  of 
roofs  and  bridges  for  himself. 

Suppose,  for  instance,  that  he  finds  a  roof,  somewhat  like 
his  own,  to  weigh — including  possible  snow,  etc. — 40  Ibs.  per 
square  foot  of  horizontal  area  covered.  Suppose  his  principals 
are  to  be  placed  8  feet  apart,  the  span  being  50  feet,  then  each 
principal  has  to  support  about 

8x50x40,  or  16,000  Ibs. 

Now,  if  Fig.  97  is  the  shape  of  his  principal,  as  A  B,  B  c, 
c  D,  and  D  E  are  all  equal,  we  may  suppose  that,  however  the 

roof  covering  may  be  sup- 
ported by  the  principals,  the 
piece  of  rafter,  A  B,  or  any 
other  of  the  divisions,  sup- 
ports 4,000  Ibs.  The  joint  B 
Fig.  97.  gets  half  the  load  on  A  B  and 

half  the  load  on  B  c ;  conse- 
quently the  load  at  the  joint  B  is  taken  to  be  4,000  Ibs., 
and  similarly  for  c  and  D.  The  joints  A  and  E  have  2,000  Ibs. 
each. 

When  the  above  vertical  loads  have  been  given  to  the  joints, 
we  have  to  consider  wind  pressure  on  one  side  of  the  roof.  If 
we  suppose,  as  we  reasonably  may,  that  40  Ibs.  per  square  foot 
is  the  greatest  pressure  of  wind  ever  likely  to  occur  on  a  surface 
at  right  angles  to  the  direction  of  the  wind,  then  the  normal 
pressure  per  square  foot  on  roofs  of  the  following  inclinations 
may  be  taken  from  the  following  table,  which  is  obtained  from 
experiment. 


APPLIED    MECHANICS. 


157 


Angle  vof  Roof 
6°        .     .     , 

Normal 
Pressure. 
.     .       5-0 

10° 

.     .      9'7 

20°        .     .     . 

.     .     18-1 

30°        .     .     , 

.    .     26-4 

40° 

,     33-3 

TABLE   I. 

Normal  Pressure  of  Wind  against  Roofs. 
Angle  of  Roof. 


50° 
60° 
70° 
80° 
90° 


Normal 
Pressure. 
38-1 
40-0 
40-0 
40-0 
40-0 


Fig.  98. 


Thus,  if  the  portion  of  one  slant  side  of  the  roof  between  two 
principals  has  an  area  of  240  square  feet,  and  if  the  inclination 
of  the  roof  is  30°,  say,  then  240  x  264,  or  6,336  Ibs., 
has  to  be  supported  by  each  bay.  Transferring 
this  to  the  joints,  we  see  that  at  B  (Fig  98),  in 
case  the  wind  pressure  is  upon  the  side  A  B  c,  we 
have  the  vertical  load  x  B,  or  4,000  Ibs,  due  to 
weight  of  roof,  snow,  etc.,  and  also  Y  B,  or  one 
half  of  6,336  Ibs.,  normal  to  the  roof,  and  due 
to  wind.  Complete  the  parallelogram,  and  evi- 
dently z  B  is  the  load  at  the  joint  B  which  we  must  use  in 
our  calculations. 

The  student  will  find  that  if  a  roof-principal  can  only  be 
supported  by  a  vertical  force  at  a  certain  end,  the  stresses  in 
the  structure  are  greatest  when  the  other  side  of  the  roof  is 
acted  on  by  the  wind. 

123.  Many  joints  in  a  real  structure  are  usually  stiff  joints, 
so  that  many  pieces  may  really  be  subjected  to  bending,  as 
well  as  to  direct  compressive  or  tensile 
stresses.  A  general  method  of  taking 
stiffness  of  joints  into  account  is  quite 
unknown ;  but  when  we  discuss  bend- 
ing we  shall  see  pretty  clearly  what  is 
the  effect  of  a  stiff  joint,  and  in  some 
cases  we  shall  be  able  to  make  calcula- 
tions on  the  subject.  It  may  generally 
be  assumed  that  the  strength  of  a 
structure  is  greater  if  the  joints  are 
stiff  than  if  they  are  merely  hinges. 
This  is  not  always  the  case,  and, 
from  the  indeterminateness  of  the 
problem  of  finding  the  stresses  in  a 
structure  whose  joints  are  stiff,  many 
large  bridge  trusses  are  made  with  Fig.  100. 


158 


APPLIED    MECHANICS. 


nearly  all  their  joints  hinged.  In  roof-principals  the  joints 
are  often  made  stiff,  rather  for  the  purpose  of  stiffening  the 
whole  structure  than  for  the  sake  of  strength.  We  shall 
presently  see  the  distinction  between  stiffness  and  strength 
in  structures.  In  a  roof  all  joints  of  struts  are  usually  made 
stiff.  What  we  shall  now  say  is  of  more  importance  in  bridges 
than  in  roofs. 

If  two  or  more  pieces  of  a  structure  are  in  a  straight  line 
with  one  another  at  joints  where  they  meet,  it  is  usual,  for 
strength,  to  make  the  joints  between  them  quite  rigid.  Thus, 


L_     J!_       •"     i-  "     k    -. 


Fig    102. 

the  pieces  A  H  and  B  I  of  Fig.  95,  or  A  B  and  B  c  of  Fig.  97 
ought  to  form  one  bar.  But  this  is  only  useful  when  the 
pieces  in  question  are  struts,  and  our  reason  for  the  continuity 
of  the  pieces  is  that  a  strut  is  stronger  when  its  ends  are  fixed 
than  when  its  ends  are  not  fixed.  Thus  the  piece  B  i  (Fig.  95) 
will  resist  a  greater  thrust  if  it  is  continuous  with  A  H  and  c  K 
than  if  it  were  hinged  with  these  pieces.  (See  Art.  372.)  It  is 
not  good  in  all  cases  to  fix  the  end  of  a  strut  by  rivets,  etc., 
instead  of  a  hinge ;  because  the  benefit  due  to  fixing  an  end 
may  be  more  than  counterbalanced  by  the  evil  effects  of  bending 
introduced  to  the  strut  through  the  joints  by  a  tendency  to 
change  the  angle  which  the  strut  makes  with  the  piece  to  which 
it  is  fixed.  The  common-sense  of  the  engineer  will  always 
enable  him  to  decide  as  to  the  judiciousness  of  fixing  the  end 
of  a  strut. 


APPLIED    MECHANICS. 


159 


Pig.  108. 


124.  Sections  of  Structures. — It  is  often  of  considerable  im- 
portance to  find  immediately  the  forces  in  pieces  of  a  structure 
which  are  not  near  the  ends.  If  wo  can  draw  any  surface 
which  will  cut  through  the 
pieces  in  question,  we  can 
calculate  the  stresses  in 
these  pieces  directly,  sup- 
posing the  pieces  are  only- 
three  in  number.  Thus, 
the  section  ACE  (Fig.  103) 
cuts  the  pieces  B  A,  B  c,  and 
D  E.  Now,  not  only  the 
whole  structure,  but  every  part  of  it  is  kept  in  equilibrium. 
What  forces  keep  the  part  A  H  E  in  equilibrium  ?  They  are 
the  known  forces  at  B,  F,  and  H,  together  with  three  unknown 
forces  whose  directions  are  B  A  A',  B  c  c',  and  D  E  E'.  Given 
the  directions  of  three  forces  which  equilibrate  a  number 
of  known  forces,  we  know  that  they  may  be  determined 
in  magnitude  by  the  link-polygon  method  (Art.  101).  Some- 
times the  link-polygon  method  is  more  troublesome  than 
the  following  : — To  find  the  push  or  pull  in  D  E  E'.  We 
know  (Art.  98)  that  the  moment  of  the  force  in  E  E'  about 
the  point  B  is  equal  to  the  sum  of  the  moments  about  B 
of  all  the  external  forces  (for  the  forces  in  the  directions 
A  A'  and  c  c'  have  no  moment  about  B,  since  they  pass 
through  it).  Let  the  algebraic  sum  of  the  moments  of  the 
external  forces  be  actually  calculated,  multiplying  numerically 
each  force  by  its  perpendicular  distance  from  B.  This  sum, 
divided  by  the  perpendicular  distance  from  B  to  E  E',  will  give 
the  force  in  E  E'.  If  the  algebraic  sum  gives  a  moment  tending 
to  turn  the  structure  about  B  against  the  direction  of  the 
hands  of  a  watch,  the  force  in  E  E'  is  a  pulling  force  acting 
from  E  towards  E',  and  therefore  the  piece  D  E  is  a  tie.  The 
engineer  ought  to  practise  this  common-sense  way  of  applying 
our  fundamental  principles.  He  will  regret  it  if  he  fetters 
himself  to  graphical  statics  methods  of  working. 

It  will  be  observed  that  if  we  wish  to  know  the  forces  at 
any  section  of  any  loaded  structure,  we  must  consider  that  the 
parts  of  the  structure  on  any  one  side  of  this  section  are  in 
equilibrium.  Thus,  if  A  and  B  are  the  two  parts  of  the  structure, 
consider  the  equilibrium,  say,  of  B.  Now,  B  is  kept  in  equili- 
brium by  the  external  forces  or  loads  which  act  on  B,  and  by 


160 


APPLIED    MECHANICS. 


the  forces  which  act  on  B  at  the  section.  Of  course,  it  is  A 
which  causes  these  forces  to  act  on  B  through  the  section ;  but 
in  calculations  concerning  them  we  do  not  need  to  consider  A 
or  the  loads  on  A. 

125.  Loaded  Links. — Let  A  c,  c  D,  D  E,  and  E  B  be  four 
links  hinged  together  at  c.  D,  and  E,  and  supported  somehow  by 

hinges  at  A  and  B,  and, 
neglecting  the  weights  of 
the  links  themselves,  let  x, 
y,  and  z  be  forces  acting  at 
the  three  joints,  so  as  to 
make  the  links  take  the 
positions  shown  in  Fig.  104. 
Take  any  point,  o  (Fig.  105), 
and  draw  lines  o  m,  on,  o  pt 
and  o  q  parallel  to  the  links, 
and  from  any  point,  m,  in 
o  m,  draw  m  n  parallel  to 
the  force  z,  np  parallel  to 
the  force  y,  and  p  q  parallel 
to  the  force  x :  then  it  is 
easy  to  prove  that  the  lengths 
of  the  lines  m  n,  n  p,  and  p  q  are  proportional  to  the  forces  z,  y, 
and  x,  and  the  tensile  forces  in  the  links  are  proportional  to  the 
lengths  of  the  lines  om,on,op,  and  o  q.  For  it  is  evident  that 
the  three  forces  at  E,  keeping  the  joint  in  equilibrium  as  they 
do,  must  be  proportional  to  the  sides  of  the  triangle  omn.  If 
you  put  arrow-heads  on  om  and  on  circuital  with  the  one 
already  on  m  n,  you  will  see  that  the  bars  B  E  and  D  E  do  not 
push  the  joint  E  ;  they  pull  it  and  are  tie-bars.  Thus,  then, 
the  lengths  of  the  lines  in  Fig  105  represent,  to  some  scale,  all 
the  forces  acting  at  the  joints,  c,  D,  and  E. 

All  so  easy  as  it  is  to  prove  that  the  above  proposition  is 
correct,  it  is  well  to  illustrate  its  truth  in  the  laboratory. 
Let  A,  c,  D,  E,  B  be  a  string  fastened  to  a  vertical  board  at  A 
and  B,  with  loads  x,  y,  and  z  applied  at  c  and  D  by  means  of 
strings  passing  over  pulleys.  The  string,  instead  of  being 
fastened  at  A  and  B,  may  there  pass  over  pulleys  with  balancing 
weights.  Let  the  vertical  board  be  covered  with  paper ;  with 
a  pin  prick  points  on  the  paper  showing  the  directions  of  the 
string  everywhere,  and  transferring  the  paper  to  another  board, 
find  what  is  the  pull  in  A  c,  C  D,  D  E,  and  A  B.  We  have  no 


Pig.  105.. 


APPLIED    MECHANICS. 


161 


actual  test  of  the  pulls  in  c  D  and  D  E,  unless  an  ingenious 
student  can  introduce  very  light  spring  balances  whose  own 
weights  are  negligible. 

The  student  may  compare  this  construction  with  exactly 
the  same  construction  in  Art.  97.  He  will  see  that  in  Fig.  104 
the  resultant  of  the  forces  E  B  and  z  is  in  the  direction  D  E  ;  the 
resultant  of  the  forces  E  B,  z  and  y  is  in  the  direction  c  D ;  the 
resultant  of  the  forces  E  B,  z,  y  and  x  is  in  the  direction  A  c. 
Fig.  104  shows  the  positions,  and  Fig.  105  shows  the  amounts  of 
these  successive  resultants.  B  E  D  c  A  is  what  we  called  a  line 
of  resistance. 

126.  Loaded  Chain. — If  we  want  to  find  the  pull  in  every 
part  of  one  chain  of  a  suspension  bridge,  and  to  draw  the 
shape  of  the  chain,  it  is  first  necessary  to  know  the  weight  of  the 
bridge  at  every  place.  This  weight  is  probably  supported  by 
two  chains,  so,  as  we  have  only  one  chain  to  deal  with,  we  only 
take  half  the  weight  of  the  bridge.  We  shall  suppose  that  there 
is  no  long  girder  or  other  support  for  the  bridge  but  the  chain. 
It  is  usual  to  suspend  the  supporting  beams  of  the  roadway 
from  the  chain  by  vertical  iron  rods,  placed  at  equal  horizontal 
distances  from  one  another.  We  may  imagine  the  roadway  to 
be  as  heavy  at  one  place  as  another,  so  that  the  pulls  in  all  the 
rods  will  be  the  same.  Suppose  there  are  ten  rods,  and  in  each 
a  pull  of  20  tons.  Draw 
ten  equidistant  vertical 
lines  (Fig.  106)  to  repre- 
sent the  rods.  We  must 
get  another  condition 
before  we  can  draw  the 
chain.  Let  it  be  this,  that  the 
chain  in  the  middle,  where  it  is 
horizontal,  shall  be  capable  of  with- 
standing a  pull  of  200  tons.  Now 
draw  o  H  horizontally  (Fig.  107), 
and  make  its  length  on  any  scale 
represent  200  tons.  Make  H  A 
and  H  B  on  the  same  scale  represent 
each  100  tons  (if  your  chain  is  to 
be  symmetrical),  and  divide  them 
up,  so  that  each  portion  represents  20  tons — that  is,  the 
vertical  load  communicated  to  the  chain  by  each  tie-rod. 
Now  join  o  with  each  point  of  division  in  A  B.  Suppose 


Fijr.  106. 


Pig.  107. 


162  Al'PLIED    MECHANICS. 

that  P  (Fig.  106)  is  one  point  of  support  of  the  chain,  draw 
p  a  (Fig.  106)  parallel  to  o  A  (Fig.  107),  a  c  parallel  to  o  c, 
c  d  parallel  to  o  D,  and  so  on  till  you  reach  the  point  Q, 
which  I  suppose  to  be  on  the  same  level  as  p.  Of  course, 
the  points  of  support,  P  and  Q,  may  be  anywhere  on  the 
lines  a  P  and  m  Q.  It  is  quite  evident  from  what  you  have 
already  learnt  that  the  pull  in  any  part  of  the  chain  is 
represented  by  the  length  of  the  line  from  o,  which  is  parallel 
to  it  in  Fig.  107,  and  it  is  also  evident  that  the  chain  will  take 
this  shape  without  any  tendency  to  alter.  Note  that  when  all 
the  loads  on  a  chain  are  vertical,  the  horizontal  component  of 
any  of  the  sloping  forces  is  the  same  as  that  of  any  other,  being 
o  H.  This  is  always  called  the  horizontal  pull  of  the  chain. 

127.  We  began  by  assuming  a  pull  of  200  tons  in  the  party* h, 
where  the  chain  is  horizontal.     We  might  have  assumed  a  pull 
of  300  tons  infh ;  this  would  have  caused  the  chain  to  hang  in 
a  flatter  curve.     Assuming  a  pull  of  100  tons  infh,  we  should 
have  obtained  a  greater  difference  of  level  between  p  and  h. 

It  will  be  found  that  in  the  present  case,  where  the  load  is 
supposed  to  be  uniformly  distributed  along  the  horizontal,  the 
links  would  just  circumscribe  the  curve  called  a  parabola. 
With  any  other  distribution  of  load  they  will  fit  some  other 
curve  than  a  parabola,  but  in  any  case  you  know  now  how  to 
draw  the  shape  of  such  a  chain,  and  to  determine  the  pull  in 
any  part  of  it. 

128.  Arched  Rib. — If  instead  of  a  hanging  chain  you  wanted 
to  use  a  thin  arched  rib  to  support  your  roadway,  then  if  you  have 
numerous  vertical  rods  by  which  to  hang  your  load  to  the  rib, 
and  if  the  distribution  of  the  load  is  known,  you  can  draw  the 
curve  of  the  rib  in  exactly  the  same  way,  but  it  will  now  be 
convex  upwards,  of  course.     With  uniform  horizontal  distribu- 
tion of  your  load  you  will  get  a  parabolic  rib.     The  difference 
between  the  two  cases  is  this  :  a  slight  inequality  in  your  loads 
or  a  temporary  alteration  will  only  cause  the  chain  to  take  a 
slightly  different  position  for  the  time,  and  it  will  get  back  to 
its  old  shape  when  the  old  loading  is  returned  to ;  whereas  the 
arch  is  in  a  state  of  unstable  equilibrium,  and  as  it  is  very 
thin,  so  that  it  cannot  resist  any  bending,  a  slight  change  of 
loading  will  very  materially  alter  its  shape  and  it  will  get 
destroyed.     Such  a  rib  or   series  of   struts  is  either  stayed 
with    numerous   diagonal    pieces   or   else   it   is    made   very 
massive,  so  that  should  the  line  like  p  h  Q  (inverted),  which 


APPLIED    MECHANICS. 


163 


is  supposed  to  pass  everywhere  along  its  axis,  deviate  a  little 
from  this  position,  the  rib  may  resist  alteration  of  shape  by 
refusing  to  bend. 

129.  The  load  carried  by  an  arch  may  either  be  hung  from 
it  by  means  of  tie-rods,  or  else  it  may  rest  on  the  top  of  the 
arch,  the  weight  being  carried  from  the  different  parts  by  means 
of  struts  or  pillars  of  iron,  stone,  or  brick,  or  the  arch  may  be 
levelled  up  to  the  roadway  by  means  of  a  solid  mass  of  masonry, 
or  merely  by  one  or  two  pillars  of  masonry,  the  roadway  being 
carried  on  little  arches  from  one  to  the  other ;  or  there  may  be 
a  filling-in  of  earth.  It  is  rather  difficult  in  a  stone  or  brick 
bridge  to  say  exactly  what  is  the  load  on  every  portion  of  the 
arch,  but  it  is  guessed  at,  and  a  curve  or  line  of  resistance,  such 
as  P  h  Q,  Fig.  106  (inverted),  drawn.  It  is  shown  in  Art.  368, 
that  in  a  stone  or  brick  arch  it  is  dangerous  to  have  the  arch 
so  thin  that  the  line  P  A  Q  (inverted)  passes  anywhere  outside 
the  middle  third  of  the  arch  ring.  Thus,  in  Fig.  108  we  have  a 
section  of  a  stone  arch,  the 
various  stones  or  voussoirs,  as 
they  are  called,  being  separ- 
ated by  joints  of  mortar  or 
cement.  Now  divide  each 
joint  into  three  equal  parts 
and  draw  two  polygons,  m  m  in 
and  n  n  n,  marking  out  the 
middle  third  of  every  joint. 
Let  us  suppose  we  know  the 
weight  which  each  voussoir 
supports,  including  its  own 
weight  (it  is  usual  to  consider 
the  arch  as  one  foot  deep  at 
right  angles  to  the  paper),  and 
let  these  weights  be  the 
weights  wlt  w2,  etc.,  shown 
in  Fig.  108.  I  have  taken  a 
case  in  which  these  loads  are 
symmetrical  to  right  and  left 
of  the  crown,  Now  draw  the 
force  polygon,  Fig.  109  ;  it 


Fig.  109. 


happens     to    be    all 


in    one 

vertical  line,  the  forces  being  all  vertical.  And  now  we 
come  to  the  drawing  of  our  line  of  resistance,  but  we  are 
stopped  at  the  outset  by  not  knowing  what  is  the  thrust 


164 


APPLIED    MECHANICS. 


at  the  crown  of  the  arch.  The  pull  at  the  middle  of  our 
suspension  bridge  chain  was  quite  definite,  but  the  thrust 
at  the  crown  of  the  arch  may  be  what  we  please,  and  the 
arch  will  remain  stable  if  the  link  polygon  which  we 
draw  never  passes  outside  the  middle  third  of  any  of  the 
joints.*  Suppose  we  draw  any  symmetrical  link  polygon 
to  begin  with,  by  bisecting  a  k  in  H  (Fig.  109),  draw  H  o 
horizontal,  and  take  O  anywhere  we  please,  o  H  will  be  the 
thrust  in  the  crown  of  our  arch,  if  this  link  polygon  is  the 
correct  one.  Join  o  a,  o  1  2,  o  2  3,  etc.  Start  from  any  con- 
venient point  in  w^  Fig.  108,  say  E,  within  the  space  which 
contains  the  middle  thirds  of  all  the  joints.  Draw  E  D,  Fig.  108, 


Fig.  110. 

parallel  to  o  4  5,  Fig.  109 ;  draw  D  c,  C  B,  B  A,  A  p,  in  succes- 
sion parallel  to  the  corresponding  lines  in  Fig.  109,  and  so  also 
for  E  F,  etc.,  to  K  Q.  If  any  of  the  lines  so  drawn  passes  out- 
side the  space  mm,  n n,  we  must  choose  some  other  point  E  to 
begin  at,  and  if  we  find  that  no  choice  of  E  will  allow  the  link 
polygon  to  lie  altogether  within  the  space  mm,  un,  then  we 
must  choose  another  pole,  o,  in  Fig.  109,  until  at  length  we  find, 
as  in  the  figure,  a  link  polygon,  p  E  Q,  which  cuts  within  the 
middle  third  of  every  joint.  The  lengths  of  the  lines  in 
Fig.  109  tell  us  the  forces  acting  at  the  joints  of  Fig.  108. 
Thus  oa,  Fig.  109,  is  the  force  PA,  Fig.  108,  the  resistance 
of  the  abutment  of  the  bridge.  Again,  the  length  of  o  4  5  is 
the  force  acting  in  the  direction  E  D  between  the  stones  E 
and  D. 

130.  Professor  Fuller  has  made  the  work  of  drawing  such  a 
link  polygon  very  easy.     In  case  the  loads  are  all  parallel  to  one 

*  It  is  obvious  also  that  the  link  polygon,  wherever  it  crosses  a  joint, 
must  make  an  angle  so  near  a  right  angle  with  the  joint  that  there  can  be  no 
slipping  or  rupture  by  shearing  there. 


APPLIED    MECHANICS.  165 

another,  it  can  be  shown  that  if  a  number  of  link  polygons  are 
drawn  in  Fig.  108  for  different  lengths,  o  a,  Fig.  109,  then  the 
vertical  distances  between  the  points  A,  B,  c,  D,  etc.,  are  in  the 
same  proportion  in  all  the  link  polygons.*  Beginning  with  the 
first  load  wlt  draw  (Fig.  110)  A  1'  2'  3'  4'  5'  E,  the  half  of  any 
link  polygon  corresponding  to  P  A  B  c  D  E  in  Fig.  108.  Divide  the 
horizontal  A  s  into  any  number  of  equal  parts ;  I  choose  six. 
Erect  perpendiculars  at  A,  1,  2,  3,  4,  5,  s.  Draw  horizontal 
lines  AST  and  from  1',  2',  3',  etc. ;  draw  any  inclined  straight 
line  of  convenient  length,  E  T  ;  draw  vertical  lines  from  T,  1", 
2",  3",  etc.  From  the  points  where  the  verticals  A  A',  1  1',  2  2', 
etc.,  cut  m  m  and  n  n,  draw  horizontals  to  cut  the  correspond- 
ing verticals  from  T  1",  2",  3",  etc.  Join  the  points  so  found  by 
the  curves  m  m"  and  n  n" ;  then,  just  as  the  straight  line  E  T 
represents  a  link  polygon,  m  m"  n  n"  represents  the  area  bound- 
ing the  middle  third  of  all  the  joints,  and  any  link  polygon  will 
be  represented  on  the  right  hand  side  by  a  straight  line.  Now 
draw  a  straight  line  lying  altogether  within  the  space  m  m"  n  n". 
If  you  can  draw  several,  then  draw  that  one  which  is  steepest, 
in  this  case  c  T.  Project  this  over  to  the  left  hand  side,  and 
you  will  find  that  you  have  the  link  polygon,  which  supposes 
the  least  thrust  at  the  keystone.  The  corresponding  force 
polygon  has  its  o  H  less  than  the  o  H  of  A  E  in  the  proportion 
s  0  fco  s'  E.  The  proof  of  this  is  easy. 

In  the  figure  the  possible  line  A'  s'  if  is  so  close  to  A  s  T  that  it  is 
not  shown.  The  proposition  that  the  line  of  resistance  must  lie 
within  the  region  of  middle  thirds  will  be  dealt  with  in  Art.  368. 
Fig.  109  shows  the  amounts  of  thrust  between  the  voussoirs,  and  it 
is  worth  while  comparing  these  forces  with  the  areas  of  the  joints. 
In  Art.  368  the  strength  of  such  joints  is  more  particularly  studied 
It  is  also  worth  while  to  consider  whether  the  line  of  resistance 
may  not  be  made  to  pass  outside  the  region  of  middle  thirds  by  a 
possible  unsymmetrical  load. 

But  we  cannot  give  a  proof  of  the  proposition  on  which  the 
whole  work  depends.  If  any  line  of  resistance  may  be  drawn  to 
pass  inside  the  region  of  middle  thirds,  the  arch  is  stable,  and  the 
steepest  of  such  lines  is  the  actual  line.  We  have  a  lecture  model 
in  which  a  number  of  rounded  blocks  of  wood  CDEFQHIJ 
represent  voussoirs,  the  fixed  parts  A  and  B  being  abutments.  We 
can  load  these  voussoirs  in  all  sorts  of  ways.  When  the  loading  is 
changed,  the  voussoirs  will  be  seen  to  roll  on  one  another  into  new 

*  See  Art.  349.  Proof.  Each  link  polygon  is  really  a  diagram  of  bending 
moment,  supposing  the  loads  were  acting  on  a  horizontal  beam,  and  the  scale 
of  each  diagram  is  proportional  to  the  length  of  O  H. 


166  APPLIED    MECHANIC8. 

positions ;  and  if  the  points  of  contact  are  joined,  we  have  the  line 
of  resistance. 

When  the  loading  is  symmetrical,  loads  on  the  haunches  cause 
the  line  to  rise  at  the  haunches  and  get  lower  at  the  crown,  and  a 
load  on  the  crown  reverses  this  effect.  A  student  may  see  on  this 
model  how  change  of  loading  causes  such  a  yielding  in  the  arch  as 
produces  pressures  at  the  abutments  just  suited  to  equilibrium, 
provided  only  that  a  line  of  resistance  cuts  all  the  joints.  But  we 
must  confess  that  the  sudden  step  in  the  reasoning  from  this  to  the 
mere  statement  of  the  above  proposition  does  not  satisfy  us.  At 
the  actual  plane  joints  of  an  arch,  changes  occur  when  the  load  is 
changed,  but  these  changes  are  very  different  from  the  changes 
that  occur  in  the  model;  and  it  seems  to  us  that  the  only 
legitimate  method  of  study  is  that  of  assuming  that  a  masonry 
arch  behaves  exactly  like  an  arched  rib  of  iron  fixed  at  the  ends,  and 


Fig.  111. 

this  is  studied  in  Art.  380.  The  line  of  resistance  being  found  as 
for  an  iron  arch,  the  masonry  arch  must  be  so  designed  that  this 
line  is  kept  within  the  middle  third. 

In  a  few  cases  in  Germany  an  attempt  has  been  made  to  build 
masonry  arches  hinged  at  the  ends,  and,  indeed,  also  with  a  hinge 
at  the  crown.  This  can  be  done  by  bedding  the  masonry  at  these 
places  upon  iron.  A  quasi  hinge  has  also  been  employed  by 
introducing  plates  of  lead  at  the  two  joints  at  the  springings,  and 
one  at  the  crown,  the  lead  only  extending  over  the  middle  thirds  of 
the  joints. 

131.  Buttresses. — To  find  the  force  which  acts  from  one 
stone  to  another  in  a  buttress,  it  is  necessary  to  know  the  force 
acting  on  every  stone  from  the  outside,  and  also  the  weight  of 
the  stone.  Find  the  resultant  of  these  two  forces  for  each 
stone,  and  draw  the  link  polygon  whose  first  side  is  the  force 
on  the  top  stone.  In  Fig.  112rABKpis  the  link  polygon  so 


APPLIED    MECHANICS 


167 


drawn.  Each  side  of  it  shows  the  resultant  of  the  forces 
acting  at  every  joint,  and  the  length  of  the  corresponding  line 
Fig.  113  shows  its  amount.  Thus  the  resultant  of  the 


in 


forces  acting  at  the  joint  s  T  is  shown  by  the  position  of  the 
line  B  K,  and  its  amount  is  shown  by  the  length  of  the  line  o  t 
in  Fig.  113. 

If  we  see  that  any  of  the  sides  of  the  link  polygon  passes 
outside  the  middle  third  of  the  corresponding  joint  between 
two  stones,  we  know  that  part  of  that  joint  will  be  subjected 


Fig.  112. 


Fig.  118. 


to  tension,  a  condition  to  which  we  suppose  that  a  common 
masonry  joint  ought  not  to  be  subjected.* 

The  student  ought  to  work  examples  in  which,  besides  the 
force  F  acting  on  the  top  stone,  there  are  forces  acting  on  the 
other  stones,  due,  say,  to  the  pressure  of  water  (Art.  173),  or 
to  the  pressure  of  earth  (Art.  293). 

132.  To  return  to  the  hanging  chain.  If  the  total  weight  of  a 
chain  and  all  the  vertical  loads  upon  it  is  w,  and  if  its  ends  are  sup- 
ported at  two  points,  A  and  B,  and  at  these  points  it  makes  angles 
a  and  /3  with  the  horizontal,  the  tensions  there  are  represented  by 
the  two  sides  of  a  triangle,  which  are  parallel  to  the  directions 
there,  the  third  side,  representing  w,  being  a  vertical  line.  Let 
the  student  sketch  examples. 

*  In  many  cases  it  will  be  found  well  to  magnify  all  the  horizontal 
components  of  all  the  forces,  magnifying  the  horizontal  dimensions  of  all  the 
stones  in  the  same  proportion.  In  this  way  the  points  in  which  each  side  of 
the  link  polygon  cuts  each  joint  may  be  found  more  accurately. 


168  APPLIED    MECHANICS. 

The  tension  T  at  A  is  evidently  w  cos.  ft/sin,  (o  +  0),  and  the 
tension  at  B  is  w  cos.  a/sin,  (o  +  )8). 

If  at  B  the  chain  is  horizontal,  then  T  at  A  is  w/sin.  a,  and  the 
tension  at  B  is  w  cot.  a.  Call  this  TO. 

If  x  is  horizontal  distance  of  any  point  in  the  chain  from  B 
(where  the  chain  is  horizontal)  and  y  is  vertical  height  above  B  ;  if 
the  load  on  the  chain  between  B  and  any  point  A  is  wx  where  w  is 
the  load  per  unit  length  of  horizontal  projection  of  the  chain  [here 
I  assume  that  the  load  is  uniformly  spread  horizontally,  and  this  is 
nearly  the  case  in  any  very  flat  chain  or  telegraph  wire],  then,  as 
before,  if  a  is  the  angle  of  inclination  of  the  chain  at  A,  we  have 

TO  =  wx  cot.  a  ;  or,  in  the  language  of  the  calculus,  TO  -~-  =  wx  ; 

and  it  follows  that  the  shape  of  the  chain  is  y  =  £  —  x2  .  .  .  .  (1), 

To 

a  parabola  with  vertical  axis  whose  vertex  is  at  B,  the  lowest  point. 
Given  the  heights  of  the  two  points  of  support  of  such  a  chain 
above  its  lowest  point,  and  also  the  horizontal  distance  between 
them  and  w,  it  is  easy  to  calculate  the  value  of  TO  and  everything 
else. 

EXERCISE. 

In  a  suspension  bridge  of  800  feet  span  and  80  feet  dip,  the  weight  of 
the  platform,  chains,  and  rods,  etc.,  is  2£  tons  per  foot  run,  what  will  be 
the  horizontal  tension  in  each  of  the  two  chains,  and  the  tension  in  each 
at  the  point  of  support  at  the  piers  ?  Am.,  1125  tons  ;  1212  tons. 

133.  In  a  telegraph  wire  of  span  /,  the  points  of  support  being  at 
the  same  level,  if  the  dip  (supposed  to  be  small  compared  with  I)  is  a, 
the  weight  of  wire  being  w  or  wl  nearly,  it  is  easy  to  show  that  the 
tension,  which  I  shall  here  call  P,  is  wfc/Sa  or  w//8«  at  the  lowest 
point,  and  the  tension  elsewhere  is  not  much  greater.  Now  sup- 
pose s,  the  whole  length  of  a  wire,  to  be  b  (1  +  k6  +  ftp)  where 
b  is  its  length  at  the  lowest  temperature,  and  this  we  shall  call 
the  zero  of  temperature,  under  no  tension,  0  being  the  temperature 
Centigrade  at  any  other  time  above  the  zero,  k  the  coefficient  of 
expansion,  and  /3  a  constant  easily  determined  when  one  knows 
Young's  modulus  for  the  material  and  the  cross-section  of  the  wire, 
just  as  w  is  known  from  the  material  and  cross-section.  If  a  is  the 
dip,  find  the  tension  P,  and  find  the  length  of  the  wire.  The 
length  of  a  parabolic  curve  is  easily  found  by  integration,  but 
x  +  2  y2/3  x  gives  tne  length  from  the  origin  to  any  point  with 
sufficient  accuracy  for  such  purposes  as  the  present.  [Check 
this  statement  when  you  have  leisure.]  So  that  length  of  wire  is 

I  +  8  a2/3  l- 

If  an  is  the  dip  at  the  zero  of  temperature  we  are  led  to  the 
result 


where  a  =  a/I,  o0  —  ajl. 

If,  now,  the  span  I  and  w  (the  weight  per  unit  length  of  a  wire) 
are  given,  and  if  PO  be  taken  as  the  greatest  pull  to  which  the  wire 


APPLIED    MECHANICS.  169 

ought  to  be  subjected,  this  will  be  at  the  lowest  temperature, 
which  we  here  call  zero.  Then  o0  =  wl/8  PO. 

Choose  now  some  greater  value  of  o,  and  from  this,  by  (1), 
calculate  0,  and  note  also  that  p  =  wl/8  a ;  so  that  it  is  easy  to 
make  out  a  table  showing  a  and  P  for  various  temperatures. 

This  calculation  may  be  regarded  as  a  mere  exercise  for 
students,  and  yet  there  may  be  occasion  for  its  use  in  practice. 

The  linesman  who  puts  up  telegraph  wire  pays  very  little 
attention  to  the  result  of  such  a  calculation.  He  gives  such  a  dip 
to  his  wire  as  seems  good  to  him,  making  it  somewhat  less  in 
winter  than  he  would  in  summer.  However  much  he  may  have 
killed  the  wire  already,  it  will  permanently  stretch  a  little  more  if 
a  very  cold  night  comes,  without  real  hurt  to  its  material. 

If  a  chain  is  uniform,  and  is  loaded  only  with  its  own  weight, 
and  if  we  desire  an  exact  answer  instead  of  the  above  approximate 
answer  for  flat  chains  or  wires,  we  let  w  =  w  s  where  w  is  the 
weight  of  unit  length  of  the  chain  and  *  is  the  length  of  chain 
,  from  B,  its  lowest  point,  to  any  point  A.  Then  TU  =  w  s  cot.  a 

leads,  by  the  calculus,  to  the  result  y  =  -  L>x/c  +  e^/c) ,    where 

W  c  =  TO,  and  c  is  the  vertical  height  of  B  above  the  origin.  This 
curve  is  called  the  catenary. 

As  we  have  T  =  TO  tan.  a,  it  is  easy  to  find  T.  It  is  also  easy 
to  find  s  in  terms  of  x  and  y.  The  properties  of  this  curve  give 
very  easy  exercises  in  the  calculus,  and  so  they  are  well  known. 

FIDDLE  STRING. — Neglect  the  weight  of  a  string  whose  mass  is  m 
per  unit  length.  When  vibrating  in  a  plane  at  any  place  at  the  distance  x 
from  one  end  let  the  displacement  be  y  from  the  equilibrium  position. 
At  the  ends  of  a  short  length  SI,  the  resultant  force  tending  to  bring 
the  mass  m  Si  .back  to  the  equilibrium  position  is  T.  50  if  80  is  the  angle 
between  the  tangents  at  the  two  ends,  or  T.  81  x  curvature.  The  string 
being  everywhere  very  nearly  straight,  we  may  take  the  curvature  to  be 

- — ^  •  (See  Art.  25.)     This  force,  divided  by  the  mass  m  SI,  is  acceleration, 

JO    . . 

;  and  hence 

I  was  once  at  a  great  loss  to  know  how  to  measure  the  tension  in  each 
span  of  a  tight  telpher  line ;  rods  of  round  f  inch  steel,  each  200  feet  long. 
At  length  I  saw  that  I  could  use  the  fiddle-string  principle  and,  if  the  rod 
is  set  vibrating  in  the  simplest  manner  in  a  vertical  plane,  the  time  of  jt 

complete   (up  and  down)  oscillation  is  very  nearly,  t  seconds  =  21  A/  -. 

This  principle,  borrowed  from  acoustics  by  engineering,  enabled  labourers 
to  test  the  line,  span  by  span,  every  day.  If  there  were  less  than  17,  or 
more  than  19  complete  oscillations  in  a  quarter  minute,  the  span  needed 
to  be  tightened  or  slackened. 

G* 


170 
CHAPTER  IX. 

HYDRAULIC     MACHINES. 

134.  Hydraulics. — Hydraulic  machines  are  very  wonderful 
to  people  who  observe  their  action  for  the  first  time.     Ancient 
drawings  show  armies  of  slaves  dragging  on  ropes  to  lift  a  single 
weight.    Three  hundred  years  ago,  Fontana  raised  an  obelisk  at 
Rome  with  40  capstans,  worked  by  960  men  and  75  horses.     A 
few  years  ago  a  similar  obelisk  was  raised  in  London  by  four 
little  100-ton  hydraulic  jacks,  each  of  which  can  be  worked  by  one 
man.     Large  modern  guns  would  be  almost  impossible  to  work 
with  any  other  machinery.     If  you  go  to  any  large  docks  you 
will  see  how,  by  the  manipulation  of  a  few  handles,  a  boy  can 
remove  heavy  objects  rapidly  from  ships  and  place  them  on  the 
dock  by  means  of  an  hydraulic  crane.     Visit  any  large  steel 
works,   and  you  will  see  great  armour-plates  and  Bessemer 
converters  and  their  appliances  passed  about  nearly  as  readily 
as  small  objects  are  moved  by  blacksmiths  and  moulders.     The 
steam-hammer,  powerful  as  it  is,  is  giving  place  to  hydraulic 
forging  and  squeezing  machinery,  because  the  new  forces  are 
enormously  greater,  and  their  effects  can  be  more  uniformly 
distributed  over   large  masses  of   metal,   leaving   them  more 
homogeneous.     Visit    the  Victoria  Docks,  and  you   will    see 
vessels  of  3,000  tons  raised  out  of  the  water  on  a  floating  grid- 
iron and  towed  off  for  repairs.     Visit  the  River  Weaver,  in 
Cheshire,  and  you  will  see  sections  of  a  canal  rising  and  falling 
50  feet  with  canal  boats,  which  are  no  longer  delayed  for  hours 
in  floating  through  a  flight  of  locks.     Instead  of  bringing  great 
iron  girders  near  a  riveting-machine,  we  now  take  little  riveting 
machines  to  the  girder,  and  work  them  in  any  position  through 
small  flexible  pipes  from  a  distant  steam-engine  and  pumps. 

135.  Hydraulic  Press. — An  hydraulic  press  is  a  machine 
which    enables  large  weights  to  be  lifted  or  great  pressures 
exerted,  but  in  which,  instead  of  levers  and  wheels,  we  use  water 
to  transmit  the  energy.     In  Fig.  114  we  have  rams  of  large 
and  small  cross  sections  A  and  a  square  inches,  and  weights 
(neglecting  their  own  weights)  R  and  E.     Because  the  vessel  c  is 
full  of  water  which  cannot  escape  past  the  rams  on  account  of 
two  leather  collars  or  other  packing  at  D  and  G,  the  velocity 
ratio  is  A/a.    Thus  if  A/a  is   100,  and  if  a  falls  through  100 


APPLIED    MECHANICS. 


171 


Pig.  114. 


inches  it  displaces  a  x  100  cubic  inches  of  water.  We  assume 
that  the  water  cannot  escape,  and  that  all  the  room  needed  is 
procured  by  the  lifting  of  A;  then  A  must  lift  one  inch  to  make 
the  necessary  room. 

In  actual  specimens,  instead  of 
one  large  vessel  c,  we  usually  have 
vessels  round  A  and  a  slightly  larger 
than  themselves,  called  a  press  and 
pump-barrel  with  a  pipe  connecting 
them.  The  friction  at  a  leather  collar 
is  sometimes  as  little  as  one  per  cent, 
of  the  weight  on  the  ram,  and  some- 
times as  much  as  5  per  cent.  Notice 
that  the  amount  of  the  axial  force  R 
which  may  be  exerted,  neglecting 
friction  and  weights  of  water,  does  not 
in  any  way  depend  upon  the  shape  of 
the  end  of  its  ram,  or  where  it  is,  or  the  direction  of  the 
ram's  motion, 

136.  The  internal  construction  of  an  hydraulic  press  is  shown 
in  Fig.  115.  Three  men  press,  each  with  a  force,  say,  of  60  Ibs., 
on  the  end  of  the  lever  G,  whose  mechanical  advantage  is,  say 
20,  and  hence  the  plunger,  Ea,  is  pressed  downwards  with  a 
force  of  3,600  Ibs.  Just  consider  for  an  instant  what  is  the 
condition  of  things  before  E  moves.  There  is  a  ram,  R,  which 
carries  a  heavy  weight,  the  weight  to  be  lifted.  Observe  that 
this  ram  is  wanting  to  fall,  but  it  can  only  fall  into  the  vessel  D. 
Now  the  space  between  the  vessel,  or  press,  and  the  ram,  and 
all  the  space  in  the  tubes,  T,  is  filled  with  water  which  has  no 
means  of  getting  away.  It  might  get  away  by  the  little  valve 
F,  but  that  valve  can  only  open  upwards,  and  the  more  the 
water  tries  to  escape,  the  more  it  really  closes  the  valve,  just  as 
a  closely-packed  crowd  in  a  panic  keeps  an  inward-opening 
door  closed,  by  which  they  might  otherwise  escape  from  a 
theatre.  There  is  no  escape  for  the  water  on  the  pump  side ; 
there  is  just  as  little  on  the  other  side,  for  you  see  that  the 
water,  if  it  escapes  into  the  the  space  N,  finds  that  it  has  still 
to  get  past  the  leather  collar,  which  is,  however,  so  placed  and 
shaped  that  the  greater  the  water  pressure  the  tighter  the 
leather  fits  the  ram.  Fig.  117  shows  such  a  leather  collar. 
There  is  then  no  escape  for  the  water,  and  when  this  is  the  case, 
no  matter  what  weight  is  placed  on  the  top  of  the  ram,  it 


172 


APPLIED    MECHANICS. 


cannot  fall.  The  falling  of  the  ram  would  mean  some  escape 
of  the  water ;  but  as  there  is  no  escape  for  the  water,  the  ram 
will  fall  no  more  than  if  it  were  supported  on  some  quite  rigid 
material. 

137.  I  have  been  supposing  that  a  certain  quantity  of  water 
will  absolutely  refuse  to  occupy  a  smaller  space,  but  this  is  not 
quite  correct.  We  know  that  if  it  were  air  that  filled  the  space  N, 


--B 


Fig.  115. 

instead  of  water,  and  there  were  no  escape  for  it,  the  ram  would 
fall  when  a  greater  weight  was  placed  on  it ;  for  although  the 
air  cannot  escape,  it  is  contented  to  occupy  a  smaller  bulk. 
I  have  been  supposing  that  this  does  not  occur  in  water,  and 
that  water  will  refuse  to  go  into  a  smaller  space,  whatever  the 
pressure.  This  was  an  old  notion  which  people  deduced  from 
the  famous  Florentine  experiment.  A  hollow  globe  of  gold 
was  quite  filled  with  water,  and  was  hermetically  sealed.  It 
was  then  beaten  to  diminish  its  cubic  contents,  and  the  result 
was  that  drops  of  water  made  their  appearance  on  the  surface, 
having  oozed  out  through  the  pores  in  the  gold  rather  than 
submit  to  the  lessening  of  the  total  bulk. 


APPLIED    MECHANICS.  173 

I  am  told  that  a  lecturer  at  Chatham  subjected  the  water 
inside  a  cast-iron  shflll  to  so  much  pressure  that  it  came  through 
the  pores  of  the  iron,  and  appeared  as  a  fine  spray  or  mist  on 
the  outside,  and  soon  afterwards  the  shell  burst,  or,  rather,  fell 
gently  in  pieces.  But  with  a  piezometer  it  is  easy  to  show 
that  water  and  all  other  substances  will  submit  to  a  diminu- 
tion of  their  bulk  when  subjected  to  a  pressure ;  and  we  find 
that  this  diminution  for  water  is  l-20,000th  of  its  total  bulk 
for  a  change  of  pressure  of  one  atmosphere,  or  l-70th  of  its 
volume  for  a  pressure  of  2  tons  per  square  inch,  such  as  we 
find  in  hydraulic  presses — that  is,  70  cubic  inches  become  69. 
Now  this  diminution  in  bulk  is  far  too  insignificant  to  be 
of  much,  practical  importance  in  hydraulic  machines,  and 
we  may  take  it  for  granted  that  whatever  weight  we  place 
on  the  ram,  it  will  not  perceptibly  fall,  because  the  water 
refuses  to  become  smaller  in  bulk,  and  because  it  cannot 
escape  anywhere. 

We  understand  that  it  tries  to  get  away  ;  it  is  trying  to 
burst  the  thick  cast-iron  press ;  it  is  trying  to  burst  the  pipes 
and  the  purnp.  Before  it  will  burst  the  pump,  it  will  open  the 
safety  valve  H,  and  escape  ;  but  we  can  assume  for  the  present 
that  the  pressure  never  reaches  the  bursting  pressure  of  the 
arrangement. 

Now  the  labourer  acts  on  the  lever,  forcing  down  the 
plunger  E  a.  The  water  in  the  pump-barrel  is  just  like  the  water 
in  the  press ;  it  tries  to  escape,  it  tries  to  burst  the  pump-barrel, 
it  resists  the  motion  of  the  plunger.  It  tries  to  escape  through 
the  valves  F  and  H,  and  it  will  open  the  valve  p,  and  pass 
through,  if  the  labourer  acts  sufficiently  on  his  lever  ;  but  if  the 
water  passes  through  p,  the  ram  A  R  must  rise,  however  great 
the  weight  may  be  that  is  pressing  it  down.  The  question  is, 
then,  what  force  on  the  plunger  E  a  is  sufficiently  great  to  cause 
motion — that  is,  to  cause  the  water  to  pass  through  the  valve 
p,  and  so  make  the  ram  A  R  rise  ? 

Suppose  that  the  plunger  E  a  is  one  square  inch  in  section, 
and  that  one  inch  more  of  its  length  is  forced  into  the  pump ; 
evidently  the  metal  takes  up  the  place  of  an  equal  bulk  of  water, 
or  one  cubic  inch.  This  cubic  inch  of  water  has  found  one 
cubic  inch  of  room  for  itself  somewhere  else.  As  we  suppose 
no  greater  compression  of  the  water,  and  no  yielding  of  the 
sides  of  the  press,  it  is  evident  that  one  cubic  inch  of  the  ram 
must  leave  the  press  to  give  the  water  the  space  it  must  have. 


174  APPLIED   MECHANICS. 

Now,  if  the  ram  is  100  square  inches  in  section,  then  l-100th 
of  an  inch  of  its  length  contains  one  cubic  inch  in  volume.  If 
the  ram  lifts  through  the  distance  1-1 00th  of  an  inch,  it  will 
therefore  leave  one  cubic  inch  of  room  behind  it. 

We  are  not  concerned  with  the  shapes  of  the  ends  of  the 
plunger  and  ram  ;  we  know  that  if  one  more  inch  of  the 
plunger  enters  the  water,  1-1 00th  of  an  inch  of  the  ram  must 
leave  the  water;  that  is,  the  relative  speeds  of  plunger  and 
ram  are  as  one  inch  to  l-100th  of  an  inch,  or  as  100  to  1.  The 
plunger  must  move  100  times  as  quickly  as  the  ram,  and  by  the 
law  of  work,  if  there  was  no  friction,  a  force  of  one  pound  on 
the  plunger  would  balance  a  force  of  100  pounds  on  the  ram. 
We  know  the  mechanical  advantage,  then,  if  there  were  no 
friction,  in  the  portion  of  this  machine  from  plunger  to  ram,  if 
we  know  how  many  times  greater  is  the  area  of  the  ram  than 
the  area  of  the  plunger. 

138.  Different  experimenters  give  different  results  as  to  the 
loss  of  energy,  and  therefore  of  mechanical  ad  vantage,  in  friction. 
Rankine  states  that  there  is  20  per  cent,  of  loss.  Mr.  Hick 
found  results  less  than  one-tenth  of  this.  It  is  known  that  in 
accumulators,  the  pressure  which  is  great  enough  to  lift  a 
load,  being,  say  1,010  Ibs.  per  square  inch,  if  the  pressure  is 
reduced  to  990,  the  same  load  will  fall ;  thus  only  one  per  cent, 
of  the  energy  seems  to  be  wasted  in  friction  at  the  leather 
collar,  when  the  motion  is  slow.  It  is  probable  that  there  is 
always  less  than  20  per  cent,  of  loss  of  energy  altogether  in  large 
hand- worked  presses.  As  an  instance  of  this  greater  efficiency 
of  hydraulic  machines,  a  platform  weighing  12  tons  had  to  be 
lifted,  and  a  little  hydraulic  jack  was  placed  under  one  corner, 
a  screw  jack  under  the  other.  One  man  was  told  off  to  work 
the  hydraulic  jack  and  three  men  to  the  screw  jack.  The  man 
at  the  hydraulic  jack,  with  one  hand  in  his  pocket,  would  pump 
a  few  strokes  and  then  quietly  wait  a  little,  whereas  the  three 
men  were  hard  at  work  all  the  time. 

My  students  in  Japan  used  to  employ  a  little  hydraulic 
press  to  crush  bricks,  stones,  and  wood,  and  it  was  roughly 
assumed  that  the  friction  at  the  glands  was  insignificant.  Of 
course,  we  only  made  this  assumption  when  we  wanted  rapidly 
to  get  a  rough  idea  of  the  relative  strengths  of  materials  to 
resist  crushing.  Still,  it  was  a  sort  of  thing  that  we  should 
not  have  been  able  to  do  with  any  other  except  a  lever 
machine,  and  I  am  inclined  to  think  that  there  is  more 


APPLIED    MECHANICS. 


176 


inaccuracy  with  many  lever-testing  machines  than  there  was 
with  my  hydraulic  press. 

Example. — A  three-ton  hydraulic  jack,  like  Fig.  116,  not  in 
very  good  order,  was  examined  in  my  laboratory ;  weights  A 
were  hung  at  the  end  of  a^  lever,  and  we  calculated  R  the 
equivalent  weight  which  the  jack  L  was  actually  supporting. 
The  handle  was  replaced  by  a  wooden  sector,  to  which  the 
weight  B  gave  a  slow  motion,  after  one  or  two  strokes  had 
been  made  by  the  experimenter  to  get  things  into  a  steady 
condition.  The  weight  B,  which  would  slowly  overcome  a 
given  R  being  noted,  the  experiment  was  repeated  many 


Fig.  116. 

times  with  other  loads,  and  the  values  of  R  and  B  tabulated 
and  plotted  on  squared  paper.  We  found  the  result : 
B  =  -024  R  +  2-2  where  B  is  the  effort,  in  pounds,  applied 
at  the  end  of  the  handle,  and  R,  in  pounds,  is  the  load  on 
the  jack,  not  including  the  weight  of  the  ram  itself.  The 
mechanical  advantage  of  the  lever  was  1475  ;  the  ram  was 
2  inches  in  diameter  and  the  pump  plunger  1  inch,  so  that  the 
total  velocity  ratio  was  14-75  x  4,  or  59.  .  If  there  had  been 
no  friction  the  effort,  B1,  would  have  been  R  -r  59 ;  hence  the 
efficiency  is — 


B 


"59 

1 


1-42  R  +  130 


1-42  +  130 

R 


Thus  when  R  is  3  tons,  or  6,720  Ibs.,  the  efficiency  is  0-69. 
There  can  be  no  doubt  that  the  loss  of  energy  mainly  occurs  by 


176 


APPLIED    MECHANICS. 


solid  friction,  and  so  we  are  led  to  ask,  where  does  friction 
occur  in  the  hand- worked  hydraulic  press  1     It  occurs  at 

1.  The  rubbing  surface  at  the  fulcrum  of  the  lever.     This 

is  the  friction  of  solids. 

2.  The  rubbing  surfaces  of  the  two  glands;  one,  that  of  the 

plunger,  the  other,  that  of  the  ram.     Here,  again,  we 
have  friction,  as  if  between  solids. 

3.  Everywhere  in  the  water  where  there  is  motion.     This 

is  fluid  friction,  which  is  quite  different  from  that  of 
solids. 

139.  Figs.  117,  117A,  show  sections  of  the  leather  collar 
used  in  presses.    It  is  made  from  the  best  leather,  softened  in  hot 


Fig.  117. 

water,  and  pressed  between  cast-iron  moulds  to  its  present 
shape,  the  pressure  gradually  increasing.  It  is  then  left  for 
several  days  under  pressure  in  the  mould.  When  it  is  in  its 
place  in  the  press  the  water  gets  behind  the  leather,  and  presses 
it  tightly  against  the  ram.  The  friction  seems  mainly  to  occur 
at  the  part  A,  and  increases  as  the  pressure  of  the  water 
increases.  There  does  not  seem  to  be  much  friction  at  the 
portion  between  A  and  B,  and  the  efficiency  of  the  press  is 
but  little  altered  by  making  A  B  greater  or  less.  It  is, 
however,  asserted  by  some  makers  of  presses  that  the  dis- 
tance A  B  is  of  importance ;  but  I  rather  think  that  this  is 
on  account  of  deterioration.  The  part  B  A  is  constantly  being 
in  states  of  tension  and  compression,  and  is  liable  to  crack. 
When  the  leather  deteriorates,  much  time  is  wasted  in 
renewing  it. 


OF 

MECHANICS.  177 


140.  Hemp-packing  is  invariably  used  instead  of  the  leather 
collar  at  low  pressures  ;    some  manufacturers  never  use  hemp 
or  cotton  when  the  water  has  a  greater  pressure  than  700  Ibs. 
per  square  inch,  but  others  use  hemp  to  2,000  Ibs.  per  square 
inch. 

Our  subject  is  too  large  to  allow  me  to  enter  into  many 
details  as  to  the  advantages  and  disadvantages  of  leather, 
india-rubber,  and  gutta-percha  for  packing  purposes.  There  is 
a  great  divergence  of  opinion,  and  I  believe  that  much  of  the 
evidence  against  one  or  another  material  is  based  on  the  bad 
preparation  of  the  materials  against  which  the  evidence  is 
given. 

141.  In  Fig.  117s  you  will  see  the  form  of  india-rubber  cup 
used  in  hydraulic  plungers,  jacks,  and  bears.     It  is  moulded  in 
the  form  shown,  and  is  fastened  to  the  end  of  the  ram  by  means 
of  a  screwed  bolt  and  metal  washer,  as  shown  in  the  figures. 
The  water  pressure  keeps  it  tight  against  the  cylinder  at  the 
edge,  and  this  is  where  the  friction  occurs. 

The  quasi-solid  friction  between  lubricated  leather  or  hemp 
or  rubber  and  metal,  follows  the  laws  of  solid  friction,  and 
we  have  now  described  what  friction  of  this  kind  occurs  in 
hydraulic  presses.  The  remaining  source  of  loss  of  energy  is 
the  fluid  friction. 

The  flow  of  water  even  in  the  narrow  passages  past  the 
valves  may  be  very  slow,  and  therefore  the  fluid  friction  may 
be  as  small  as  we  please.  So  small,  indeed,  is  the  fluid  velocity 
that  when  oil  is  used  the  loss  of  energy  is  much  the  same. 
Probably  honey  or  tar  would  not  give  very  different  results  ; 
but  the  more  viscous  mixtures  of  tar  and  pitch  would  be 
unsuitable,  because  in  these  the  fluid  friction,  at  even  such 
small  velocities  as  we  have  to  consider,  would  be  considerable. 
Even  solid  pitch  would,  however,  act  as  a  fluid,  and  might  give 
the  same  mechanical  advantage  as  water  if  we  made  only  one 
stroke  of  the  plunger  in  a  month. 

142.  In  applying  the  law  of  work  to  our  machines,  equating 
the  energy  given  to,  and  the  energy  given  out  by,  each  machine, 
we  assume  no  store  of  energy  in  the  machine.     In  the  press,  even 
if  we  disregard  the  compression  of  the  water  and  the  elastic 
yielding  of  the  press,  we  must  remember  that  the  ram  itself  is 
lifted,  and  also  that  some  water  is  lifted  in  level.     The  lifting 
of  the  ram  is  easy  to  take  into  account  ;  consider  it  part  of  the 
weight  to  be  lifted.      In  ordinary  presses  it  is  of  no  great 


178 


APPLIED    MECHANICS. 


ff 


consequence.  But  when  the  dead  weight  of  the  ram  and  other 
things  is  greater  than  the  weight  lifted  usefully,  as  in  ware- 
house and  hotel  hoists,  it  cannot  be  neglected,  and  it  is  usually 
easily  balanced.  In  very  high  hoists  the  weight  of  water 

which  changes  its  level 
becomes  important.  (See 
Art.  169.) 

143.  Fig.  118  shows  a 
section  of  a  lifting  jack 
with  its  ram  B,  lifting  any 
weight  which  may  rest  on 
J  or  i.  Notice  how  the 
ram  R  fits  the  press  M,  and 
is  made  water-tight  by  the 
india-rubber  dish  L.  A 
handle  or  lever  is  attached 
at  H,  and  when  worked 
causes  great  pressure  on 
a  projection  or  cam  K, 
which  communicates  with 
the  plunger  Vj  of  the 
pump,  and  gives  it  a  down- 
ward motion,  pressing  the 
water  in  the  pump-cham- 
ber through  the  valve  T2 
and  the  passage  in  R  to 
the  press  M,  which  rests 
on  the  ground.  A  small 
pressure  on  the  lever  thus 
forces  more  and  more 
water  into  the  press,  and 
necessitates  the  upward 
movement  of  the  ram  R. 
The  upward  motion  of  the 
lever  causes  a  partial 
Fis- 118-  vacuum  in  the  pump- 

chamber   as   the    plunger 

vl  is  withdrawn,  and  the  pressure  of  the  air  and  liquid  in 
the  cistern  c  overcomes  the  resistance  of  the  spring  on  the 
inlet  valve  vp  which  is  really  a  part  of  the  pump  plunger, 
and  thus  effects  a  passage  for  the  water  into  pump-chamber. 
During  the  downward  stroke  the  previously-described  operation 


APPLIED    MECHANICS. 


179 


is  repeated.  If  we  want  to  lower  the  weight  we  open 
the  lowering  screw  s,  and  allow  the  water  to  return  from  the 
press  M  to  the  cistern  c,  the  ram  falling. 

144.  The  punching-bear,  shown  in  section  in  Fig.  119,  is 
similar  in  construction.  Theplunger,  pump,  valves,  etc.,  are  much 
the  same  as  in  the  last  case.     The  upper  part  of  the  stout  ram  F 
terminates  in  an  india- 
rubber  dish,  which  is 

fastened  by  the  washer 
and  a  bolt  passing 
through  the  middle 
of  the  washer  into  the 
ram. 

As  we  work  the 
top  lever  A,  the  ram 
p  holding  the  punch 
is  pressed  down  by 
water  through  a  valve 
arrangement  exactly 
similar  to  that  in  the 
lifting  jack. 

145.  We  are  now 
well  aware  of  the  way 
in  which  the  water  in 
these   machines   acts. 
It    is    nearly    incom- 
pressible, and  tries  to 
find  an  outlet  in  every 
direction.       Consider 
what  occurs  from  par- 
ticle to  particle  of  the      .  Fig.  119. 
water.     Each  particle 

presses  on  all  its  neighbours,  because  they  all  press  in  upon 
it,  and  it  presses  equally  in  every  direction. 

Wherever  the  water  comes  in  contact  with  a  solid  surface, 
it  presses  normally  against  the  surface.  There  can  be  no  such 
thing  as  oblique  pressure  in  water  when  the  water  is  at  rest,  for 
oblique  pressure  means  a  force  partly  along  the  surface,  and 
this  would  imply  some  frictional  resistance  to  sliding,  which  we 
know  cannot  occur  in  water  at  rest.  It  is  evident  from  our 
discussion  of  the  hydraulic  press  as  a  machine  in  which  there 
is  no  store  of  energy,  that  on  every  square  inch  of  the  solid 


180 


APPLIED    MECHANICS. 


surface  touched  by  the  water  there  must  be  the  same  force 
acting,  the  water  tending  to  escape  everywhere ;  and  when  we 
consider  the  whole  case  mathematically,  we  find  that  every  little 
interface  separating  any  two  portions  of  water  is  acted  on  by 
this  same  pressure  per  square  inch. 

146.  We  shall  be  perfectly  safe  in  all  our  notions  of  fluid 
pressure  if  we  consider  each  particle  of  water  to  be  a  very  small 
being,  greased  all  over,  so  that  it  cannot  possibly  resist  sliding 
past  its  neighbours.  It  can  press  normally  against  a  wall  or 
against  any  surface,  but  there  cannot  possibly  be  a  tangential 
or  fractional  pressure  between  it  and  a  wall, 
because  it  is  well  greased.  All  the  water  is 
trying  to  escape,  and  the  total  pressure  on 
any  surface  is  evidently  proportional  to  the 
number  of  water  particles  pressing  against  the 
surface.  Hence,  if  we  have  a  piston  A  and 
a  piston  B  (Fig.  120),  the  total  pressures  on  A 
and  B  are  simply  proportional  to  the  areas  of 
the  cylindric  tubes  in  which  they  can  move. 
Pig.  120.  Evidently  it  is  of  no  importance  whether  the 

inner  surface  of  a  piston  has  projections  or  not. 
Thus  there  is  the  same  total  vertical  pressure  on  piston  M  and 
on  piston  N,  Fig.  121,  if  the  cross-sectional  areas  of  their 
cylinders  are  the  same. 

Everything  depends  on  this :  will  they  leave  the  same  empty 
space  behind  them  if  each  of  them  moves  one  inch1? — and  we 


Fig.  121. 

know  that  in  each  case  the  empty  space  is  simply  the  volume 
of  one  inch  of  length  of  the  cylinder. 

147.  In  the  same  way,  although  the  end  of  the  ram  may 
be  curved,  as  in  Fig.  122,  and  is  therefore  being  acted  on  by  a 
series  of  pressures  normally  to  the  curved  surface,  as  in  the 
figure,  it  is  easy  to  show  that  the  resultant  action  of  these  is  in 
the  direction  A  B,  and  is  really  the  same  as  if  the  ram  had  a 


APPLIED    MECHANICS. 


181 


Fig.  122. 


flat  end.     Every  one  of  these  forces  has  a  horizontal  tendency, 

more  or  less,  and  when  we  leave  out  of  account  these  horizontal 

actions,  we  get  the  same  vertical  result  for  all 

shapes  of  ends. 

You  will  understand  this  better,  perhaps, 

if  we  consider  a  vessel,  A,  B,  c  (Fig.  123),  to 

be  filled  with  fluid  at  such  a  great  pressure 

that  we  can  neglect  the  pressure  due  to  the 

fluid's  own  weight.     We  are  assuming  now 

that  we  can  do  this  in  the  hydraulic  press  and 

in  steam  boilers.    The  pressure  on  the  surface 

everywhere  is  shown  by  the  arrows. 

It  is  evident  that  the  total  horizontal  force  on  the  curved 

surface  A  c  B  is  exactly  equal  and  opposite  to  the  total  hori- 
zontal force  on  the  flat  surface  A  B,  because  if  there  were  on 
the  whole  more  force  on  one  than  the  other,  the 
vessel  would  move  bodily,  an  idea  which  is  absurd. 
Hence,  when  we  want  to  find  the  total  horizontal 
force  on  the  curved  surface,  we  never  dream  of 
going  into  the  long  calculation  which  you  might 
think  necessary,  for  it  is  simply  equal  to  the  area 
in  square  inches  of  the  flat  surface  A  B,  multiplied 
by  the  pressure  per  square  inch. 

148.  Suppose  we  want  to  find  the  horizontal  burst- 
ing tendency  of  the  egg-ended  boiler  M  N,  Fig.  124, 

that  is,  say,  the  force  tending  to  burst  it  by  direct  pull  of  the 

iron  at  the  section  A  B — we  do  not  trouble  ourselves  with  the 

shape   of    the   boiler   anywhere   ex- 
cept at  A  B  itself.    The  bursting  force 

is  the  inside  area  of  A  B  in  square 

inches,  multiplied  into  the  pressure 

per  square  inch.  The  area  of  the  iron 

in  the  section  A  B  is  exposed  to  this 

pull.     These  are  the  two  important 

facts   to  be  remembered.     Consider 

any  section  whatsoever  of   a  boiler, 

or  ram,   or  pipe.       Remember   that 

the  fluid   pressure  is  calculated  over   the   whole   area.     The 

resistance   of    the   iron    is   only    calculated    over   the   actual 

sectional  area  of  the  metal. 

Thus,  if  we  want  to  find  the  tendency  to  burst  along  such 

a  section  as  A  B,  we  take  the  total  inside  area  of  the  section 


Fig.  123. 


182  APPLIED    MECHANICS. 

multiplied  by  the  pressure,  and  this  is  equal  to  the  stress 
in  the  iron  all  along  this  section,  multiplied  by  the  whole 
sectional  area  of  the  iron. 

In  a  cylindric  boiler  or  press,  which  is  everywhere  of  the 
same  thickness,  it  is  easy  to  show  that  if  we  neglect  the  effect 
of  the  ends,  the  tendency  to  burst  laterally  is  twice  as  great 
as  the  tendency  to  burst  endwise. 

Thus  in  the  endwise  bursting,  if  p  is  the  bursting  pres- 
sure in  pounds  per  square  inch,  f  the  tensile  strength  of  the 
material  in  pounds  per  square  inch,  r  the  radius  of  the  boiler, 
and  t  the  thickness  of  metal,  the  total  force  tending  to  produce 
bursting  is  the  area  of  the  circular  cross  section,  3 '14  r2,  multi- 
plied by  p,  and  the  total  force  resisting  fracture  is  the  circum- 
ference of  the  circular  cross  section,  6 '28  r,  multiplied  by  t  and 
by/  Hence — 

3-14  r*p  =  6-28  rft, 

or  the  bursting  pressure  p  =  2ft  +  r. 

Again,  if  a  boiler  is  I  inches  long,  the  total  force  tending  to 
burst  the  boiler  laterally  is  the  area  I  times  the  diameter  of  the 
boiler,  multiplied  by  p,  or  2  r  I  />,  and  the  total  force  resisting 
fracture,  if  we  neglect  the  ends,  is  the  area  of  the  iron  2  I  tt 
multiplied  by/  Hence — 

2  rip  =  2  ltft 
or  the  bursting  pressure  p  =ft  -5-  r. 

Hence  it  would  take  twice  as  much  pressure  to  burst  the 
boiler  if  we  assumed  it  to  burst  endwise.  We  always  calculate 
the  strength  of  a  pipe  or  boiler  on  the  second  assumption 
therefore,  and  we  have  the  rule  : — The  bursting  pressure  in 
pounds  per  square  inch  is  equal  to  the  tensile  strength  of  the 
metal  in  pounds  per  square  inch,  multiplied  by  the  thickness 
of  the  metal  in  inches,  divided  by  the  radius  of  the  boiler  or 
pipe  in  inches. 

149.  When  the  press  is  thick,  as  we  find  it  in  an  hydraulic 
machine,  it  is  rather  more  difficult  to  calculate  the  bursting 
pressure,  because  the  tensile  strain  is  not  distributed  uniformly 
over  the  section  at  which  there  is  a  tendency  for  rupture  to 
occur.  The  inside  portions  of  the  metal  near  the  water  are 
subjected  to  more  pulling  forces  than  the  outer  portions. 

The  mathematical  part  of  this  subject  will  come  before  us 
in  Art  275.  The  subject  is  important  and  needs  to  be  intro- 


APPLIED    MECHANICS.  185 

duced  here  also.  The  result  of  the  mathematical  investigation 
is  this : — Neglecting  the  strength  of  the  ends,  the  bursting 
pressure,  multiplied  by  the  sum  of  the  areas  of  the  outer  and 
inner  circles  of  the  cross  section  of  the  cylinder,  is  equal  to  the 
tensile  stress  which  the  metal  will  stand  multiplied  by  the  area 
of  the  metal  exposed  in  such  a  cross  section.  This  rule  you 
will  find  applicable  to  thin  cylinders  as  well ;  it  is  the  same 
rule  as  the  one  already  given  for  thin  cylinders.  You  see  that 
if  the  fluid  pressure  is  equal  to  the  tensile  strength  of  the  metal, 
no  thickness  of  the  cylinder  can  prevent  its  bursting. 

The  investigation  might  perhaps  have  some  use  if  iron 
when  it  leaves  the  foundry  had  all  through  its  thickness  the 
sama  qualities,  and  a  perfect  absence  from  strain.  It  is  good  to 
remember  that  an  iron  casting  when  it  leaves  the  foundry, 
although  not  quite  so  curiously  strained  as  a  Prince  Rupert's 
drop  or  toughened  glass,  is  yet  in  a  state  of  strain  which  we 
know  very  little  about. 

The  above  investigation  shows  that  in  a  thick  cylinder  there 
is  an  exceedingly  great  difference  in  the  tensions  of  the  inner  and 
outer  portions  of  the  metal  of  a  gun  or  hydraulic  press.  For 
the  purpose  of  producing  a  more  uniform  state  of  stress,  the 
inside  portions  of  the  metal  are  often  chilled  ;  that  is,  the  metal 
in  the  inside  is  very  quickly  cooled  after  it  is  cast.  The  result 
is  that  these  portions  are  virtually  in  a  state  of  great  compres- 
sion, and  hence,  when  the  regular  strain  occurs  through  water 
pressure,  there  is  a  considerable  tensile  yielding  in  the  inner 
portions  of  metal  before  the  bursting  tensile  stress  is  reached 
there ;  and  hence,  with  a  proper  amount  of  chilling,  it  is 
possible  to  have  the  tensile  stress  in  the  metal  nearly  uniform 
when  fracture  is  ready  to  occur.  In  such  a  case  as  this,  the 
calculation  of  the  bursting  pressure  of  a  press  is  just  the  same 
as  if  it  were  thin.  We  shall  return  to  this  subject  in  Art.  275. 

It  will  sometimes  surprise  a  student  to  see  how  much 
pressure  is  withstood  by  a  wooden  pipe  wound  round  its  outside 
with  hoop-iron,  or  a  hollow  cylindric  masonry  or  concrete 
reservoir  with  strong  tightened  chains  or  rings  of  iron  on  its 
outside.  [The  engineer  need  not  fear  deterioration  of  iron 
imbedded  in  cement.  My  brother,  who  has  great  experience, 
tells  me  that  burying  even  one  end  of  a  bar  of  iron  in  cement 
has  the  power  of  preserving  all  of  it  from  ordinary  weathering 
or  galvanic  action.] 

150.  Cast  steel  is  now  getting  common  for  the  cylinders  of 


184 


APPLIED    MECHANICS. 


small  jacks  and  bears.  The  large  presses  used  in  warehouses 
are,  however,  still  made  of  loam-moulded  cast  iron,  as  all  presses 
used  to  be  in  my  apprenticeship  days.  But  in  a  great  deal  ef 
large  press- work  cast  steel  has  come  largely  into  fashion,  its 
tensile  strength  being  so  much  greater  than  the  strength  of  cast 
iron  as  to  make  a  most  extraordinary  difference  in  the  outside 


Pig.  125. 

sizes  of  the  cylinders.  Thus,  for  example,  in  a  certain  cotton - 
press,  the  use  of  steel  enabled  three  cylinders  to  be  placed  side 
by  side  in  a  space  which  would  only  have  allowed  one  to  be 
used  if  it  had  been  made  of  cast  iron. 

Fig.  125  is  a  drawing  of  an  hydraulic  press,  which  gives  a 
fairly  good  idea  of  the  press  used  in  warehouses  for  bales  of 
linen  and  Manchester  goods,  linen-yarn,  etc.  The  ram  is  usually 
about  10  inches  in  diameter,  and  the  working  pressure  of  the 


APPLIED    MECHANICS.  18f) 

water  from  2  to  3  tons  per  square  inch.  As  the  area  of  the 
cross  section  of  the  ram  is  78  square  inches,  this  means  a  total 
pressure  of  upwards  of  230  tons.  When  used  for  packing  hay, 
for  expressing  oil  from  seeds,  and  for  other  purposes,  alterations 
are  made  in  the  shape,  length  of  standards,  and,  indeed,  in  the 
size  of  all  the  parts.  In  expressing  oil,  for  example,  there 
would  be  an  alteration  in  the  arrangement  of  the  table  or 
platten  and  the  head.  Sometimes  the  table  becomes  a  piston, 
fitting  into  a  cylinder  attached  to  the  head.  Usually,  however, 
the  arrangement  is  what  you  see.  If  yarn  has  to  be  pressed,  it 
is  placed  in  a  great  oak  box,  bound  with  iron,  running  on 
wheels.  It  is  run  in  between  the  columns.  The  pumps  work, 
a  movable  bottom  of  the  box  rises  up,  the  pressure  on  the  yarn 
gets  greater  and  greater,  and  when  about  230  tons  is  the  total 
force,  the  yarn,  very  much  diminished  in  bulk,  is  tied  up,  the 
ram  descends,  the  great  box  is  rolled  out,  and  another  loosely- 
filled  one  is  rolled  behind  it  to  undergo  the  same  process. 

For  warehouse  use  the  pumps  are  often  worked  by  hand, 
but  even  thirty  years  ago  I  remember  that  orders  were  gene- 
rally for  pumps  worked  by  power  •  that  is,  driven  from  shafting 
and  worked  by  cranks.  In  this  case  it  was  common  to  attach 
a  series  of  these  presses  to  one  set  of  pumps,  all  the  presses 
in  a  warehouse  being  fed  from  a  long  pipe,  each  having  its 
separate  valve. 

151.  You  will  see  that,  although  in  hydraulic  jacks  for  lifting 
heavy  weights  the  weight  on  the  ram  is  the  same  during  the 
whole  of  any  operation,  this  is  not  the  case  in  baling  presses. 
Thus,  for  example,  in  the  Indian  cotton  trade,  governed  as  it  is 
by  the  Suez  Canal  regulations,  it  is  necessary  to  press  cotton  so 
compactly  that  it  is  like  a  piece  of  oak,  and  it  can  be  planed  up 
like  oak. 

During  the  early  part  of  the  operation  the  pressure  on  the 
ram  is  therefore  small,  becoming  very  great  towards  the  end ; 
and  it  is  the  greatest  pressure,  of  course,  which  decides  the 
relative  sizes  of  plunger  and  ram.  If  the  rani  were  made  to 
rise  quickly  at  the  beginning,  and  more  slowly  towards  the  end, 
it  is  obvious  that  there  would  not  only  be  a  saving  of  time,  but 
a  more  regular  doing  of  work. 

In  hand  presses  it  is  usual  to  change  the  fulcrum  of  the 
lever,  so  that  a  labourer  may  work  more  rapidly  at  the  begin- 
ning. It  is  also  common  to  use  a  large  pump  plunger  in  the  early 
part  of  a  pressing  operation,  changing  it  to  &  small  one  at  the 


186  APPLIED    MECHANICS. 

end ;  or  to  use  two  of  equal  size,  throwing  one  of  them  out  of 
gear  towards  the  end  of  the  operation ;  or  to  use  water  from  an 
accumulator  (Art,  152)  in  the  early  part  of  the  operation, 
finishing  with  water  from  an  intensifier  (Art.  152). 

In  an  early  form  of  cotton  baling  machine,  instead  of  using 
one  large  ram,  three  were  used.  At  first  only  the  centre  press 
was  connected  with  the  pumps,  and  of  course  it  rose  quickly ; 
meanwhile  the  other  two  were  filling  with  water  from  a  tank, 
but  the  pressure  of  the  water  in  them  was  insignificant.  As 
the  operation  proceeded  one  of  the  side  presses  was  discon- 
nected from  the  tank,  and  was  fed  from  the  pump.  The 
operation  proceeded  more  slowly  now,  as  the  pump  had  to 
supply  two  presses  instead  of  one;  but  the  possible  total 
force  was  doubled.  Towards  the  end  of  the  operation  the 
third  press  was  disconnected  from  the  tank,  and  was  connected 
with  the  pump;  the  operation  proceeded  more  slowly  still, 
although  the  pump  might  be  working  at  much  the  same  speed 
as  in  the  beginning.  But  the  possible  total  force  was  just 
three  times  what  it  would  have  been  with  only  one  ram. 

In  a  later  form  there  are  twelve  pump  plungers  attached 
directly  to  the  cross-heads  of  the  steam-engines.  At  the 
beginning  all  twelve  are  working,  as  the  pressure  is  small 
Then  as  the  pressure  gets  greater,  one  set  of  four  pumps  is 
detached,  so  that  they  do  not  pump  water  into  the  press,  but 
merely  pump  water  back  to  the  cistern  from  which  they 
draw  it.  Thus  eight  pumps  are  now  pumping,  forcing  into  the 
press  less  water  than  the  twelve  did  before ;  but  as  they  have 
the  whole  steam  piston  force  acting  on  them,  they  are  able  to 
force  the  water  in  against  a  very  much  greater  ram  pressure. 
Later  on  in  the  operation  four  more  pumps  cease  to  act. 

A  further  improvement  is  to  be  noticed.  The  head  or 
platten  is  made  with  two  long  columns  attached  to  its  under 
side,  hanging  down.  When  the  first  operation  is  finished,  the 
bottoms  of  these  columns  are  just  above  the  base,  and  may  be 
locked  firmly,  so  that  the  head  cannot  fall  back  again.  Now 
the  finishing  stroke  is  made;  two  19-inch  rams  are  pressed 
downwards  on  the  upper  end  of  the  bale  with  a  much  greater 
pressure  than  it  was  possible  to  apply  with  the  bottom  1 1  -inch 
ram. 

Other  forms  of  cotton-press  are  used.  Sometimes  fewer 
pumps  are  used,  and  two  bottom  presses  and  rams,  instead  of 
one  with  larger  rams,  for  pressing  downwards  from  the  top  of 


APPLIED    MECHANICS.  187 

the  press  to  finish  the  operation.  In  some  cases  the  finishing 
pressure  operation  may  be  performed  when  the  bottom  ram  is 
being  withdrawn,  so  that  one  bale  is  being  finished  in  the 
upper  part  of  the  press  when  the  box  is  being  filled  with  cotton 
for  a  new  bale.  Diagrams  have  been  obtained  giving  the 
nature  of  the  pressures  to  which  a  cotton  bale  is  subjected. 
In  one  which  is  before  me,  when  the  lower  ram  has  risen  11 
feet,  the  force  is  only  8  tons;  but  in  another  foot  the 
force  increases  to  16  tons;  in  another  foot,  to  33  tons;  and 
in  another  to  59  tons.  The  upper  rams  begin  to  operate  when 
the  total  force  is  about  160  tons;  but  on  moving  through  3 
inches,  they  have  to  exert  200  tons ;  three  inches  further,  300 
tons ;  three  inches  further,  500  tons  ;  and  three  inches  further 
— that  is,  when  the  bale  was  finished — they  were  exerting  a 
force  of  900  tons. 

We  see,  then,  that  capability  of  working  very  rapidly  when 
the  forces  are  small,  and  working  very  slowly  when  the 
forces  are  great,  on  the  supposition  that  the  steam-engine  is 
always  working  at  the  same  speed — these  are  the  important 
things  to  be  looked  for  in  hydraulic  presses. 

152.  There  are  hydraulic  power  companies  now  in  many 
towns,  whose  4-  and  6-inch,  and  other  sizes  of  pipes  are  laid  along 
the  streets  like  gas-pipes,  so  that  any  customer  may  be  supplied 
with  pressure  water  to  work  pressing,  lifting,  and  other 
machinery,  paying  merely  for  the  amount  of  water  used.  The 
force-pumps  are  usually  worked  directly  from  the  cross-heads  of 
steam-engines ;  they  must  be  capable  of  delivering  the  maximum 
supply  required  by  customers.  If  the  area  of  the  pump  plunger 
is  a  square  inches,  the  pressure  p  Ib.  per  square  inch,  the  force 
on  the  plunger  must  be  ap  Ibs.  As  there  ought  always  to  be 
a  fly-wheel  on  the  engine,  it  is  the  average  effective  pressure  of 
the  steam,  minus  f rictional  forces,  which  is  determined  by  this. 
If  the  length  of  a  stroke  is  I  feet,  and  'there  are  n  effective 
strokes  per  minute,  the  horse-power  expended  in  the  pump  is 
plan  -r-  33,000.  Usually  there  are  three  pump  plungers  working, 
so  that  the  flow  of  water  shall  be  more  nearly  uniform.  There 
is  also  always  an  accumulator,  or  heavy  weight  w,  of  perhaps 
112  tons,  resting  on  the  top  of  a  ram,  of  20  inches  diameter, 
its  press  communicating  freely  with  the  hydraulic  mains.  In 
such  a  case,  the  pressure  in  the  accumulator  would  be  800  Ibs. 
per  square  inch.  "When  there  is  a  small  demand,  w  will  be 
seen  to  rise,  its  press  taking  the  surplus  supply.  When  the 


188 


APPLIED    MECHANICS. 


demand  is  great,  w  will  be  seen  to  fall,  its  press  providing  for 
the  extra  demand.  Very  often  the  engines  will  be  seen  work- 
ing fast ;  w  will  be  seen  rising.  When  w  reaches  a  certain 
height  it  acts  on  a  lever  which  shuts  oft 
steam  from  the  engines;  these  work  less 
rapidly,  and  w  will  be  seen  to  fall.  "When 
w  descends  far  enough  it  moves  a  lever,  and 
admits  more  steam.  The  total  store  of  energy 
in  w  lifted  is  often  not  more  than  half  a 
minute  maximum  supply,  yet  it  is  sufficient 
for  regulation.  In  the  above  case  the  maxi- 
mum store,  if  the  maximum  lift  is  12  feet, 
is  112  x  2,240  x  12  foot-pounds.  The  same 
answer  will  be  obtained  if  we  remember 
that  every  pound  of  water  means  a  store  of 
2-3  x  800  foot-pounds.  Or,  again,  every 
cubic  foot  means  a  store  which  is  numeri- 
cally equal  to  the  pressure  in  pounds  per 
square  foot.  When,  as  on  board  ship  and 
elsewhere,  a  great  weight  would  be  incon- 
venient, an  intensifler  (Fig.  126)  is  used.  If 
water  from  a  tank  at  the  pressure  p  communi- 
cates with  one  side  of  a  piston  A  of  area  A,  it 
acts  just  like  a  weight  of  the  amount  p  A.  c 
is  the  accumulator,  whose  ram  a  is  so  much  smaller  than  the 
piston  that  the  pressure  P  in  the  accumulator  is  that  necessary 
in  the  hydraulic  mains  D  E.  When  a  tank  of  water  is  not 
available  for  A,  the  pressure  of  steam  is  employed. 

Example.  — If  we  wish  to  get  700  Ibs.  pressure  from  a  ram 
5  inches  diameter,  and  we  have  a  supply  from  a  tank  88  feet 
high,  what  size  of  piston  is  needed  in  the  intensifier?  Here 
p  is  88  -4-  2-3,  or  38-3  Ibs.  per  square  inch.  (See  Art.  410.) 
The  total  force  needed  is  700  x  52  x  7854,  or  13,744  Ibs.; 
so  that  the  area  of  the  piston,  or  large  ram,  is  13,744  -^  38*3, 
or  359*13  square  inches ;  or  it  is  21*4  inches  in  diameter. 

Exercise. — A  press  has  two  stuffing  boxes  in  line,  and  a 
vertical  rod  protrudes  through  both,  the  lower  end  being  the 
smaller.  The  diameters  being  8  inches  and  5  inches,  and  the 
water  pressure  1,400  Ibs.  per  square  inch,  what  is  the  load  on 
this  differential  accumulator  ?  If  it  may  rise  10  feet,  what  is  its 
store  of  energy,  and  what  is  its  store  of  water  in  cubic  feet,  and 
in  gallons'!  Ans.,  42,882  Ibs.;  428,820  foot-lbs.;  2-127;  13-25. 


Fig.  126. 


APPLIED    MECHANICS.  189 

153.  Pressure  water,  as  sold  by  the  hydraulic  companies, 
would  be  cheaper  if  a  better  load  factor  (average  output  + 
maximum  output  from  the  central  station)  existed,  as  the 
engines  could  work  continuously.  They  charge  about  2d.  per 
horse-power  per  hour,  and  the  cost  of  production  is  about  0-6 
pence.  The  cost  of  the  actual  power  given  out  by  the  motors 
used  by  customers  varies  greatly. 

Exercise. — At   a   pressure   of   800  Ibs.    per  square    inch, 


what   is  the   charge   for    1,000    gallons    of  water  at  2d.  per 
horse-power  per  hour?  Ans.,  18-67  pence. 

154.  A  water-pressure  engine  may  be  looked  upon  as  the  in 
verse  of  a  reciprocating  pump.  If  we  neglect  the  shocks,  which 
are  always  due  to  imperfect  construction,  when  a  water-pressure 
engine  or  pump  works  at  a  certain  speed,  the  loss  of  energy  by 
friction  in  the  engine  seems  to  be  the  same  at  high  and  at  low 

Pressure,  and  hence  there  is  greater  efficiency  at  high  pressure. 
n  all  cases,  wherever  kinetic  energy  is  produced  in  water,  it 
is  almost  altogether  wasted  except  in  turbines,  and  to  some 
extent  in  certain  punching-machines. 

Fig.  127  shows  one  of  the  simplest  forms  of  engine.    Water 
enters    the  arrangement  by  the   pipe  A  whei*  the  cock   B   ia 


190  APPLIED    MECHANICS. 

opened  by  means  of  the  handle  N.  There  are  three  rams  here, 
in  oscillating  presses,  gearing  on  the  same  crank  pin;  but  I 
mean  to  confine  my  attention  to  one  of  them.  The  supply  of 
water  enters  at  A,  and  finds  its  way  to  the  space  P,  by  means 
of  a  passage  in  the  framework  of  the  machine.  There  is 
another  passage  leading  to  the  exhaust  spaces  G,  and  allowing 
water  to  flow  from  these  spaces  through  the  discharge  pipe  Q. 
By  reversing  the  handle,  F  may  be  made  the  exhaust  space, 
and  G  the  supply  space,  and  when  the  handle  is  in  the  middle 
position  it  acts  as  a  brake,  so  that  by  means  of  this  handle  we 
can  make  the  engine  work  in  opposite  directions,  or  stop  it 
altogether. 

Water  enters  at  L,  and  fills  the  space  J,  and  it  presses  on 
the  ram  c,  causing  it  to  leave  more  empty  space  behind  it. 
You  know  now  how  to  calculate  the  force  with  which  the 
plunger  is  being  pressed.  The  plunger  cannot  go  out  without 
turning  the  crank,  HK.  Observe,  too,  that  as  there  is  no 
connecting  rod,  the  cylinder  j  turns  just  like  the  oscillating 
cylinder  of  certain  steam-engines.  When  the  crank  reaches  its 
dead  point,  the  plunger  can  go  out  no  further ;  but  when  this 
happens  the  orifice  L  is  just  ceasing  to  let  water  enter  from  F, 
and  is  beginning  to  let  the  water  escape  into  G.  As  three 
plungers  act  on  the  same  crank  pin,  the  crank  does  not  stop 
anywhere,  and  as  it  moves  on,  the  plunger  c  comes  back  again, 
driving  the  water  from  J  through  L  into  G  and  away  by  Q. 

This  is  all  exceedingly  simple.  The  quantity  of  water  used 
in  one  stroke  of  c  is  simply  the  volume  of  c  which  leaves  J  in 
one  stroke,  assuming  that  the  water  can  come  in  quite  freely. 
Each  pound  of  this  water  has  a  certain  amount  of  pressure 
energy,  which  it  gives  up  to  the  plunger,  and  which  you  may 
take  to  be  the  pressure  energy  in  F  minus  the  pressure  energy 
in  G.  This  is  only  true  if  we  assume  that  the  water  has  no 
kinetic  energy  where  it  is  in  contact  with  the  plunger ;  but  as 
the  plunger  is  itself  moving,  we  know  that  we  are  here  making 
a  small  error.  This  is  the  same  as  saying  that  the  pressure  on 
the  plunger  is  less  than  the  pressure  in  the  mains.  We  are 
also  assuming  no  loss  by  friction  at  the  passage,  L,  which  is 
again  an  error. 

Neglecting  friction,  if  p  is  the  total  pressure  difference 
between  the  supply  and  exhaust,  and  a  is  the  area  of  cross 
section  of  the  plunger,  then  p  a  is  the  force  upon  it  in  pounds. 
If  I  is  twice  the  length  of  the  crank  in  feet,  and  there  are  n 


APPLIED    MECHANICS. 


191 


revolutions  per  minute,  each  plunger  gives  out  p  la  n  -f-  33,000 
horse-power,  and  the  three  plungers  give  out  three  times  this. 
Notice  that  this  engine  is  almost  too  simple;  there  is  con- 
siderable wire-drawing,  and  therefore  loss  of  pressure  due  to 
friction  at  the  valves. 

155.  Fig.   128  is  a  section  of  another  three-throw  water- 


Fig.  128. 

pressure  engine.  The  pistons  in  the  three  cylinders,  ABC,  are 
trunked  and  connected  to  one  crank-pin,  D,  o  being  the  centre  of 
the  crank-shaft.  The  method  of  packing  is  clearly  shown.  The 
water  presses  only  on  one  side  of  each  piston,  and  as  between 
the  rods  and  crank-pin  only  pushing  forces  act,  the  bearing  is 
only  on  a  small  portion  of  the  circumference  of  the  pin.  It 
can  easily  be  imagined  how  the  pressure  water  is  admitted  to 
each  cylinder  at  the  end  of  the  stroke  of  its  piston,  and  how  it 
is  allowed  to  escape  at  the  other  end.  The  engine  will  start 


192  APPLIED    MECHANICS. 

from  any  position,  and  the  flow  of  water  to  it  is  fairly  regular, 
because  three  cylinders  are  used. 

156.  It   is   evident  to  anyone  who  has  studied   air-  and 
steam-engines  that  water- pressure  engines  may  be  of  a  very 
great  number  of  shapes. 

In  most  engines  of  this  kind  the  work  to  be  done  per  stroke 
may  be  very  different  at  different  times,  and  yet  the  pressure 
water  used — that  is,  the  energy — is  always  the  same,  and  so  there 
is  considerable  loss.  One  method  for  remedying  this  evil  is  to 
shorten  the  crank  as  the  work  being  done  is  less. 

Another  method  which  has  been  suggested  is  that  of 
admitting  pressure  water  for  less  than  the  whole  stroke,  simply 
taking  water  from  the  discharge-pipe  for  the  remainder.  When 
engines  have  a  fixed  sort  of  duty,  there  is  no  need  for  any 
adjustment. 

The  common  construction  of  water-pressure  engines  will  be 
readily  understood  if  you  understand  the  construction  of  the 
steam-engine.  Remember,  however,  that  the  velocity  of  water 
ought  never  to  be  great  in  the  engine  or  pipes.  Wire-drawing 
leads  to  serious  loss  by  friction  in  the  steam-engine ;  it  is  far 
more  serious  in  water-pressure  engines.  In  these  the  valves 
ought  to  be  quite  open,  giving  a  very  large  passage  for  water 
to  flow  through  almost  immediately.  Hence,  although  the  slide 
arrangements  of  Fig.  127  are  allowable  in  small  engines,  they 
cannot  be  used  in  large  economical  engines  working  constantly. 
Remember,  too,  that  all  frictional  losses  are  made  much  greater 
by  quick  motion,  and  by  reversals  of  motion,  and  hence  that  it 
is  very  important  to  have  a  long  stroke  of  piston  or  plunger. 

Lastly,  remember  that,  although  there  ought  to  be  no  waste 
space  between  steam-piston  and  cylinder  at  the  end  of  the 
stroke  (very  little  clearance),  this  is  of  almost  no  importance 
in  water-pressure  engines,  because  of  the  incompressibility  of 
water. 

157.  We  have  described  (Art.  136)  the  force-pump  used  by 
hydraulic  engineers.     All  force-pumps  are  much  the  same  in 
principle,  but  in  ordinary  force-pumps  for  water  (see  Art.  406), 
and  for  use  in  feeding  boilers  there  is  usually  less  trouble  with 
the   valves   and   packing  than  there  is  in  hydraulic  pumps. 
Students  will  pay  attention  in    Chap.  XXIII.   to  the  blow 
caused  by  sudden  stoppage  of  the  flow  of  water. 

Direct-acting  steam-pumps  are  much  used  in  draining  mines. 
Sometimes  water-pressure  engines  have  been  used  instead  of 


APPLIED    MECHANICS.  193 

steam — that  is,  water  from  an  accumulator  on  the  surface  goes 
down  the  mine  to  a  water-pressure  engine  which  drives  a  pump; 
the  mine  water  and  the  exhaust  water  of  the  engine  are  sent 
to  the  surface.  In  such  a  case,  pressures  are  very  great,  and 
the  effects  due  to  sudden  stoppage  of  flow  in  pipes  may  produce 
damage  unless  the  valves  are  so  arranged  that  when  they  close 
or  open  the  general  circulation  is  not  much  affected. 

The  common  lift-pump  is  described  in  Art.  406.  Its 
form  as  an  air-pump  is  usually  described  in  books  on 
the  steam-engine,  and,  indeed,  it  is  in  such  books  that 
pumping  machinery  more  naturally  finds  a  place  than  in 
books  on  applied  mechanics.  In  Art.  433  we  find  that  in 
a  stream  line  the  total  energy  of  one  pound  of  fluid  remains 
constant  if  there  is  no  friction ;  a  pump  adds  to  this  store  ;  a 
turbine  or  water-wheel  takes  away  from  this  store ;  friction 
also  diminishes  the  store,  converting  mechanical  energy  into 
heat. 

158.  What  now   are  the   conditions   under  which   trans- 
mission of  power  by  hydraulic  action  is  most  suitable  ? 

1st.  Intermittent  action,  because  the  accumulator  is  so 
nearly  perfect,  giving  out  energy  simply  in  proportion  to  the 
quantity  of  water  used,  and  yet  allowing  an  engine  of  small 
power  to  be  storing  continually. 

2nd.  Action  requiring  not  a  very  great  quantity  of  power. 

3rd.  Action  of  a  comparatively  slow  kind,  the  water  never 
being  allowed  to  flow  so  fast  that  its  store  of  kinetic  energy  is 
great,  since  the  kinetic  energy  is  nearly  all  wasted ;  slow  action, 
with  considerable  force. 

4th.  Action  which  is  greatly  continuous  in  one  direction, 
not  requiring  much  stoppage  or  reversal  of  the  water  motion. 

You  will  see  from  this  that  the  conditions  required  in 
pressing  machinery,  cranes,  hoists,  and  lifts  are  better  satisfied 
by  hydraulic  transmission  of  power  than  they  can  be  by  any 
other  method  of  power  transmission  which  is  known  to  us. 

159.  In  Fig.  129  we  have  one  specimen  of  a  hydraulic  crane, 
whose  action  it  is  very  easy  to  understand.     Suppose  water  at 
700  Ibs.  pressure  per  square  inch  admitted    to  the  cylindric 
space  A,  and  that  the  space  B,  on  the  other  side  of  the  piston, 
although  filled  with  water,  has    only  a  comparatively  small 
pressure,  and  communicates  with  a  low-lying  tank ;  neglecting 
the  small  pressure  in  B,  we  see  that  the  piston  is  pushed  for- 
ward with  a  force  of  700  x  A  Ibs.  if  A  is  the  area  of  the  piston 

H 


194 


APPLIED    MECH  \NICS. 


in  square  inches.  Now  the  motion  of  the  piston  is  multiplied 
eight  times  by  the  chain,  which  passes  over  blocks,  each  contain- 
ing four  sheaves,  attached  at  M  and  N.  The  block  at  N  gets 
the  motion  of  the  piston,  and  the  chain  at  P  must  be  drawn  in 
eight  times  more  quickly  than  this.  You  know  that  the  pull 
in  the  chain  may  be  one- eighth  as  much  as  the  total  force  on 
the  piston,  and  it  can  therefore  lift  through  eight  times  the 
distance  a  weight  of  one-eighth  the  amount,  or  700  A  -f-  8  Ibs. 


Fig.  129. 

This  is  the  original  form  of  Lord  Armstrong.  Whatever 
defect  there  is  lies  in  the  use  of  chains  passing  over  numerous 
sheaves  giving  rise  to  a  great  amount  of  friction.  Cranes  require 
so  little  horse-power  to  work  them,  however,  that  mere  economy 
of  coal  is  barely  worth  considering,  and  the  risk  of  accident,  which 
might  be  done  away  with  very  greatly  by  direct  hydraulic  action, 
is  not  important  either.  You  see  that  if  A  and  B  both  receive 
pressure  water  there  is  less  water  used  than  before;  for  although 
as  much  comes  into  A  as  before,  B  is  sending  water  back  to 
the  pump  or  into  A.  In  fact,  the  total  pressure  on  the  piston 
is  700  Ibs.  multiplied  by  the  difference  between  the  areas  of  the 
two  sides  of  the  piston  exposed  to  pressure — that  is,  the  mere 
area  of  cross-section  of  the  thick  piston-rod.  Hence,  we  can  work 


APPLIED    MECHANICS. 


195 


this  crane  so  that  it  lifts  heavy  loads  or  light  loads— that  is,  it  is 
double-powered      Unfortunately,  however,  when  working  any 


Fig  130 


heavy  load,  it  is  consuming  as  much  energy  as  if  it  were  lifting 
the  heaviest  load  it  is  capable  of  lifting.  When  lifting  on  its 
second  power,  and  lifting  a  light  load,  it  is  using  as  much  energy 


196  APPLIED    MECHANICS. 

as  if  it  were  lifting  the  heaviest  load  this  second  power  is 
capable  of  lifting. 

Thus,  let  us  suppose  it  lifting  eight  tons  to  a  height  of  20  ft., 
and  that  one  cubic  foot  of  water  is  used  in  the  operation.  Then, 
on  the  same  power,  suppose  it  to  be  lifting  four  tons  to  the  same 
height,  it  still  uses  one  cubic  foot  of  water,  and  in  both  cases 
there  is  the  same  energy  used  for  the  pumps.  Of  course,  by 
a  combination  of  cylinders,  it  would  be  possible  to  vary  the 
work  expended  as  the  load  varied,  but  the  expedient  is  not 
of  a  very  practical  character. 

Such  a  crane  as  this  has  usually  another  cylinder  and  chain 
f  of  turning  it  round.  In  many  cranes  the  action  is  more  direct ; 
a  two-sheaved  pulley-block,  or  one  with  a  single  sheave,  or 
even  no  pulley-block  at  all,  being  used.  (See  Fig.  130.)  In 
many  cases  loads  are  pushed  up  on  the  top  of  a  ram,  in 
others  they  are  pulled  up  by  a  piston  rod  directly ;  in  fact, 
there  is  an  endless  variety  in  the  arrangements,  but  the 
principles  of  the  hydraulic  press  parts  are  the  same  in  all. 

160.  Now,  supposing  it  is  easy  to  get  a  supply  of  pressure- 
water,  what,  besides  cranes,  hoists,  and  pressure-engines,  can 
be  worked  by  means  of  it  1  All  forging  and  welding  machines, 
which  with  moulds  and  dies  properly  shaped  and  pressed 
together  with  enormous  steady  force,  seem,  for  objects  of 
settled  shapes,  to  be  a  very  great  improvement  on  any  method 
of  hammering ;  stamping  machines,  for  all  sorts  of  purposes ; 
and  bending  machines,  for  joggling  and  bending  angle-irons, 
rails,  and  beams. 

Students  must  examine  such  bending-machines  for  them- 
selves, and  notice  how  the  travel  of  the  press-block  is  deter- 
mined by  tappets,  which  open  and  close  the  valve  for  water 
supply  at  any  point  in  the  stroke  we  please.  By  means  of  such 
a  bending-machine  any  number  of  curves  may  be  made  identi- 
cally the  same.  In  many  punching,  forging,  stamping,  and 
shearing  machines  we  also  have  the  inlet  and  exhaust  valve 
worked  by  hand  or  foot  levers,  the  stroke  being  regulated 
with  great  nicety  by  tappet  motions.  Thus,  in  a  riveter,  a  short 
ram  of  eight  inches  in  diameter  may  be  used.  Used  with  an 
accumulator  and  pumps  of  its  own,  the  pressure  is  usually 
1,400  Ibs.  per  square  inch  ;  so  that  the  total  force  available  is 
31  tons. 

Next  we  have  tools  which  are  easily  portable.  In  Fig.  131 
the  ram  gets  a  motion  about  its  centre,  its  axis  and  line  of  actior 


APPLIED    MECHANICS. 


197 


being  an  arc  of  a  circle.  By  this  compact  arrangement  we  do 
away  with  connecting-rods  and  many  other  complications,  and 
we  get  a  great  increase  in  stiffness.  Instead,  then,  of  bringing 
work  weighing  many  tons  to  a  machine,  we  bring  a  little 
machine,  weighing  5  cwt.,  to  the  work ;  we  can  punch  holes 
and  finish  the  riveting ;  that  is,'  we  can  finish  most  of  it — for 
there  must  always  remain  rivets  in  difficult  positions  which 


Fig.  131. 

require  to  be  done  by  hand — with  no  other  intermediate 
gearing  between  this  little  riveter  and  the  steam-engine  than  a 
small  jointed  pipe,  the  two  forms  of  the  universal  joint  which 
are  used  being  made  water-tight  by  leather  collars.  Instead  of 
a  jointed  pipe,  a  pipe  formed  into  a  spiral  forms  a  good  yielding 
connection.  It  is,  perhaps,  in  looking  at  these  little  riveters, 
rather  than  in  any  other  examples,  that  you  will  be  struck  by 
the  simplicity  of  hydraulic  working.  The  flexible  pipe  trans- 
mits power  more  faithfully  than  huge  beams  and  cog-wheels 
would  do. 


198 


APPLIED   MECHANICS. 


161.  The  machines  which  I  have  described  are  well  suited 
to  hydraulic  work.  A  punching,  or  shearing,  or  stamping,  or 
riveting  machine,  driven  by  shafting,  repeats  its  stroke  at  regular 
intervals,  and  the  workman  cannot  arrest  a  stroke.  He  wastes 
time  in  waiting  for  a  stroke,  but  when  a  stroke  is  being  made, 
and  he  sees  that  his  plate  has  been  wrongly  placed,  or  has  shifted 
its  position,  he  must  just  as  patiently  watch  the  inevitable 


1'22-vru  hole,. 
1' 18m.' plate 
Iron/ 


Irorv  Bcur 


Fig.  132. 


Fig.  133. 


*'iWO(Jbs  per  so  w 
aw 


p.  sq.  TJTI 
-1500 


tf, 


Fig.  134. 

completion  of  the  stroke.  With  the  hydraulic  punching- 
machine,  on  the  contrary,  the  workman  can  stop  the  motion 
at  any  instant,  even  if  the  punch  has  made  its  mark  on  the 
surface  of  the  plate.  He  can  start  instantly  from  a  condition 
of  rest ;  he  has  not  to  wait  till  he  pulls  the  belt  on  to  the  fast 
pulley,  or  starts  the  donkey-engine,  and  he  knows  that  when 
there  is  no  stroke  being  made  there  is  no  power  being  wasted. 
Then,  again,  all  the  shafting  and  other  machinery  of  a  large 
shop  have  not  to  be  set  in  motion,  or  varied  in  their  motion, 
to  punch  a  five-eighth  inch  hole.  A  man  wastes  just  the  same 
energy  in  punching  this  one  hole  as  if  it  were  one  of  a  hundred 


At»PtlEl>   MECHANICS.  199 

he  was  punching.  Suppose,  again,  that  a  man  carelessly  puts 
a  1^-inch  plate  under  a  punch  arranged  for  a  five-eighth  inch 
plate.  The  sudden  blow  of  an  ordinary  punching-machine, 
with  its  fly-wheel,  would  produce  a  fracture  somewhere.  In  a 
hydraulic  machine  there  is  a  simple  stoppage.  Think,  too,  of 
the  strength  of  roof  and  columns  needed  to  carry  shafting ;  of 
the  trouble  in  the  use  of  overhead  cranes  when  there  are 
many  shafts  and  belts ;  and,  above  all,  think  of  the  noise,  in 
comparison  with  the  invisibility  of  pressure  mains,  and  the 
dead  silence  of  hydraulic  tools.  Observe,  too,  that  these 
hydraulic  machines  need  but  little  foundation. 

The  diagrams  of  Figs.  132, 133,  and  134  (p.  198),  showing  the 
water-pressure  in  the  cylinders  of  various  tools  at  Toulon  during 
their  stroke,  are  exceedingly  interesting.  Fig.  132  shows  a 
curve  from  a  punching-machine.  The  pressure  is  not  equal  to 
that  in  the  accumulator,  unless  the  tool  has  so  much  resistance 
to  overcome  that  it  is  moving  very  slowly  indeed.  The  sudden 
rise  shows  what  occurs  when  the  punch  is  just  beginning  to  act 
on  the  plate.  In  practice  you  must  understand  that  this  early 
part  of  the  diagram  has  no  existence ;  Mr.  Tweddell's  tappet 
motion  prevents  such  waste.  The  area  of  each  diagram  is 
roughly  the  amount  of  energy  utilised.  The  area  of  the  rect- 
angle in  Fig.  134  shows  the  energy  taken  from  the  accumu- 
lator. In  all  these  diagrams,  then,  it  must  appear  to  you  that 
there  is  a  large  amount  of  waste.  This  is  greatly  reduced  by 
the  tappet  motion,  and  in  any  case  it  is  very  much  greater  in 
such  tools  when  driven  by  shafting. 

Fig.  133  shows  a  diagram  from  a  plate-shearing  machine. 
The  uniformity  of  the  pressure  is  due  to  the  angle  which  the 
edges  of  the  shears  make  with  one  another. 

The  riveting-machine  diagrams  are  most  interesting. 
Neglecting  the  useless  part  of  the  stroke  in  Fig.  134,  we  see 
the  rise  of  pressure,  E  F,  due  to  the  setting  up  of  the  rivet,  F  H, 
the  clinching  of  the  rivet  and  closing  of  the  plates.  The  sudden 
stoppage  of  motion  of  the  water  gives  a  blow,  shown  by  the 
pressure  becoming  half  as  much  again  as  the  accumulator 
pressure,  and  we  rely  on  this  blow  as  perfecting  the  filling  of 
all  cavities. 

162.  In  Fig.  135  two  rams,  A  and  a,  are  equal,  one  carrying 
a  gun  which  is  to  be  quickly  lifted  and  lowered,  the  other  a  a 
counterweight.  As  it  is  not  convenient  to  vary  the  counter- 
weight, and  as  there  is  really  balance  only  in  one  position  (for 


200 


APPLIED  MECHANICS. 


A 

i 

_A1_! 

a 

r 

I                        r 

£q  c 

c 

*0 

rxj 

J                        l 

Fig.  135. 


either  A  or  a  getting  lower  causes  the  pressure  acting  on  it 
upwards  to  get  greater),  a  plan  taken  is  this :— A  and  a  have 

two  presses  communicating 
by  means  of  a  small  pipe; 
and  E,  which  is  greater  than 
R,  is  partly  supported  on  a 
small  extra  ram  of  area  a  in 
a  press  of  its  own,  which  may 
either  be  made  to  communi- 
cate with  the  other  two 
presses,  or  with  a  neighbour- 
ing tank  where  the  pressure 
is  small.  In  the  one  case, 
E  rests  on  what  is  practic- 
ally a  ram  of  area,  a  -}-  a', 
and  it  is  lifted ;  in  the  other 
case  on  a  ram  of  area  a,  and 
it  falls.  The  supply  of  pressure  water  to  a'  is  effected  by 
means  of  a  pump  and  accumulator.  When  pressures  are 
already  very  great,  change  of  pressure  due  to  rise  and  fall  of 
a  ram  is  not  very  important,  but  in  long  rams  it  is  often 
important. 

163.  In  most  of  the  machines  which  we  have  described, 
although  water  usually  changes  its  level  during  the  action, 
this  change  of  level  has  been  so  small  as  to  be  negligible. 
But  in  nearly  all  lifting  operations  we  have  to  consider  the 
work  done  in  the  lifting  of  water  as  the  ram  rises. 

We  all  know  the  conditions  required  in  an  ordinary  hotel  or 
chambers  hoist ;  those  conditions  are  absolutely  the  same  for 
warehouse  hoists,  because  a  hoist  which  carries  goods  occasion- 
ally carries  men  in  charge  of  these  goods.  Long  ago,  I  had 
some  designing  and  carrying  out  of  mill  hoists,  in  which  the 
cage  was  lifted  by  a  rope  passing  over  an  elevated  pulley, 
driven  from  the  main  shafting  of  the  mill,  and  stopped  at  any 
point  of  ascent  or  descent  by  automatic  disengaging  apparatus 
which  also  braked  the  pulley.  The  cage  was  balanced  by 
counter- weights,  as  a  window  is  balanced.  Our  greatest  trouble 
was  in  the  arrangement  of  safety  apparatus,  which  would  stop 
the  cage  in  falling  should  the  rope  break.  Now,  it  is  well 
known  that  such  safety  apparatus  can  never  be  thoroughly 
depended  upon,  however  ingenious  its  design  may  be,  because 
the  ordinary  working  of  the  hoist  does  not  keep  the  safety 


APPLIED   MECHANICS. 


201 


apparatus  in  action ;  immunity  from  accidents  causes  it  to  be 
neglected,  and  when  an  accident  does  happen  it  will  not  work. 

There  is  nothing  so  safe  as  a  hoist  whose  rapid  motion  is 
resisted  by  a  considerable  amount  of  friction.  But,  unfor- 
tunately, if  the  friction  is  that  of  solids  oil  one  another,  there 
is  as  much  frictional  resistance  to  the  ordinary  working  of 
the  hoist  as  there  is  when  an  accident  occurs,  and  hence 
assurance  of  safety  by  friction  means  tremendous  loss  of 
power  at  all  times. 

Now,  you  remember  that  the  frictional  resistance  of  water 
was  of  quite  a  different  kind.  There  is  almost  no  resistance 
to  the  flow  of  water,  if  the  flow  is  slow.  There  is  only  a 
moderate  loss  of  power  in  the  ordinary  use  of  a  hydraulic  hoist; 
but  the  motion  cannot  become  too  rapid  for  safety,  for  the 
frictional  resistance  is  exceedingly  great  at  high  speeds. 

Although,   therefore,   serious  accidents  cannot    bb  wholly 
prevented — water  may  leak  away  by  valves  and  leave  the  cage 
wholly  unsupported,  for  example — yet   there  is  more  safety 
possible  with  hydraulic  than  with  other 
hoists. 

164.  In  a  great  many  hydraulic 
hoists  the  action  is  precisely  the  same  as 
in  Armstrong's  cranes.  Fig.  136  shows 
such  a  construction,  used  by  Armstrong 
himself.  A  is  a  pressure  cylinder,  with 
its  ram  carrying  at  B  the  movable 
block  with  sheaves,  which  pull  the 
chain  or  wire  rope,  M,  K.  There  is  a  loss 
of  effect,  due  to  the  altering  weight  of 
the  chain,  as  the  cage  rises  or  falls. 
This  difficulty  may  be  got  rid  of  by 
letting  the  ram  move  vertically,  when 
the  altering  weight  of  the  ram  itself 
may  be  made  to  balance  the  alter- 
ing weight  of  the  chain.  All  such 
hoists  as  this  can  be  readily  balanced, 
so  that  the  dead  weights  may  balance 
at  all  points  in  the  ascent  and  de- 
scent. They  are,  however,  subject 
to  the  risks  inseparable  from  the  use 

of  chains  or  ropes,  and  must  be  regarded  as  unsatisfactory 
for  this  reason.       That  the   lifting  of  every  load  means  the 


:M 


Fig.  136. 


202  APPLIED    MECHANICS. 

expenditure  of  the  same  amount  of  energy  is  not  a  considera- 
tion of  any  importance  in  these  hotel  hoists.  Of  course, 
there  is  a  slightly  greater  speed  when  the  load  is  small,  as 
the  water  pressure  is  capable  of  lifting  the  heaviest  probable 
loads  ;  but  you  know  enough  already  about  water  friction  to 
see  that  the  increase  of  speed  must  be  insignificant.  This  con- 
dition is  the  same  for  all  hydraulic  hoists  hitherto  constructed. 
Very  often  we  use  a  direct  acting  hoist.  Here  the  ram  moves, 
pushing  the  cage  up  directly.  When  the  pressure  of  water  is 
very  considerable,  say  200  Ibs.  per  square  inch,  and  the  lift  is 
not  too  high,  this  form  of  hoist  is  good;  for  although  rather 
wasteful  of  power,  it  is  exceedingly  simple.  The  press  is 
sunk  so  far  beneath  the  basement  that  there  is  room  for 
the  whole  length  of  the  ram  when  the  cage  is  in  its  lowest 
position. 

165.  It  is  necessary  now  to  consider  the  diminution  of 
lifting  force  as  a  ram  rises,  and  we  return  to  Fig.  135. 

Let  the  total  load  upon  the  ram  of  area  A,  including  its 
own  weight,  be  called  R,  and  let  the  total  load  on  the  ram  of 
area  a,  including  its  own  weight,  be  called  E,  and  for  easier 
calculation  assume  the  ends  of  the  rams  to  be  flat  and  horizontal. 
When  a  is  just  about  to  descend,  let  the  end  of  A  be  h  feet 
above  a.  When  E  falls  one  foot  we  found  how  high  R  must  rise 
if  we  could  only  neglect  the  weights  of  the  water.  But  we  shall 
now  consider  the  weight  of  the  water,  and  take  the  areas  a 
and  A  to  be  in  square  feet,  and  the  forces  R  and  E  to  be  in 
pounds. 

When  a  falls  a  very  short  distance,  x  feet,  further  displacing 
ax  cubic  feet  of  water,  or  62*3  a  x  Ibs.  of  water,  this  water  is 

lifted  through  h  feet.     Now  A  will  rise  x   x  —  feet,  and  the 

A 

work  done  by  E  in  falling  being  E  x,  is  equal  to  the  work 
R  x  x  —  done  in  lifting  R,  and  also  to  62  -3  a  x  h,  the  work 

done  in  lifting  the  water.  Writing  this  down  algebraically, 
we  find  — 


or  the  pressure  at  the  end  of  a,  which  is  E  -4-  a,  is  greater  thai? 
that  at  A,  which  is  R  -f-  A,  by  the  amount  62-3  h  Ibs.  per  square 
foot.  We  find  this  to  be  an  increase  of  2,11  6  Ibs.  per  square  foot, 


APPLIED    MECHANICS. 


203 


\ 


or  14'7  Ibs.  per  square  inch,  or  one  atmosphere  as  it  is  called, 
for  every  34  feet  of  difference  of  level. 

When  a  ram  rises  34  feet,  if  the 
supply  pressure  to  the  press  remains  the 
same,  the  lifting  force  on  the  ram 
diminishes  14-7  Ibs.  per  square  inch, 
and  for  a  68  feet  rise  the  diminution 
of  pressure  would  be  twice  as  much. 
These  changes  of  pressure  are  not  very 
important  when  we  are  dealing  with 
pressures  in  the  press  of  200  or  300  Ibs. 
per  square  inch,  but  they  may  be  very 
important  at  lower  press  pressures, 
and  in  some  cases  we  make  the  supply 
pressure  increase  as  the  ram  rises. 
Notice  that  you  may  look  upon  the  phe- 
nomenon in  two  seemingly  different 
ways.  You  may  either  say  to  yourself, 
"  As  a  stone  is  lighter  when  sur- 
rounded by  water,  so  this  ram  is  lighter 
when  it  is  at  the  bottom,  for  more  of  it 
is  surrounded  by  water  in  the  press  " ; 
or  you  may  put  it  in  this  form,  "The 
pressure  on  the  bottom  of  the  ram  must 
just  balance  the  weight  of  ram,  cage, 
etc., but  as  the  bottom  of  the  ram  rises, 
this  means  that  we  ought  to  have  a  con- 
stant pressure  at  the  bottom  of  the  ram 
wherever  it  may  be,  and  consequently 
a  gradually  increasing  pressure  in  the 
cylinder  everywhere  as  the  ram  rises." 
Now,  I  do  not  care  which  of  these  two 
views  you  take,  but  you  must  not  mix 
them,  and  say  that  "  not  only  does  the 
ram  get  heavier,  but  it  needs  a  greater 
pressure  at  its  lower  end  as  its  lower 
end  rises."  I  prefer  always  to  say, 
"The  ram  appears  to  get  heavier  just 
in  proportion  to  the  amount  it  has 
been  raised,  and  this  must  be  balanced 
by  increasing  the  pressure  of  the 
supply  water,"  Figf  m 


204 


APPLIED   MECHANICS. 


Now,  remember  that  our  supply  water  in  the  direct- 
acting  hoist  is  at  a  constant  pressure, 
and  you  will  see  that  it  is  quite 
impossible,  with  such  a  simple 
arrangement,  to  have  perfect  uniform- 
ity of  action,  although  it  is  approxi- 
mated to  more  and  more  nearly  as  the 
pressure  is  greater.  In  this  kind  of 
hoist  it  is  usual  to  let  the  water  escape 
from  the  cylinder  to  a  discharge  cistern 
considerably  above  the  cylinder,  so  that 
in  its  descent  the  ram  and  cage  may 
not  fall  too  rapidly.  Here,  again,  we 
have  the  same  want  of  uniformity  of 
action,  since  the  apparent  weight  of 
the  ram  gets  less  as  it  falls. 

166.  The  usual  practice  has  been  to 
nearly  balance  the  dead-weight  of  ram 
and  cage  by  a  weight,  as  in  Fig.  137 
(p.  203),  so  as  not  to  require  too  high 
a  lift  in  the  discharge-pipe,  and  to  so 
arrange  that  the  varying  weight  of  chain 
shall  just  balance  the  apparent  change  of 
weight  of  the  ram.  It  is  evident  that 
if  the  ram  rises  one  foot,  the  counter- 
weight increases  by  the  weight  of  two 
feet  of  chain  ;  hence,  the  weight  of  two 
feet  of  the  chain  ought  to  be  equal  to 
that  of  water  occupying  the  volume  of 
one  foot  length  of  the  ram.  Unfortu- 
nately, these  chains  and  counterweights 
destroy  the  simplicity  and  absolute 
safety  of  the  hydraulic  hoist.  If  the 
ram  were  to  break  near  its  upper  end, 
the  cage  would  be  drawn  violently 
upward  by  the  chain.  The  upper  part 
of  the  ram  is  in  tension,  and  the  lower 
part  in  compression. 

It  is  obvious,  then,  that  there  must, 
for  a  complete  and  perfect  hydraulic 
lift,  be  such  a  regulation  of  the  pressure 
Fig.  IBS  of  the  water  as  it  enters  the  cylinder 


APPLIED    MECHANICS.  205 

of  a  hoist  that  the  only  force  to  be  overcome  shall  be  the 
variable  weight  placed  in  the  cage,  whether  that  of  passengers 
or  goods,  together  with  the  necessary  friction.  The  hydraulic 
balance  hoist  satisfies  this  condition.  It  can  be  worked  with 
either  high  or  low  pressure  water ;  the  ram  is  always  in  com- 
pression, supporting  the  load,  and  no  part  of  the  machinery  is 
above  the  cage,  and  there  is  no  part  of  the  machinery  likely  to 
break  in  such  a  way  as  to  cause  an  accident. 

This  hydraulic  balance  lift  is  shown  in  Fig.  138  (p. 
204).  The  hydraulic  cylinder,  ram,  and  cage  are  as  usually 
made,  except  that  the  ram  is  somewhat  smaller  in  diameter. 
Its  size  is  determined  by  the  strength  required  to  carry  the 
load,  and  not  by  the  working  pressure  of  water  available.  The 
lift  cylinder  is  in  hydraulic  connection  with  a  second  and 
shorter  cylinder,  E,  below  which  there  is  a  cylinder,  c,  of  larger 
dimensions.  There  is  a  piston  in  each,  connected  by  the  rod. 
The  capacity  of  the  annular  space  E  below  the  upper  piston 
is  equal  to  the  displacement  of  the  lift  ram.  The  annular  area 
of  the  lower  piston  is  sufficient,  when  subject  to  the  working 
pressure  of  the  supply  water,  to  overcome  friction  and  lift  the 
net  load ;  and  the  full  area  of  the  upper  piston  is  sufficient, 
when  subjected  to  the  same  pressure,  to  balance  within  a  small 
amount  the  unalterable  weight  of  the  ram  and  cage. 

Assuming  the  cage  at  the  bottom  of  its  stroke,  the  valve 
is  opened  by  a  man  in  the  cage  pulling  on  a  rope,  by  a  system 
of  levers,  and  pressure  water  is  admitted.  The  pressures  on 
the  two  pistons  cause  them  to  descend,  forcing  water  from  the 
annular  space  to  the  hoist  cylinder.  The  hoist  ram  ascends, 
and  in  doing  so  gets  heavier,  but  the  pistons  are  descending, 
and  the  total  pressure  on  them  is  getting  greater  just  in  the  same 
proportion.  When  the  ram  reaches  the  top  of  its  stroke  the 
valve  is  closed  and  the  lift  stops.  Now  open. the  exhaust  valve, 
which  lets  the  water  pass  away  from  c — only  from  c,  re- 
member— and  the  weight  of  the  ram  and  cage  presses  the 
water  from  the  lift  press  into  E,  causing  the  pistons  to  rise. 

To  make  good  any  possible  leakage,  provision  is  made  for 
admitting  pressure  water  under  F,  and  so  raising  it,  the  lift 
ram  being  at  the  bottom  of  its  stroke,  that  water  will  flow  into 
the  space. 

167.  We  see  that  the  hydraulic  hoist  has— (1)  The  great 
element  of  safety  from  the  absence  of  possible  breakage  of  chains 
or  ropes.  But  in  some  forms  it  is  not  without  a  danger  of  its 


206  Ai»PLitei> 

own — namely,  the  danger  that  when  the  cage  is  remaining 
caught  in  a  fixed  elevated  position  for  a  time,  the  cylinder 
may  be  emptying  of  water  through  a  leakage  of  the  valves. 
(2)  That  the  expenditure  of  energy  depends  very  little 
upon  the  dead  load.  But  there  is  still  the  drawback  that 
every  load,  however  small,  requires  the  same  expenditure  of 
energy  as  the  greatest  load  which  the  hoist  can  lift.  This 
drawback  is  common  to  all  hydraulic  hoists  such  as  I  have 
been  describing. 

168.  You  can  now  have  no  difficulty  in  understanding  the 
construction  of  all  warehouse  and  hotel  lifts.     The  hydraulic 
principles  involved  in  all  lifts  are  the  same,  but  with  larger 
weights   to  be  raised   there  are  peculiarities  of   construction 
which  ought  to  be  studied  from  actual  drawings  of  such  lifts, 
as,  for  example,  the  pair  of  lifts  at  the  Seacombe  Pier,  on  the 
Mersey,  to  take  carts  and  waggons  from  the  floating  landing- 
stage  to  the  high  level.     The  height  of  lift  is  32  feet,  and  the 
net  load  20  tons.     Here  we  have  simply  direct  acting  rams  and 
presses  sunk  in  the  river-bed,  the  presses  being  surrounded  by 
cast-iron  protecting  cylinders.     In  this  case  there  is  no  attempt 
to  balance  the  increasing  weight — that  is,  the  displacement  of 
the  ram  as  it  rises.     Each  cage,  or  platform,  is  supported  on 
one  ram,  and  the  designer  has  to  take  care  that  the  weight  of 
any  waggon  shall  be  so  carried  on  the  top  of  the  ram  that  no 
part  of  the  structure  is  unduly  strained.     The  waggon  rests  on 
a  platform,  which  can  slide  on  the  bottom  of  the  cage,  so  that 
although  the  cage  rises  vertically,  the  platform  may  everywhere 
be  in  the  position  in  which  it  would  be  on  the  landing-stage,  if 
the  landing-stage  rose  and  fell.     The  landing-stage  is  attached, 
as  we  know,  by  girders,  which  do  not  alter  in  length,  so  that 
it  does  not  rise  and  fall  vertically,  but  really  in  arcs  of  circles. 
In  this  case,  the  lift  is  of  a  variable  amount,  depending  on  the 
height  of  the  tide.     There  is  a  connecting  valve  between  the 
two  presses,  so  that  a  descending  load  in  one  lift  may  raise  a 
less  load  in  the  other  when  necessary. 

169.  This  idea  of  having  two  lifts  side  by  side,  so  that  the 
lowering  of  one  may  cause  the  rise  of  the  ether,  you  will  under- 
stand better  in  the  original  example  of  its  use,  the  canal  lift 
of  Messrs.  Clark  and  Standfield,  on  the  River  Weaver,  in 
Cheshire.     Figs.  139  and  140  show  the  canal  and  the  river, 
one  50  feet  above  the  other.     We  want  to  raise  or  lower  canal 
boats  from  the  one  to  the  other,  and  we  want  to  avoid  the 


APPLIED    MECHANICS. 


207 


expenditure  of  water,  and  the  delay  which  occurs  when  there 
is  a  chain  of  locks. 

There  are  two  great  wrought-iron  troughs,  each  75  feet  long 


Fig.  139. 


and  15  J  feet  wide,  each  being  carried  on  the  top  of  a  ram  3  feet 
in  diameter.  Now,  when  I  tell  you  this,  you  will  at  once  see 
how  incomplete  my  description  is.  Each  tremendous  trough 


208 


APPLIED    MECHANICS. 


is  carried  easily  by  one  ram  acting  at  its  centre.  You  can,  iu 
your  imagination,  go  into  all  the  details  of  girder- work  necessary 
for  the  safe  carrying  of  such  a  load  in  such  a  manner.  The 
weight  of  each  trough,  with  the  water  and  barges  in  it,  is  240 
tons,  and  this  gives  a  pressure  of  about  one  quarter  of  a  ton 
per  square  inch  in  the  press.  At  each  end  of  each  trough  there 
is  a  gate. 

When  the  trough  is  up,  and  gate  A  is  lifted,  the  trough 
formo  part  of  the  aqueduct.  A  barge  floats  into  it  from  the 
canal,  and  the  gate  is  closed  again ;  also 
the  aqueduct  end  is  itself  closed  with  a 
gate.  Now  the  trough  is  lowered  con- 
taining the  barge,  and  when  it  reaches 
the  lower  level,  gate  B  is  lifted,  when, 
of  course,  the  trough  really  forms  part 
of  the  river.  You  must  describe  to  your- 
selves how  this  press  is  sunk — how  we 
have  a  tunnel  which  enables  us  to 
examine  the  packing  of  the  cylinders, 
how  these  great  columns  are  firmly  sup- 
ported, so  that  we  may  have  guides  to 
prevent  the  tilting  of  the  troughs.  All 
this  is  constructive  detail  which  you  can 
read  about  in  the  Proceedings  of  the 
Institution  of  Civil  Engineers.  You 
Pig.  140.  might  fear  also  that  the  joints  of  the 

gates  at  the  end  of  the  trough  might  be 

leaky,  and,  above  all,  that  the  joint  of  the  trough  with  the 
aqueduct  end  might  be  leaky ;  but  even  from  the  figure  you 
can  see  how  perfectly  these  difficulties  can  be  got  over. 

Suppose  I  have  a  boat  in  any  vessel  of  water,  and  I  know 
that  the  water  is  at  a  certain  level,  and  I  take  out  this  boat 
and  put  in  another  boat,  and  add  water  or  take  it  away  until 
the  level  is  just  what  it  was  before.  You  know  that  the  weight 
of  boat  and  water  is  always  the  same.  The  total  weight  simply 
depends  on  the  level  of  the  water.  The  weight  of  the  boat 
alone  is  always  equal  to  that  of  the  water  it  displaces,  and, 
therefore,  the  trough  filled  to  a  certain  height  with  water,  if 
there  is  no  boat,  will  just  weigh  the  same  as  if  there  is  a  boat 
in  it  and  the  water  is  at  the  same  level  as  before. 

Now,  suppose  that  the  trough  M  is  part  of  the  canal,  and 
there  is  or  is  not  a  barge  in  it,  and  suppose  that  N  is  down, 


APPLIED   MECHANICS. 

and  is  in  communication  with  the  river,  and  there  is  or  is 
not  a  barge  in  it.  Now  close  the  gates.  Suppose  there 
is  five  feet  of  water  in  M,  and  that,  if  there  is  not  four  feet 
six  inches  of  water  in  N,  we  let  water  into  or  out  of  it  from 
the  river  till  the  water  is  at  this  level.  Now  let  the  valve  be 
opened ;  water  will  flow  from  one  press  to  the  other,  for  M 
is  heavier  than  N,  and  M  will  fall,  causing  N  to  rise.  Suppose 
the  lifts  were  very  high  indeed,  it  is  evident  that  this  fall- 
ing will  not  stop  till  M  is  well  below  the  level  of  N;  in 
fact,  till  the  lightening  of  the  ram  M,  as  it  displaces  more 
water,  and  the  increased  weight  of  ram  N  as  it  leaves  the  water 
in  its  press — till  these  two  added  together  are  equal  to  the  real 
difference  in  the  loads  of  M  and  N.  In  the  present  case,  the  lift 
is  not  great  enough  for  such  an  effect  to  occur.  But  as  soon  as 
M  enters  the  river  it  gets  very  much  lighter  in  weight.  Indeed, 
as  soon  as  it  sinks  six  inches  into  the  river,  the  weights  of  M 
and  N  are  equal,  so  that  leaving  out  of  account  the  displace- 
ments of  the  rams,  N  cannot  be  raised  any  further  by  the  falling 
of  M.  There  is  exact  balance,  and  no  further  motion,  then, 
when  N  and  M  have  still  more  than  four  feet  six  inches  further 
to  travel.  The  valve  is  now  closed  ;  there  is  no  further  com- 
munication between  the  presses.  The  press  of  N  is  put  in 
communication  with  an  accumulator,  which  lifts  N,  pressing  it 
home  at  its  water-tight  joint  with  the  aqueduct.  The  water  in 
N  is  six  inches  lower  in  level  than  it  ought  to  be,  however. 
Water  is  allowed  to  pass  from  the  aqueduct  through  a  valve 
into  the  space  between  the  two  gates  of  the  trough  and  of  the 
aqueduct,  when  the  latter  are  easily  lifted ;  the  former  are  also 
easily  lifted  now,  and  water  passes  freely  into  N,  as  it  is  now 
part  of  the  aqueduct,  and  the  barges  can  be  taken  off  into  the 
canal.  Meanwhile  we  have  left  M  still  resting  in  great  part 
on  its  ram.  The  water  from  its  press  is  now  allowed  to 
escape,  and  M  sinks  in  the  river,  its  river-gate  is  lifted,  and 
the  barge  is  taken  out. 

Twelve  siphons  in  M  were  put  in  working  order  during  its 
immersion,  and  when  it  is  being  lifted  again  they  empty  it 
down  to  the  level  of  their  free  ends,  which  are  adjusted  to  leave 
exactly  four  feet  six  inches  of  water  in  M.  Thus  there  is  an 
automatic  adjustment  of  the  water-levels  in  M  and  N,  so  that 
whatever  be  the  weights  of  the  floating  barges  they  contain, 
the  total  weight  is  always  the  same  for  every  operation. 

The  waste  in  an  operation  comprises,  first,  six  inches  depth 


210  APPLIED  MECHANICS. 

of  water  in  one  of  the  troughs.  The  falling  of  this  50  feet  is 
1,800,000  foot-pounds.  Secondly,  the  accumulator  raises  N  four 
feet  six  inches,  and  as  the  weight  of  N  is  about  240  tons,  the 
waste  here  is  2,419,200  foot-pounds.  The  total  waste  is,  then, 
about  4,219,200  foot-pounds. 

Now,  suppose  the  same  operation  were  performed  down  a 
flight  of  ordinary  canal  locks,  under  favourable  circumstances, 
they  would  require  that  14J  feet  depth  of  water  of  the  area  of 
one  lock  should  fall  50  feet,  or  there  would  be  an  expenditure 
of  51,500,000  foot-pounds,  or  twelve  times  as  much  as  with  the 
lift.  If,  however,  a  canal  has  a  plentiful  supply  of  water,  this 
is  not  of  such  great  importance  as  it .  seems.  It  becomes  of 
great  importance  when,  as  in  the  present  case,  there  is  but  a 
small  supplyof  water.  The  advantage  of  such  a  system  as  the 
present  is  rather,  in  my  opinion,  in  the  fact  that  the  operation 
is  finished  in  eight  minutes,,  whereas  in  a  similar  ascent  or  de- 
scent, but  by  means  of  a  flight  of  locks,  at  Kuncorn,  the  opera- 
tion of  letting  one  barge  through  requires  one  hour  and  a  half. 

170.  We  are,  however,  more  interested,  just  now,  in  the 
hydraulic  question,  the  saving  of  energy,  the  saving  of  water 
from  the  canal,  and  accumulator  energy,  and  so  we  may  consider 
a  somewhat  similar  canal  lift  which  has  been  constructed  at 
Fontinette,  Belgium,  to  replace  a  flight  of  five  locks,  with  a  total 
fall  of  43  feet.  The  troughs  are  of  double  the  length,  and  are  of 
greater  depth  than  the  last.  Each  ram  is  six  feet  six  inches  in 
diameter,  and  there  is  the  improvement  that  we  have  what  are 
called  compensating  reservoirs.  Water  flows  from  one  of  them 
to  the  descending  trough,  thus  increasing  the  weight  of  it,  just 
in  proportion  as  its  ram  becomes  immersed  in  its  press ;  and 
water  flows  back  again  from  the  ascending  trough  to  the 
reservoir,  just  in  proportion  as  its  ram  comes  out  of  its  press. 
Thus  the  ram  displacement  is  balanced.  But  there  is  a  further 
improvement :  the  descending  trough  does  not  descend  into 
water,  for  this  made  it  get  light  too  soon ;  it  descends  into  a 
dry  chamber,  and  only  becomes  a  portion  of  the  lower  canal 
when  the  gates  are  lifted.  Thus  the  falling  of  one  trough 
can  lift  the  other  all  the  way  to  the  upper  level,  and  the 
accumulator  is  only  needed  to  supply  leakage  from  the  presses. 
A  single  operation  causes  a  loss  of  20  tons  of  water  from  the 
upper  level,  or  less  than  2,000,000  foot-pounds  of  energy 
altogether,  and  yet  the  troughs  have  more  than  twice  the 
capacity  of  those  at  Anderton. 


ALLIED  MECHANICS.  211 

171.  We  have  noticed  here  one  of  the  methods  adopted  for 
balancing   the   change  due   to  displacement  of   rams.       But 
many  other  methods  have  been  adopted  to  suit  special  cases, 
in  which  the  falling  of  an  accumulator  ram  causes  the  rise 
of  another  ram.    Thus,  for  example,  we  let  a  ram  rise  vertically 
above  the  accumulator,  passing  water-tight  into  a  tank  of  water. 
Its  increased  apparent  weight  (it  is  always  covered),  as  the 
accumulator  ram  descends,  compensates  for  the  displacement 
effects  of  all  the  other  rams  in  the  arrangement.     Again,  we 
may  have  a  tank  of  water  as  part  of  the  load  of  the  accumu- 
lator,   and   its   water-level   is   kept   the   same   as   that   of   a 
neighbouring  fixed  tank,  by  means  of  a  siphon,  so  that  the 
weight  increases  as  the  accumulator  ram  sinks.     By  properly 
shaping  this  tank  we  are  able  to  get  any  variation  of  pressure 
that  is  wanted  during  an  operation,  for  such  purposes  as  cotton- 
pressing,  etc.     It  is  easy  to  see  that  we  have  here  a  means  of 
balancing  accurately  the  dead  weight  of  any  platform  and  its 
ram,  which  has  to  be  raised  and  lowered,  at  every  point  during 
an  operation,  by  the  use  of  an  ordinary  accumulator. 

172.  Another  very  important  improvement  in  lifts  is  this. 
Suppose  that  a  bridge  has  to  be  lifted  by  rams  at  its  ends. 
Suppose  the  presses  of  these  rams  are  connected  with  an  ac- 
cumulator, they  are  not  likely  to  rise  equally  fast,  and  the 
bridge  would  be  tilted ;  indeed,  one  of  them  may  not  rise  at 
all,  and  the  other  may  be  rising  quickly.     Again,   suppose 
we  have  a  separate  accumulator  for  each  separate  press,  so 
that  there  is  no  water  communication  between   them,    great 
care  has  to   be   taken   at  the  valves   to    make  the   ends    of 
the  bridge  rise  equally  fast.     This  difficulty  is  got  over  by 
having  two  presses  and  two  rams  on  the  accumulator.     Now, 
let  each  accumulator  press  be  connected  with  each  bridge  press. 
The  frame  carrying  the  accumulator  weight  is  guided  so  that 
it  cannot  tilt ;  each  equal  ram,  therefore,  falls  the  same  amount 
as  the  other,  and  the  equal  bridge  rams  must  rise  one  as  much 
as  the  other. 

When  a  large  structure  has  to  be  lifted,  the  lifting  presses 
which  get  this  synchronous  action  are  so  distributed  in  sup- 
porting the  structure  that  there  is  no  tilting  of  it  possible. 

Fig.  141  shows  an  hydraulic  grid.  The  strong  girder, 
with  its  projecting  ribs,  rests  on  the  ends  of  a  number  of  rams, 
whose  presses  are  sunk  in  the  bed  of  the  dock  or  tidal  river. 
The  keel  of  a  vessel  is  brought  directly  over  it,  and  secured  iu 


APPLIED    MECHANICS. 


position  by  the  bilge  blocks,  and  side -shoring  frames;  the 
presses  are  worked,  and  the  rams  lift  the  grid  and  the  vessel 
above  the  level  of  high  water.  Now  a  number  of  struts,  which 
were  previously  held  up  horizontally,  are  liberated,  and  hang 
from  the  grid  alongside  the  rams  \  on  lowering  the  rams,  the 
lower  ends  of  these  struts  fit  into  the  tops  of  the  presses, 


Fig.  141. 

forming  a  support  for  the  grid,  and  the  rams  are  withdrawn 
into  their  presses. 

There  are  only  a  few  guide  columns  needed  for  the  grid  to 
slide  against  as  it  rises  and  falls,  because  the  presses  are 
arranged  in  three  equal  groups  on  the  above-mentioned  princi- 
ple, supplemented  by  an  automatic  safety  valve,  which  lets  the 
water  escape  from  a  press  when  its  ram  has  risen  more  than 
the  others.  Hence  the  grid  must,  when  rising  and  falling, 
remain  perfectly  level. 


APPLIED   MECHANICS.  213 

EXERCISES. 

1.  Hydraulic    jack;    velocity  ratio    of    lever,   30;    ram,   2£  inches 
diameter  ;  pump -plunger,  f  inch  diameter.     If  it  is  experimentally  found 
that,  E  being  effort  on  handle,  w  the  weight  lifted,  there  is  a  straight-line 
law  connecting  E  and  w ;  and  if,  when  w  =  1,605  Ibs.,  E  =  10  Ibs. ;  when 
w  =  6,805  Ibs.,   E  =  50  Ibs. ;    show  that  w  =  305  +  130  E.      If  when 
w  =  7,000  Ibs.  the  pressure  of  the  water  is  found  by  a  pressure-gauge  to 
be  1 ,932  Its.  per  square  inch,  what  is  the  loss  by  friction  at  the  leather  ? 
Assuming  the  same  percentage  loss  at  the  two  leathers,  what  is  the  law 
connecting  E  and  the  force  P  with  which  the  lever  acts  on  the  plunger  ? 

Ans.,  8'9  per  cent. ;  P  — 40-5  +  17'3  E. 

2.  A  steel  hydraulic  press  13  inches  internal  diameter  is  3  inches 
thick.     What  is  the  greatest  tensile  stress  when  there  is  a  fluid  pressure 
of  2|  tons  per  square  inch  ?  Ans.,  7 '6  tons  per  square  inch. 

If  a  steel  pipe  is  1  inch  in  diameter  inside,  and  the  greatest  tensile  stress 
is  to  be  5  tons  per  square  inch  when  there  is  a  fluid  pressure  of  3  tons  per 
square  inch,  what  is  the  thickness  of  the  metal  ?  Ana.,  ±  inch. 

3.  In  an  accumulator  the  average  pressure  is  to  be  700  Ibs.  per  square 
inch  ;  ram,  12  inches  diameter.     What  is  the  necessary  weight  ?     If  the 
ram  rises  10  feet,  what  energy  is  stored  up  ?    Neglect  change  of  pressure 
due  to  lifting.     If  the  pressure  is  found  to  fluctuate  between  720  and 
680  Ibs.  per  square  inch  between  slow  lifting  and  slow  falling  of  the 
weight,  what  is  the  force  of  friction  ?     If  the  pressures  are  settled  for  a 
position  half-way  up,  what  is  the  real  fluctuation,  taking  friction  and 
change  of  level  into  account  ? 

Ans.,  79,200  Ibs. ;  792,000  ft.-lbs. ;  2,260  Ibs. ;  2,505  Ibs. 

4.  In  a  three-cylinder  single-acting  pressure  engine  like  that  shown  in 
Fig.  128  each  piston  is  4  inches  diameter  and  3  inch  stroke  ;  the  average 
acting  pressure  is  700  Ibs.  per  square  inch.     What  is  the  indicated  horse- 
power at  50  revolutions  per  minute,  assuming  an  average  back-pressure 
due  to  friction,  etc.,  of  200  Ibs.  per  square  inch  ?    The  coil  of  a  rope  on  a 
drum  on  the  crank-shaft  is  18  inches  diameter  ;  what  is  the  average  pull 
in  the  rope  if  the  brake  horse-power  is  0*7  of  the  indicated  ?    If  the  pull 
is  to  be  only  300  Ibs.,  what  ought  the  stroke  to  be  altered  to  ? 

An*.,  7-14;  700  Ibs.  1-3  inch. 

5.  The  area  of  the  piston  of  a  hydraulic  crane  is  90  square  inches  on 
one  side  and  40  on  the  other  ;  it  pushes  a  three-sheave  pulley -block.     If 
the  water-pressure  is  700  Ibs.  per  square  inch,  what  weights  can  be  lifted 
— (1)  when  pressure-water  is  admitted  on  one  side  only,  (2)  when  ad- 
mitted on  both  sides.     Take  the  efficiency  of  the  "hydraulic  parts  as  0-9, 
and  of  the  pulley  and  crane  parts  as  0-4.     If  the  full  loads  in  the  two 
kinds  of  working  are  being  lifted,  what  work  is  done  per  cubic  foot,  per 
pound,  and  per  gallon  of  water? 

Ans.,  2,100  Ibs. ;  3,780  Ibs. ;  36,290  ft.-lbs. ;  582-6  ;  5,826  ft.-lbs. 

6.  A  hydraulic  punch  has  a  ram  8  inches  diameter ;  f -inch  holes  are 
being  punched,  each  requiring  a  force  of  70,000  Ibs.     What  is  the  water- 
pressure  ?     This  water  comes  from  a  steam  intensifier ;  the  area  on  which 
the  steam  acts  is  300  square  inches ;  the  area  of  the  ram  is  30  square 
inches.     What  is  the  pressure  difference  of  the  steam,  neglecting  friction  ? 

Ans.,  1,393  Ibs.  per  square  inch  ;  139*3  Ibs.  per  square  inch. 

7.  In  a  pipe  from  which  a  press  is  supplied,  the  pressure  is  in  one  cuse 
400  Ibs.  per  square  inch  and  in  another  it  is  100  Ibs.  per  square  inch  ;  the 


214 


APPLIED    MECHANICS. 


end  of  the  ram  of  a  hoist  is  70  feet  below  the  level  of  the  pipe,  and 
gradually  rises  to  be  10  feet  above  the  pipe.  What  is  the  change  of 
pressure  ?  If  the  rams  are  3  and  6  inches  in  diameter  respectively,  what 
are  the  lifting  forces  at  the  bottom  and  top  ?  What  are  the  fractional 
changes  in  the  two  cases  ?  If  chains  go  from  the  cages  vertically  over  a 
pulley  to  a  counter- weight,  what  ought  to  be  the  weight  of  the  two  chains 
per  foot  of  their  length  ? 

An*.,  34-8  Ibs.  per  square  inch ;    3,043  Ibs. ;  2,797  Ibs. ;    3,687  Ibs. ; 
2,703  Ibs. ;  8  per  cent.,  28  per  cent.,  3'OG  Ibs.,  12-23  Ibs. 

8.  Show  that  if  a  tank  has  water  in  it  to  the  same  level,  whether  there 
are  floating  objects  in  it  or  not,  its  total  weight  is  the  same.  If  the  weights 
of  two  tanks  (with  rams)  are  160  and  144  tons,  and  they  rest  on  rams  A 
and  B,  each  3  feet  diameter,  what  are  the  pressures  at  the  bottoms  of  the 
rams  ?  If  the  presses  communicate  with  one  another,  what  is  the  difference 
of  levels  of  the  bottoms  of  the  rams  when  there  is  balance  ?    Disconnect 
the  presses;  connect  the  press  B  with  an  accumulator  whose  ram  is  21 
inches  diameter,  its  end  being  now  30  feet  below  the  end  of  B.     What  is 
the  load  needed  for  the  accumulator  if  B  is  to  be  lifted  another  5  feet  ? 

9.  The  areas  of  vertical  cross-sections  of  the  immersed  part  of  a  ship  at 
intervals  of  10  feet  are  A  square  feet,  given  in  the  following  table.     Find 
approximately  the  weight  wa  of  water  displaced  between  every  two  neigh- 
bour sections.     The  weight  of  the  portions  of  the  ship  bounded  by  the  same 
sections  are  w2  ;  what  are  the  resultant  loads  w2  —  Wj  acting  downward  on 
the  ship  between  every  two  sections  ?   Draw  a  curve  showing  these  values. 


A 

jo 

30 

75 

ino 

10) 

100 

!)S 

H 

7S 

•to 

Wl 

9,300 

33,000 

55,000 

62,000 

62,000 

61,700 

59,000 

52,000 

37,000 

12,000 

Wfl 

23,000 

37,000 

47,000 

54,000 

57,000 

56,000 

53,000 

47,000 

39,000 

30000 

1 

10.  Fifty  rams,  each  of  14  inches  diameter,  begin  to  lift  a  gridiron 
from  the,  bottom  of  a  harbour,  and  so  lift  a  ship  which  was  floating 
directly  above.  The  plane  areas  in  square  feet  bounded  by  the  water-line 
of  the  vessel,  after  the  following  numbers  of  feet  of  lift,  are  shown  as  the 


Lift  in  feet. 

0 

1 

8 

6 

9 

12 

15 

18 

21 

24 

27 

30 

A 

4,050 

4,090 

4,120 

4,100 

3,970 

3,400 

3,000 

2,200 

1,400 

600 

200 

0 

P 

250 

283 

350 

450 

547 

637 

717 

780 

824 

848 

868 

800 

numbers  A.  If  the  total  apparent  weight  of  the  gridiron  is  taken  to  be 
constant  at  860  tons,  find  the  pressure  in  the  presses  for  every  position, 
and  compare  with  the  answers  given  in  the  table. 

173.  Force  due  to  Pressure  of  Fluids. 

Exercise  1. — Prove  that  if  p,  the  pressure  of  a  fluid,  is  constant, 
the  resultant  of  all  the  pressure  forces  on  the  plane  area  A  is  A  p, 
and  acts  through  the  centre  of  the  area. 

2.  The  pressure  in  a  liquid  at  the  depth  h  being  wh,  where  w 
is  the  weight  of  unit  volume,  what  is  the  total  force  due  to  pressure 
on  any  immersed  plane  area  ?  Let  D  E  (Fig.  142)  be  the  surface  from 
which  the  depth  h  is  measured,  and  where  the  pressure  is  Q.  Let  B  o 


APPLIED    MECHANICS.  215 

be  an  edge  view  of  the  area  ;  imagine  its  plane  produced  to  cut  the 
level  siirface  of  the  liquid  c  in  D.  Let  the  angle  EDO  be  called  a  ; 
let  the  distance  D  p  be  called  a?,  and  let  D  Q  be  called  x  +  8a? ;  and 
let  the  breadth  of  the  area  at  right  angles  to  the  paper  at  p  be 
called  z.  On  the  strip  of  area  z  .  5x  there  is  the  pressure  wh  if  h  is 
p  H,  the  depth  of  p,  and  h  =  x  sin.  a  ;  so  that  the  pressure  force  on 
the  strip  is  wx .  sin.  a  .  z  .  8%,  and  the  whole  force  is 

TDC 
x  .  z  .  d x  .  .  .  .  (1). 


JDC 
X 
D  B 


Also,  if  this  resultant  acts  at  a  point  in  the  area  at  a  distance  x 
from  D,  taking  moments  about  D, 

re 
x1 .  z  .  dx  .  .  .  .  (2). 


f*n  c 

sin.  a  1        x1 .  z  . 
J  DB 
CD  c 
tl        x 

J  D  B 


Observe  in  (1)  that  1        x  .  z  .  dx  =  A  a?,  if  A  is  the  whole  area 

J  D  B 

and  x  is  the  distance  of  its  centre  of  gravity  from.  D.  Hence  the 
average  pressure  over  the  area  is  the  pressure  at  the  centre  of 
gravity  of  the  area.  ,,D  c 

Observe  in  (2)  that  I        a?2 .  z  .  dx  =  i,  the  moment  of  inertia  of 

J  DB 

the  area  about  D.  Letting  i  ==  k'2  A,  where  k  is  called  the  radius 
of  gyration  of  the  area  about  D,  we  see  that 

F  =  w  sin.  a .  A  #,          F  x  =  w  sin.  a  .  A  &2. 
Hence  x  =  k^jx.  .  .  .  (3),  the  distance   from  D   at  which    the 
resultant  force  acts. 

Example. — If  D  B  =  0  and  the  area  is  rectangular,  of  constant 
breadth  b,  then 


-J 


'DC  i 

0-2  .  <to  =  -  D  c3, 
0 


and  A  =  5  .  D  c  ;    so    that 
&2  =  ^  D  ^  aiso  3,  _  i  D  c 

Hence  x  =  f  D  c ;  that  is, 
the  resultant  force  acts  at  f  of  the  way  down  the 
rectangle  from  D  to  c,  and  the  average  pressure 
is  the  pressure  at  a  point  halfway  down. 

It  is  an  easily-remembered  relation  that  we 
find  in  (3).  For  if  we  have  a  compound  pendu- 
lum whose  radius  of  gyration  is  k,  and  if  x  is 
the  distance  from  the  point  of  support  to  its  centre  of  gravity, 
and  if  x  is  the  distance  to  its  point  of  percussion,  we  have  the  very 
same  equation  (3).  Again,  if  x  is  the  length  of  the  simple  pen- 
dulum, which  oscillates  in  exactly  the  same  time  as  the  com- 
pound one,  we  have  again  this  same  relation  (3).  These  are 
merely  mathematical  helps  to  the  memory,  for  the  three  physical 
phenomena  have  no  other  relation  to  one  another  than  a  matjae- 
matical  one. 


216 


APPLIED    MECHANICS. 


EXERCISES. 

1.  Find  the  whole  pressure  of  water  upon  a  vertical  dam  10  feet  deep, 
30  feet  wide.     What  would  be  the  pressure  of  water  of  the  same  depth 
against  a  dam  inclined  at  45°  ?  Ans.,  93,750  and  132,200  Ibs. 

2.  A  water-tank  is  13  feet  square  and  4  feet  6  inches  deep.     Find  the 
pressure  upon  one  of  the  sides  when  the  tank  is  full.  Ans.,  8,226-56  Ibs. 

3.  In  the  vertical  plane  side  of  a  tank  which  holds  water  there  is  a 
rectangular  plate  whose  depth  is  1  foot  and  breadth  2  feet,  the  upper  edge 
being  horizontal,  and  8  feet  below  the  surface  of  the  water.     Find  the 
pressure  on  the  plate.  Ans.,  1,062-5  Ibs. 

4.  A  rectangular  tank,  5  feet  square,  is  filled  with  water  to  a  height  of 
7f  feet.     A  rectangular  block  of  wood  weighing  312-5  Ibs.,  and  having  a 
cross- section  of  5  square  feet,  is  placed  in  the  tank,  where  it  floats  in  the 
water  with  this  section  horizontal.     How  much  does  the  water  rise  in  the 
tank,  and  what  is  the  increase  in  pressure  on  a  vertical  side  of  the  tank  1 

Ans.,  2f  inches,  489  Ibs. 

5.  A  sluice-gate  is  6  feet  broad  and  8  feet  deep,  and  the  water  rises  3 
feet  on  one  side  above  the  lower  edge,  and  7  feet  on  the  other  side.     Find 
the  resultant  pressure  and  centre  of  pressure  for  each  side  of  the  gate. 
What  is  the  resultant  of  these  and  its  position  ? 

An*.,  1,682  Ibs.,  9,158  Ibs. ;  2  ft.  and  4f  ft.  below  surface;  7,476  Ibs., 
4-36  ft.  from  surface. 

6.  A  vertical  wall,  10  feet  high  and  2  feet  thick,  weighing  153  Ibs.  per 
cubic  foot,  has  to  support  the  pressure  of  water  (weighing  62-3  Ibs.  per 
cubic  foot).     How  high  may  the  water  rise  against  one  side  of  the  wall 
without  causing  the  resultant  force  at  the  base  to  pass  outside  the  outer 
edge?  Ans.,  6'65  feet. 

7.  A  rectangular  opening  A  B  c  D  is  made  in  the  outer  vertical  face  of  a 
reservoir  embankment  and  closed  by  a  door,  hinged  along  the  lower 
horizontal  edge  A  B.     Find  the  pressure  to  be  applied  at  the  upper  edge 
CD   of  the  door,   in  order  to  shut  against  the  pressure  of  the  water. 
Breadth  of  door,  3  feet ;  depth,  4  feet ;  depth  of  upper  edge  below  surface, 
8  feet.  Ans.,  3,489  Ibs. 

8.  In  a  hydraulic  jack  the  ram  is  5  inches,  and  the  pump  plunger  f 
inch  diameter;   the  leverage  for  working  pump,  10  to  1;   what  is  the 
velocity  ratio  ?    Experimentally  we  find  that  a  force  of  20  Ibs.  lifts  8,200 
Ibs. ;  what  is  the  efficiency  ?  Ans.,  444£ ;  92^  per  cent. 

9.  Water  at   750   Ibs.   per  square  inch  acts  on  a  piston  8  inches 
diameter ;  the  velocity  is  multiplied  by  8  by  a  four-sheaved  pulley  block ; 
what  weight  may  be  lifted  by  the  chain  if  the  efficiency  of  the  mechanism 
is  55  per  cent  ?    In  a  lift  of  40  feet,  how  many  gallons  of  water  are  used  ? 

Ans.,  2,592  Ibs. ;  10-87. 
v,"  174.  Whirling    Fluids. — If    fluid 

D  rotates  like  a  rigid  body  about  the 
vertical  axis  o  o  (which  is  in  the 
plane  of  the  paper)  with  the  angu- 
lar velocity  a  radians  per  second,  at 
a  point  P  let  there  be  a  particle  of 
weight  w  Ibs.  ;  let  o  r  =  x  feet. 

The  centrifugal  force  is  —  o2a?  in  the 
Fig.  143.  9 


APPLIED    MECHANICS. 


217 


direction  p  D  ;  the  weight  is  w  in  the  vertical  direction  p  c  ;  and  the 
resultant  force  is  in  the  direction  p  E  of  the  amount  w  \'\  +  o4a?2/^2, 

the  tangent  of  the  angle  D  p  E  being  w  -. o2a?  or  rj\arx. 

Imagine  many  curves  drawn  such  that  their  direction  at  any  point 
represents  the  direction  of  the  resultant  force  there.    Such  curves  are 


Fig.  144. 


called  lines  of  force.  Let  the  point  P  be  y  feet  above  some  datum 
level,  and  let  us  find  the  equation  to  the  line  of  force  which  passes, 
through  P.  The  slope  of  the  curve  dy/dx  is  -tan.  DPE  (Fig.  143),, 

OT  ^=~«^;  S0  that  y=  -  J5log.»  +  fl....(l),  **«»» 
c  is  a  constant.  We  see,  then,  that  the  lines  of  force  are  logarithmic: 
curves.  If  there  is  a  curve  to  which  p  E  is  normal  at  the  point  p,, 


218  APPLtfeb   MECHANICS. 

its  slope  is  positive,  being  tan.  v  P  D,  or  ~  =  —  x ;   so  that  the 

equation  to  the  curve  is  y  =  -^-  a?2  -f  c,  where  c  is  a  constant, 

depending  upon  the  datum  level  from  which  y  is  measured.  This 
is  a  parabola,  and  if  it  revolves  about  the  axis  oo  we  have  a 
paraboloid  of  revolution.  Any  surface  which  is  everywhere  at 
right  angles  to  the  force  at  every  point  is  called  a  level  surface,  and 
we  see  that  the  level  surfaces  in  this  case  are  paraboloids  of  revolu- 
tion. These  level  surfaces  are  sometimes  called  equi-potential 
surfaces.  It  is  easy  to  prove  that  the  pressure  is  constant  every- 
where in  such  a  surface,  and  that  it  is  a  surface  of  equal  density  ; 
so  that  if  mercury,  water,  oils,  and  air  are  in  a  whirling  vessel,  their 
surfaces  of  separation  are  paraboloids  of  revolution. 

The  student  ought  to  draw  one  of  the  lines  of  force  and  cut  out 
a  template  of  it  in  thin  zinc,  o  o  being  another  ed^e.  By  sliding 
along  o  o  he  can  draw  many  lines  of  force.  Now  let  him  cut  out 
a  template  for  one  of  the  parabolas,  and  with  it  draw  many  level 
surfaces.  The  two  sets  of  curves  cut  each  other  everywhere 
orthogonally.  Fig.  144  shows  the  sort  of  result  obtainable  where 
a  a,  b  b,  c  c  are  the  logarithmic  lines  of  force,  and  A  A,  B  B,  c  c  are 
the  level  paraboloidal  surfaces. 

175.  The  engineer  seldom  deals  with  other  volumetric  forces 
than  those  due  to  gravity  and  centrifugal  force.  By  dealing  in  this 
way  with  centrifugal  force  he  is  able  to 
treat  a  rotational  problem  as  a  statical  pro- 
blem Whatever  be  the  system  of  volumetric 
forces,  we  are  supposed  to  know  the  lines 
of  force  in  the  fluid  and  a  series  of  level 
or  equi-potential  surfaces  cutting  these 
lines  at  right  angles.  In  Fig.  145,  if  ABC 
and  A'  B'  c'  be  lines  of  force,  and  B  B  and 
c  c  be  two  level  surfaces  ;  if  F  is  the  force 
on  the  fluid  at  o  per  unit  volume  and 
Fig.  145.  B  c  =  Ss ;  if  the  pressure  at  B  B'  is  p  and 

at  c  d  it  is  p  +  Sjt?,  consider  the  equilibrium 
of  the  prism  whose  end  of  small  area  a  is 

at  B  B'  and  other  end  of  equal  area  is  at  c  c'  (we  take  the  ends  of 
equal  area  because  we  afterwards  assume  a  and  8s  to  be  smaller 
and  smaller  without  limit).  Evidently,  as  the  volume  is  a .  Sst  the 
volumetric  force  is  F  .  a .  8*,  and  we  have 

p  a  -f  F  a .  5*  =  {p  -f  tip)  a  .  .  .  .  (1). 
Hence        P  .  5s  =  Sp,    or    F  =  dp/d»  ....  (2). 

Example  1.— If  gravity  alone  acts  and  5s  is  called  8  A,  so  that  h 
is  measured  vertically  downwards  from  A  towards  c,  and  we  take 
P  =  wy  the  weight  of  a  cubic  foot  of  fluid,  to  be  constant,  being 

62-3  Ibs.  per    cubic  foot  for  water,  then  J?  =  w  .  .  .  .  (3),   or 

p  =  wh  +  c  where  c  is  a  constant.  If  we  measure  h  from  a  level 
where  we  take  the  pressure  to  bu  pw  then  c  =ftp.  and  P  —  p0=  wh 


At»t>Llfil>   MECHANICS.  219 

Example  2.  —  Take  w  to  be  variable,  say  w  =  cp,  a  rule  that  holds 
for  gases  at  constant  temperature,  then  (3)  becomes  ~  =  cp,  or  —  = 

c  .  dh,  or  log.  p  =  ch  +  c  where  c  is  a  constant. 

Let  p  =  p0  where  h  =  o,  then  c  =  log.  p0,  and  we  have  log. 
p/p0  —  ch  .  .  .  .  (4).  The  actual  value  of  c  depends  upon  the 
constant  temperature  supposed  to  be  maintained. 

Example  3.  —  Take  w  to  be  variable,  say  w  =  cpVt  where 
y  =  1-414  for  air,  being  the  ratio  of  the  specific  heats.  This  is 
the  law  which  is  much  more  likely  to  hold  in  a  mass  of  gas  than 
the  constant  temperature  law. 


Then       =  eplh,  orp'1^  .  dp  =  c  .  dh,  or  p1  ~  ln/(l  -  l/7)  = 

ch  +  c.  It  is  easy  to  show  that  this  leads  to  the  result  that  the 
temperature  increases  in  proportion  to  the  increase  of  depth  in  the 
atmosphere. 

Example  4.  —  In    our    whirling    fluid  it   is    easy,   since   F  = 

w  V  1  +  «4a?2/72  and  ^  =  V  1  +  #2/«4a;2>  to  find  from  (2)  the  law 

of  y,  if  one  knows  how  to  integrate.  Take  the  simpler  case,  in 
which  x  is  so  great  that  the  lines  of  force  may  be  regarded  as 
horizontal,  and  the  level  surfaces  vertical  circular  cylinders.  Then 
letting  8s  be  called  5a?,  x  being  the  radius  of  the  path  in  which  a 
particle  revolves, 

W    9  J    W    9  3 

p  =  —  era?,  and  —a2*  .  dx  •=  dp. 


wa 
(1)  If  w  is  constant,  --^      P  +  c  where  c  is  some  constant. 

p=pQ  wh 
,2)  If  »= 


Let  p=pQ  where  x  =  a?0,  and  we  have  p  -pQ  =  ^r~  (**-  V)  «... 


If  we  take  &=—  where  y  =  l'414  for  air,  and  the  easily  found 

value  of  c  for  any  given  conditions,  we  can  "find  the  increase  of 
pressure  due  to  centrifugal  force  in  a  mass  of  whirling  gas,  as  in 
the  wheel  of  a  fan  whea  it  is  not  delivering  much  fluid. 


220 


CHAPTER   X. 

MACHINERY    IN    GENERAL. 

176.  Mechanism. — When  the  power  of  a  steam-engine  is  dis- 
tributed through  a  factory,  the  distribution  is  performed  by 
means  of  shafts,  spur  and  bevil  wheels,  belts  and  pulleys,,  and 
other  kinds  of  gearing.     As  we  are  writing  for  men  who  have 
observed  such  transmission  of  energy,  it  is  no  part  of  our  object 
to  describe  here  what  can  be  seen  in  any  workshop.     Perhaps 
no  study  is  more  useless  from  books  alone  than  the  study  of 
mechanism  ;  whereas  it  is  very  useful  and  easy  if  we  examine  the 
actual  thing,  or  make  a  skeleton  model  or  a  skeleton  drawing. 

At  the  same  time  it  is  necessary  to  read  books.  The 
present  book  deals  with  the  kinetics  of  mechanism ;  but  there  is 
another  part,  a  preliminary  part,  and  students  must  read  some 
book  giving  descriptions  of  contrivances  and  the  mere  kine- 
matics of  the  subject.  We  give  a  short  sketch  of  certain 
important  principles  here,  and  in  Chap.  XXVI.,  because  they 
are  necessary  for  the  proper  understanding  of  our  own  division 
of  the  subject. 

177.  Velocity  Ratio. — In  any  machinery  the  velocity  of  a 
point  may  be  calculated  when  the  velocity  of  any  other  point 
is  known.     The  number  of  revolutions  per  minute  made  by  a 
shaft  tells  us  the  velocity  of  any  point  on  any  wheel  or  pulley 
fixed  on  the  shaft ;  the  circumference  of  the  circle  described  by 
such  a  point,  multiplied  by  the  number  of   revolutions  per 
minute,  is  evidently  the  distance  moved  through  by  the  point 
in  one  minute.     Now,  when  one  shaft  drives  another  by  means 
of  spur  or  bevil  wheels,  or  by  two  pulleys  and  a  strap,  it  is 
evident  that  the  number  of  revolutions  per  minute  made  by  one 
of  the  shafts,  multiplied  by  the  number  of  teeth  of  the  wheel, 
or  by  the  circumference  or  diameter  of  the  wheel  or  pulley,  is 
equal  to  the  number  of  revolutions  per  minute  made  by  the 
other  shaft,  multiplied  by  the  number  of  teeth,  or  by  the  cir- 
cumference or  diameter  of  the  other  wheel  or  pulley.     This  is 
evidently  true,  supposing  that  the  strap  does  not  slip  on  the 
pulley.      Hence  the  rule — to  find  the  speed  of  a  shaft,  driven 
from  another  by  means  of  any  number  of  wheels  and  pulleys, 
multiply  the  speed  of  the  driving  shaft  by  the  product  of  the 


APPLIED    MECHANICS.  221 

diameters  or  numbers  of  teeth  in  all  the  driving  wheels  or 
pulleys,  and  divide  by  the  product  of  the  diameters  or  numbers 
of  teeth  in  all  the  driven  wheels  or  pulleys.  By  the  diameter  of 
a  spur  wheel  we  mean  the  diameter  of  its  pitch  circle.  Two 
spur  wheels  enter  some  distance  into  one  another,  and  the 
circle  on  one  which  touches  the  circle  on  the  other  (the  dia- 
meters of  these  circles  being  proportional  to  the  numbers  of 
teeth  on  the  wheels),  is  called  the  pitch  circle.  The  circum- 
ference of  the  pitch  circle,  divided  by  the  number  of  teeth, 
gives  the  pitch  of  the  teeth. 

178.  Shapes  of  Wheel  Teeth. — We  know  that  if  two  spur 
wheels  gear  together,  however  badly  their  teeth  are  formed,  so 
long  as  a  toofch  in  one  drives  past  the  line  of  centres  of  a  tooth 
in  the  other,  their  average  speeds  follow  the  above  rule.     But 
if  we  want  the  speed  ratio  at  any  instant  to  be  the  same  as  at 
any  other  instant,  it  is  necessary  to  form  the  teeth  in  a  certain 
way.     The  curved  sides  of  teeth  ought  to  be  cycloidal  curves. 
The  proof  of  this  is  not  very  difficult ;  it  is  given  in  Art.  462. 
It  is  not  usual  to  employ  these  cycloidal  curves,  for  it  is  found 
that  certain  arcs  of    circles  approximate  very  closely  to  the 
proper   curves.     The  method  of  drawing  rapidly  the  curved 
tooth  of  a  wheel  you   will  find  taught  by  every  teacher  of 
mechanical  drawing,  you  will  find  described  in  a  great  number 
of  books,  and  you  will  see  it  in  use  in  the  workshop.*     You 
must  remember  that  no  study  of  books,  and  I  may  also  say,  no 
fitter's  or  turner's  work  that  you  may  engage  in,  will  make  up 
for  want  of  the  experience  which  you  would  gain  by  actually 
drawing  to  scale  a  spur  or  bevil  wheel,  a  bracket  or  pedestal 
with  brasses,  and  a  few  other  contrivances  used  in  machinery. 
A  worm  and  worm-wheel,  that  is,  a  screw,  every  revolution 
of  which  causes  one  tooth  of  a  wheel  to  be  driven  forward,  is 
sometimes  used  when  we  wish  to  drive  a  shaft  with  a  very  slow 
speed.     If  the  worm-wheel  has  30  teeth,  it  evidently  makes 
one- thirtieth  of  the  number  of  revolutions  of  the  driving  shaft. 

179.  Skeleton  Drawings. — When  we  consider  the  relative 
motions  of,  say,  a  piston  and  the  crank  which  it  drives,  we 
come  to  something  which  it  is  not  so  easy  to  state  without  some 
little  knowledge  of  mathematics.     It  is  the  same  with  all  sorts 
of  combinations  of  link  work,  and  with  cams.     Even  a  good 
knowledge  of  mathematics  is  only  sufficient  to  give  one  a  rough 
general  idea  of  the  relative  motion  in  such  cases ;  and  for  the 

*  Consult  Professor  Unwin's  "Machine  Design  "  on  the  terth  of  wheels, 


222 


APPLIED    MECHANICS. 


study  of  any  special  case  there  is  nothing  so  good  as  a  skeleton 
drawing  or  a  model.  I  give  one  example  of  the  use  of  skeleton 
drawings — a  crank  and  connecting- 
rod.  Let  A  and  B  (Fig.  14 6)  be  the  ends 
of  a  connecting-rod.  As  A  moves 
from  ax  to  c  and  back  again,  B  de- 
scribes the  complete  circle,  61  d  b^. 
Set  off  equal  distances,  5X  &2,  62  B,  B  64, 
etc.,  and  make  62  »2,  b^  «4,  etc.,  equal 
to  the  length  of  the  connecting-rod. 
Then  the  points  a^  «2,  etc.,  and  6X  b^ 
etc.,  show  in  a  very  good  way  the 
relative  motions  of  A  and  B.  When 
you  have  finished  this  exercise,  work 
others  in  which,  with  the  same  length 
of  crank,  you  have  longer  or  shorter 
connecting-rods.  You  will  get  some 
such  results  as  are  shown  in  the  upper 
part  of  the  figure.  In  every  case,  if 
we  imagine  the  crank  to  revolve  uni- 
formly, the  motion  of  A,  the  end  of 
the  connecting-rod,  is  shown ;  the 
distance  from  any  one  point  to  the  next 
is  passed  over  by  A  in  the  same  in- 
terval of  time.  Simple  Harmonic 
Motion  (see  Chap.  XXV.)  is  the  name 
given  to  the  motion  of  the  piston- 
rod,  when  we  imagine  the  connecting- 
rod  to  be  infinitely  long ;  or  rather,  as 
we  make  the  connecting-rod  longer  and 
longer,  we  get  more  and  more  nearly 
to  this  sort  of  motion.  You  see,  then, 
that  by  skeleton  drawings  I  mean 
drawings  which  show  successive  posi- 
tions of  the  different  parts  of  a 
mechanism  whose  motions  we  want  to 
5  6  6  «  •  6?j*  study.  You  will  find  that  an  eccen- 

tric and  its  rod  may  be  regarded  as 
a  crank  (the  length  of  a  crank  is  the 

distance  between  the  axis  about  which  the  eccentric  is  revolv- 
ing and  its  true  centre),  and  a  very  long  connecting-rod  (the 
length  of  the  connecting-rod  being  the  length  of  the  eccentric 


APPLIED   MECHANICS.  223 

rod  measured  to  the  true  centre  of  the  disc).  The  advantage 
derivable  from  skeleton  drawings  will  be  more  obvious  if  you 
consider,  in  the  above  case  of  a  crank  and  a  connecting-rod, 
that  A  need  not  be  the  cross-head  at  the  end  of  the  piston- 
rod  ;  it  may  be  the  end  of  a  lever,  and  so  move  in  the  arc  of 
a  circle ;  it  may  be  a  slide  moving  in  a  slot  of  any  curved 
form.  One  of  the  most  instructive  cases  of  skeleton  drawing 
is  a  link-motion.  Taking  any  good  drawing  of  a  link-motion 
to  start  with,  find  the  relative  motions  of  piston  and  of  slide 
valve  for  various  positions  of  the  link.  In  the  study  of  the 
motion  of  a  slide  valve  it  is  much  too  usual  to  assume  that  the 
piston's  motion  is  what  is  shown  in  Fig.  146  as  simple  harmonic 
motion.  The  reason  of  this  lies  in  the  ease  with  which  it  can 
be  stated  in  mathematical  language  ;  but  it  is  incorrect. 

180.  In  this  early  part  of  our  work  we  wish  to  confine  our 
attention  to  energy  and  power  calculation  in  the  simplest  mech- 
anisms ;  the  transmission  of  power  by  rotating  shafts ;  and  the 
kinematic  principles  involved  are  very  simple.  In  Chap.  XXVI. 
we  have  something  more  to  say  about  mechanism  in  general. 

EXERCISES. 

1.  Two  parallel  shafts,  whose  axes  are  to  he  as  nearly  as  possihle  2  feet 
6  inches  apart,  are  to  he  connected  hy  a  pair  of  spur  wheels,  so  that  while 
the  driver  runs  at  100  revolutions  per  minute,  the  follower  is  required  to 
run  at  25  revolutions  per  minute.     Find  the  diameters  of  the  wheels  and 
also  the  numher  of  teeth  on  each,  if  the  pitch  is  1^  inches. 

Ans.,  48  inches,  12  inches;  120  teeth,  30  teeth. 

2.  A  main  shaft  carrying  a  pulley  of  15  inches  diameter  and  running 
at  60  revolutions  per  minute  communicates  motion  hy  a  helt  to  a  pulley 
of  12  inches  diameter,  fixed  to  a  countershaft.     A  second  pulley  on  the 
countershaft  of  8J  inches  diameter  carries  oil  the  motion  to  a  revolving 
spindle   which   is   keyed   to  a  pulley  of  4£  inches  diameter.     Find  the 
number  of  revolutions  per  minute  made  hy  the  last  pulley.      Ans.,  150. 

3.  On  the  crank-shaft  of  an  engine  there  is  a  pulley  2  feet  6  inches  in 
diameter.  By  means  of  a  helt  this  drives  a  pulley  25  Inches  in  diameter  on 
a  second  shaft,  on  which  is  another  pulley,  24  inches  in  diameter,  which 
drives  another  pulley,  15  inches  in  diameter,  fixed  on  a  third  shaft.     On 
this  shaft  is  a  pulley  25  inches  in  diameter,  which  drives  one  of  10  inches 
diameter  on  a  fourth  shaft.     On  this  shaft  is  another  pulley,  20  inches  in 
diameter,  which  drives  a  pulley  8  inches  in  diameter  fixed  on  a  dynamo 
shaft.     If  the  engine  runs  at  100  revolutions  a  minute,  what  will  he  the 
speed  of  the  dynamo  1  Ans.,  1,200  revs,  per  minute. 

4.  In  a  screw-jack  where  a  worm-wheel  is  used,  the  pitch  of  the  screw 
is  |-inch,  the  numher  of  teeth  is  20,  and  the  length  of  the  lever  which 
works  the  worm  is  12  inches.     What  is  the  velocity  ratio?     Ans.,  2413-1. 

5.  A  wheel  of  40  teeth  is  turned  hy  a  winch  handle  14  inches  long, 
and  gears  with  a  rack  having  teeth  of  1  inch  pitch.     If  the  axis  of  the 


224  APPLIED    MECHANICS. 

wheel  is  fixed,  what  is  the  travel  of  the  rack  for  two  turns  of  the  handle  P 

Ans.,  80  inches. 

6.  The  table  of  a  drilling- machine  is  raised  by  a  worm-wheel  in  com- 
bination with  a  rack  and  pinion.     The  handle  which  works  the  worm  is 
12  inches  long,  the  worm  wheel  has  30  teeth,  and  the  pitch  circle  of  the 
rack  pinion  is  4  inches  in  diameter.     What  is  the  lift  of  the  table  for  one 
turn  of  the  handle?    If  the  table  and  accessories  weigh  500  Ibs.,  what 
weight  on  the  table  would  be  balanced  by  a  force  of  12  Ibs.  applied  at  the 
handle,  if  45  per  cent,  of  the  force  applied  be  lost  ?    Ans.,  0'42  inch  688  Ibs. 

7.  If  the  two  wheels  in  the  back-gear  of  a  lathe  have  63  teeth  each, 
and  the  pinions  25  teeth,  what  is  the  reduction  in  the  velocity  ratio  of  the 
lathe  spindle  due  to  the  back-gear?  Ans.,  6'35  ;  1. 

8.  The  slide-rest  of  a  screw-cutting  lathe  moves  along  the  bed  14 
inches  while  the  leading  screw  makes  56  revolutions.  What  is  the  pitch  of 
the  screw  thread?  Ans.,  %  inch. 

9.  It  is  desired  to  cut  a  screw  of  f  inch  pitch  in  a  lathe  with  a  leading 
screw  of  4  threads  to  the  inch,  using  four  wheels.     If  both  screws  be 
right-handed,  what  wheels  would  you  employ  ? 

10.  It  is  required  to  cut  a  left-handed  screw  of  5  threads  to  the  inch  in. 
a  lathe  fitted  with  a  right-handed  guide-screw  of  \  inch  pitch.     Show 
how  the  change  wheels  might  be  arranged,  and  state  the  numbers  of 
teeth  on  them. 

181.  How  a  Shaft  transmits  Power. — I  have  refused  to 
describe  for  you  what  you  may  see  for  yourselves  at  any 
time  in  workshops — how  spur  and  bevil  wheels  and  belts 
transmit  power ;  how  there  are  arrangements  for  disengaging 
such  gearing,  and  stepped  cones  for  giving  change  of  speed 
when  belts  are  used ;  how  shafts  are  carried  near  walls 
or  columns  ;  how  machine  tools  work,  and  a  hundred  other 
matters  about  which  a  little  obser- 
vation and  drawing  are  of  more  im- 
portance than  a  large  amount  of  read- 
ing. Bub  there  are  some  matters  con- 
nected with  machinery  of  great 
interest  to  you  which  you  are  not 
likely  to  observe  unless  I  direct  your 
attention  to  them.  When  a  shaft 
Fig.  147.  transmits  power  it  is  in  a  state  of 

strain;   it  is  in  a  twisted  condition. 

The  twist  is  not  perceptible  to  the  eye,  of  course,  but  methods 
have  been  arranged  to  show  it  to  the  eye,  and  measure  it ;  and 
it  is  found  that  the  twist  in  a  shaft  is  proportional  to  the  horse- 
power transmitted  by  the  shaft  divided  by  the  number  of  revolu- 
tions per  minute.  Now  to  explain  what  I  mean  by  a  twist.  Let 
a  straight  line  be  drawn  along  the  shaft  when  power  is  not  being 


APPLIED    MECHANICS. 


225 


transmitted,  then,  if  power  be  transmitted,  the  shaft  will 
receive  a  twist,  and  this  line  will  become  a  spiral  line.  The 
inclination,  at  any  point,  of  the  spiral  line  to  its  old  position, 
is  a  measure  of  the  twist.  *  When,  instead  of  the  ordinary 
coupling,  Fig.  147,  in  which  the  two  halves  are  connected  by 
means  of  bolts,  we  use  one,  Fig.  14S,t  in  which  the  two  halves 
are  connected  by  means  of  spiral 
springs,  these  springs  get  extended 
when  the  shaft  transmits  power. 
The  yielding  of  the  springs  cannot 
be  observed  unless  we  make  some 
arrangement  like  that  shown,  where 
the  motion  of  A  relatively  to  c 
causes  the  arm  E  to  move  and  bring 
the  bright  bead  B  towards  the  axis. 
If  everything  is  made  dead  black 
except  the  bead  it  will  be  seen  de- 
scribing a  circle  of  greater  or  smaller 
radius,  and  a  scale  with  a  sliding 
pointer  enables  us  to  measure  ac- 
curately the  distance  moved  inwards 
by  the  bead.  The  reading  on  the 
scale  multiplied  by  the  number  of 
revolutions  of  the  shaft  per  minute 
tells  us  at  once  the  horse-power  actu- 
ally passing?  through  the  coupling.  J 

At  Finsbury,  the  scale  comes  out  from  a  wall,  the  shaft  is 
about  ten  feet  above  the  floor  ;  the  observer  stands  on  a  ladder 
with  a  gas-jet  behind  him.  It  is  quite  easy  to  avoid  error  due 
to  parallax,  and  to  read  with  a  very  considerable  amount  of 

*  The  best  measure  of  the  twist  is  this  angle  of  the  spiral  divided  by  the 
radius  of  the  shaft,  and  the  quotient  is  called  the  angle  of  twist.  See  Art.  294. 

f  Ayrton  and  Perry's  Dynamometer  Coupling. 

j  The  total  moment  of  the  forces  of  the  springs  in  pound-feet,  or,  as  it  has 
been  called  by  Professor  James  Thomson,  the  torque,  multiplied  by  the  angular 
velocity  in  radians  per  minute,  divided  by  33,000,  is  the  horse-power.  Suppose 
that  when  one  of  the  lengths  of  shafting  is  held  fast  we  find  the  position  of  the 
bead  when  we  hang  weights  on  levers  or  round  pulleys  or  wheels  fastened  to 
the  other  length  ;  a  torque  of  52'5  pound-feet  will  cause  the  bead  to  move 
radially  inwards  by  a  distance  which  we  call  '01  on  our  scale ;  a  torque  of  105 
pound-feet  causes  the  bead  to  move  inwards  a  distance  which  we  call  '02  on 
our  scale,  and  so  on.  Such  a  coupling  ought  to  be  graduated  by  actual  experi- 
ment. We  generally  have  done  it  statically,  and  it  is  then  necessary  to 
eliminate  friction  by  vibration,  &c.  In  actual  practice,  friction  is  eliminated, 
because  of  the  continual  vibration  of  the  parts. 


Fig.  148. 


226  APPLIED    MECHANICS. 

accuracy.  It  is  exceedingly  interesting  to  watch  the  bead  when 
the  loads  are  suddenly  altered. 

It  is  a  great  pity  that  there  should  not  be  at  least  one  such 
dynamometer  coupling  on  every  length  of  shafting  in  factories. 
We  have  it  from  a  man  who  has  made  careful  measurements 
that  the  loss  of  power  in  ordinary  shafting  is  very  great  indeed. 
If  the  fact  were  continually  before  our  eyes  great  improve- 
ments would  certainly  be  effected.  This  is  a  function  of 
measuring  instruments  (keeping  defects  prominently  before  us) 
which  is  very  important.  Mechanical  engineers  are  largely  in 
the  habit  of  treating  indicated  horse-power  of  an  engine  as  if  it 
were  the  actual  power  given  out  by  the  engine.  Steam-engine 
construction  has  improved  enormously  of  late,  mainly  because 
electricians  have  been  able  to  measure  electrical  power  with 
great  accuracy,  and  so  the  mechanical  losses  of  power  were 
brought  prominently  into  notice. 

182.  Belts. — If  the  pulley  A,  Fig.  149,  is  driven  from  B  by 
means  of  a  belt,  you  must  remember  that  there  is  a  pull  in  the 

part  of  the  belt  M,  as  well  as 
in  the  part  N.  These  two 
pulls  are  generally  pretty 
great,  as  you  know,  but  if 
you  could  measure  them  ac- 
curately you  would  find  that 
there  is  more  pull  in  N,  else 
Fig.  149.  A  would  not  turn.  It  is  the 

difference  of  these  pulls  which 

concerns  us.  You  may,  perhaps,  understand  this  better  from 
Fig.  150.  The  pull  in  A  M  is  the  weight  of  M,  say,  20  Ibs.  The 
pull  in  A  N  is  the  weight  of  N,  say,  50  Ibs.  If  N  falls  two  feet, 
M  rises  two  feet,  and  the  work  done  upon  the  pulley  and  which 
it  transmits  through  the  shaft  somewhere  else  is  50  x  2,  or  100 
foot-pounds,  minus  20  x  2,  or  40,  the  difference  being  60  foot- 
pounds. In  fact,  it  is  the  difference  of  pull  in  the  two  cords, 
30  Ibs.,  multiplied  by  the  space  passed  over  by  the  cord,  2  feet; 
result,  60  foot-pounds. 

The  horse-power  given  by  a  belt  to  a  pulley  is,  then,  the 
difference  of  pull  in  the  belt  on  the  two  sides  of  the  pulley, 
multiplied  by  the  speed  of  the  belt  in  feet  per  minute,  divided 
by  33,000. 

This  is  only  a  particular  case  of  the  general  rule.  If  M  is 
the  sum  of  the  moments  of  a  number  of  forces  tending  to  cause 


APPLIED    MECHANICS. 


227 


rotation,  about  the  axis,  in  pound-feet,  and  if  a  is  the  angular 
velocity  in  radians  per  minute,  then  Ma  is  the  work  in  foot- 
pounds per  minute,  so  that  Ma  4-  33,000  =  H,  the  horse- 
power ....  (1). 

Example. — In  our  dynamometer  coupling,  if  there  are  four 
springs,  each  exerting  a  force  of  160  Ibs.  ;  the  distance  of  the 
axis  of  each  spring  from  the  axis  of  the  shaft  being  07  foot, 
the  turning  moment  is  4  x  160  x  07,  or  448  pound-feet.  If 
the  shaft  makes  150  revolutions  per  minute,  this  means 
150  x  27T,  or  942  radians  per  minute;  and  hence,  448  x  942 
-r  33,000  =  12-8  horse-power. 

Example. — An  ordinary  flange  coupling  has  six  bolts,  at 
07  foot  from  the  axis;  what  force  is  resisted 
.by  each  bolt  (it  tends  to  break  by  shearing, 
Art.  281),  when  60  horse-power  is  being  trans- 
mitted at  120  revolutions  per  minute?  Answer. 
— If  F  is  the  force,  6  F  x  07  is  the  moment 
in  pound-feet;  this,  multiplied  by  120  x  2ir 
4-  33,000,  is  equal  to  60;  and  hence,  F  =  33,000 
x  60  -^  (6  x  07  x  120  x  2*-),  or  F=625  Ibs. 

When  we  know  the  maximum  horse-power 
at  the  minimum  speed  (observe  that  torque  or 
turning  moment  depends  upon  power  -f  speed), 
we  can  calculate  the  maximum  F  for  each  bolt, 
and  our  knowledge  of  the  strength  of  materials 
(Art.  284)  enables  us  to  say  if  the  bolt  is 
strong  enough. 

The  general  rule  appears  in  many  special  M 
forms,  of  which  we  now  have  one  example  in 
belting ;  and  just  as  it  enables  us  to  calculate 
the  strength  of  bolts  in  a  flange  coupling,  or 
the  size  of  the  shaft  itself  (Art.  296),  or  -tells 
us  how  to  order  the  springs  of  a  dynamometer 
coupling,  so  it  here  tells  us  how  to  find  the  proper  size  of 
a  belt.  We  have  here  (N— M)  v  -4-  33,000  =  H,  the  horse- 
power. The  greatest  pull  in  the  belt  is  N  Ibs.,  and  it  is  this 
which  determines  how  strong  the  belt  must  be.  Hence,  we 
must  answer  the  question  : — 

If  it  is  the  difference  of  pull  that  produces  turning,  why  is 
there  so  great  a  pull  even  in  M,  Fig.  149,  as  we  usually  find? 
liefer  again  to  Fig.  150.  If  we  want  the  difference  between  M 
and  N  to  be  30  Ibs.,  why  not  make  M  have  no  weight  at  all,  and 


Fig.  150. 


228 


APPLIED    MECHANICS. 


N  may  then  be  only  30  Ibs.  1  Evidently  we  should  not  be  able 
to  get  friction  enough,  and  the  weight  N  would  fall,  causing 
the  cord  to  slide  on  the  pulley ;  in  fact,  the  friction  between 
the  cord  and  pulley  must  be  more  than  30  Ibs.,  else  there  will 
be  slipping ;  and  to  produce  this  friction  it  is  necessary  to  have 
a  weight  at  M  as  well  as  at  N.  If  we  allowed  the  cord  to  lap 

round  more  of  the 
pulley,  the  necessary 
friction  might  be  pro- 
duced with  a  less 
weight  at  M.  To  get 
an  idea  of  the  friction 
between  a  cord  and  a 
pulley,  arrange  a  pulley, 
or  other  round  object, 
p,  as  in  Fig.  151.  Fix 
it  firmly.  Place  a 
weight  at  M,  say  1  Ib. 
Now  place  weights  in 
the  scale-pan  at  N,  until 
the  cord  just  slips 
slowly.  Say  we  find 
3  Ibs.  to  be  necessary. 
The  difference  between 
N  and  M,  or  2  Ibs.,  is  the 
friction.  Now  put  twice 
Pig.  i5i.  the  former  weight  at  M  ; 

you  will  find  that  about 

twice  the  former  N  will  j  ust  cause  slipping,  so  that  the  friction 
is  doubled.  In  fact,  we  have  our  old  law,  "  friction  is  propor- 
tioned to  load."  But  now  let  us  see  how  friction  depends  on 
the  amount  of  lapping  of  the  cord.  In  your  first  experiment 
measure  the  cord  actually  in  contact  with  the  post  p.  Suppose 
it  to  be  4  inches :  now,  keeping  M,  1  Ib.,  let  the  cord  lap  round 
more  of  the  post  P,  say  8  inches  this  time,  and.  find  the  weight, 
N,  which  will  just  produce  a  slow  sliding.  You  will  find  it 
to  be  9  Ibs.  If  the  cord  touches  on  12  inches  of  the  post 
p,  you  will  find  that  27  Ibs.  at  N  will  be  necessary  to  slowly 
overcome  the  friction.  It  is  only  by  actually  trying  this 
experiment  for  yourself  that  you  will  get  a  clear  idea  of  how 
rapidly  the  friction  increases  with  the  amount  of  lapping.  It 
is  on  this  account  that  one  man  can  check  the  motion  of  the 


APPLIED    MECHANICS. 


229 


largest  vessel   by  simply  coiling   a  rope  a   few   times   round 
a  post. 

The  apparatus,  Fig.  151,  is  so  arranged  that  any  required 
amount  of  lapping  may  be  given  to  the  cord  round  the  fixed 
post  p.  In  an  actual  experiment,  the  fixed  weight  M  was  50 
grammes.  By  means  of  the  pulleys  the  amount  of  lapping 
round  P  was  varied,  and  weights  were  placed  in  N,  in  each  case 
just  sufficient  to  overcome  the  friction  and  raise  M  slowly,  as 
above  described.  The  following  are  the  results  of  the  whole 
series  of  experiments  : — 


Number  of  times  the  cord 
laps  round. 

The  weight  required  at  N  to 
overcome  friction  and  the 
weight  of  M. 

Logarithms  of  the  ratio  of  N 

tOM. 

, 

80 

0-2041 

1 

105 

0-3222 

1 

150 

0-4771 

n 

200 

0-6021 

H 

255 

0-7076 

i| 

330 

0-8195 

2 

400 

0-9031 

tt 

500 

1-0000 

2i 

700 

1-1461 

ii 

1,000 

1-3010 

3 

1,150 

1-3617 

* 

1,500 

1-4771 

Plotting  the  first  and  third  columns  on  squared  paper,  we 
find  that  a  straight  line  passes  nearly  through  all  the  points. 
From  this  line  we  deduce  the  equation — 

n  =  2-2  log-,      .'„, :  ; 

where  n  is  the  number  of  times  the  cord  laps  round.  From 
this  it  is  easy  to  show  that  the  coefficient  of  friction,  /*,  between 
the  cord  and  the  pulley  is  -166. 

You  must  then  remember  that  the  tension  in  M,  Fig.  149,  is 
necessary  to  produce  as  much  friction  as  will  prevent  slipping. 
If  ever  the  excess  pull  in  N  is  greater  than  the  friction,  there 
will  be  slipping.  If  the  belt  slips,  there  is  energy  wasted, 
which  you  can  calculate  if  you  know  the  force  of  friction,  and 
multiply  by  the  distance  through  which  slipping  occurs. 


230 


APPLIED    MECHANICS. 


183.  The  law  is  this.  If  /i  is  the  coefficient  of  friction  between 
the  cord  or  belt  and  the  pulley ;  if  Us  the  length  of  the  cord  or  belt 
which  touches  the  pulley,  say  in  inches ;  and  r  the  radius  of  the 
pulley  in  inches,  then 


N  and  M  being  the  pulls  in  the  belt  or  cord  on  the  two  sides  of  the 
pulley,  l/r  is  the  angle  of  lapping  stated  in  radians. 

This  rule  is  arrived  at  mathematically  in  the  following  way. 

Let  A  P  Q  B  be  the  part 
of  a  pulley  touched 
by  the  belt  MA  P  Q  B  N. 
Imagine  the  pulley 
fixed,  and  slipping 
occurring  because  the 
tension  at  N  is  greater 
than  at  M.  Let  the 
angle  A  o  P  be  called  0, 
Let  the  tension  in  the 
belt  at  P  be  T.  We 
study  what  occurs  at 
the  place  P  Q,  and  we 
greatly  magnify  p  Q 
in  Fig.  153.  What 
are  the  forces  acting 

upon  the  short  piece  of  belt  P  Q  ?  We  have  a  pull  T  +  5  T  at  o., 
and  a  pull  T  at  P,  and  forces  acting  radially  from  the  pulley, 
their  resultant  being  x;  we  have  also  friction  whose  amount 
is  AI  x  if  /u  is  the  coefficient  of  friction,  and  this  friction  is  what  we 


152- 


T+ST 


Fig.  153. 


Fig.  154. 


overcome  by  the  excess  tension  5  T.  To  find  x,  assume  5  T  =  o, 
and  no  friction  (it  will  be  found  that  these  terms  become  less  and 
less  important  as  the  distance  P  Q  is  made  less  and  less).  Let  the 
angle  A  o  Q  be  called  0  +  50,  so  that  p  o  Q  =  50.  The  three  forces 
T,  —  Xj  T  being  in  equilibrium,  let  them  be  represented  in  direction 
and  sense  by  the  sides  of  the  triangle  shown  in  Fig.  154,  and  it  is 
evident  that  x  =  T  .  50.  Hence  the  force  of  friction  is  ;i  .  T  .  50, 
which  is  overcome  by  ST.  When  slipping  is  just  occurring, 
M.T.50  5T....(1); 

or  rather,  because  the  theory  is  true  only  when  50  is  thought  to  be 
smaller  and  smaller  without  limit, 

—  =/KT.  .  .  .  (2). 
This  is  an  example  of  the  compound  interest  law.     "  The  rate  of 


APPLIED    MfcCilAtftCS.  231 

increase  of  T  as  0  increases  is  proportional  to  T  itself."  In  any 
elementary  book  on  the  calculus  it  is  shown  that  (2)  is  the  same  as 

log.  T  =  fid  +  constant. 

Putting  T  =  M  when  6  =  o,  and  T  =  N  when  9  =  A  o  B,  we 
find 

log.  N  —  log.  M  =  fjt  .  A  O  B  .  .  .  .  (3). 

This  is  the  rule  given  above.  Assuming  that  there  is  a  constant  /*, 
and  that  we  know  its  value,  this  rule  enables  us  to  design  belts  and 
ropes  to  be  strong  enough  to  transmit  a  given  amount  of  horse- 
power H  at  the  belt  velocity  v  feet  per  minute.  For  we  have 
already  used  (N  —  M)  v  ~  33,000  =  H  .  .  .  .  (4).  If  the  angle 
A  o  B  be  called  0  for  shortness,  it  is  easv  to  deduce  from  (3)  and  (4) 

,_?».««-  ^/(^_"  !)....  (4). 

This  is  the  tension  in  the  belt  where  it  is  greatest,  and  it  is  from 
this  value  that  we  calculate  the  strength  of  the  belt.  Thus  if  b 
and  t  are  the  breadth  and  thickness  of  the  belt  in  inches,  and  if  / 
is  the  safe  load  per  square  inch  of  section  for  the  material,  we 
make  btf  equal  to  the  calculated  N.  We  take  /about  330  Ibs.  per 
square  inch  in  leather  belting,  and  this  leads  to  the  easily 
remembered  rule,  giving  bt  in  square  inches, 


6*  =  e^Kerf  -  1)  ----  (5). 


184.  In  a  great  number  of  cases  the  least  angle  of  lapping 
when  one  shaft  drives  another  by  belting  is  -f  ths  of  the  circum- 
ference, and  fj.  may  be  taken  as  0*3  ;  and  so  we  are  led  to  the 
common,  easily  remembered  shop  rule,  b  t  =  200  H  -~  v  .  .  (6). 
It  is  to  be  remembered  that  b  is  the  breadth,  t  the  thickness  of 
the  belt  in  inches  ;  H  is  the  horse-power  transmitted,  v  the 
velocity  of  the  belt  in  feet  per  minute;  p.  the  coefficient  of 
friction,  and  0  the  angle  of  lapping  in  radians. 

Leather  belting  is  seldom  more  than  4  feet  wide,  even  in 
America,  where  the  widest  belting  is  employed.  A  single 
leather  belt  is  about  a  quarter  of  an  inch  thick,  and  the 
material  will  stand  a  pull  of  from  700  to  1,200  Ibs.  per  inch  of 
its  width,  before  breaking.  A  single  belt  at  an  ordinary  laced 
joint  may  be  taken  to  stand  a  greatest  working  pull  of  about 
one-third  of  the  strength  of  the  leather.  An  ordinary  laced 
joint,  with  splice,  has  about  twice  the  strength  of  an  ordinary 
grip-fastened  joint,  and  not  quite  twice  that  of  a  butt  laced 
joint.  Many  engineers  take  80  to  90  Ibs.  per  inch  of  width  of 
a  single  belt  as  the  usual  pull  on  the  tight  side. 

Strips  from  the  best  parts  of  the  tanned  hide  are  cemented 
at  long  splices  to  form  belts  ;  vulcanised  indiarubber  alone,  or 
with  intermingled  canvas  cotton,  and  various  other  kinds  of 


APPLIED    MECHANICS. 

waterproof  belting  are  used.  There  is  always  slipping  or 
creeping,  so  that  it  is  not  quite  accurate  to  take  the  velocity 
of  the  belt  as  being  equal  to  that  of  the  pulley.  (See  Art.  70.) 

Belts  are  in  use  which  are  made  of  a  great  number  of  little 
links  of  thick  leather  on  wire  pins,  forming  a  chain.  Hemp  and 
cotton  ropes  are  now  often  used,  instead  of  belting.  Many  V 
grooves  in  the  rim  of  a  drum  receive  each  its  own  rope,  which 
is  usually  lying  wedged  in  the  groove,  with  not  much  tension 
when  there  is  no  motion,  except  what  is  due  to  its  own  weight, 
the  span  being  20  to  60  feet.  As  in  belts  conveying  a  con- 
siderable amount  of  power,  the  tight,  or  N  side,  is  underneath, 
as  this  gives  more  lapping.  The  strengths  of  ropes  and 
chains  are  given  in  Art.  266,  but  in  rope-driving  our  ability 
necessitates  some  such  rule  as  (see  Art.  183) 

N  -  M  =  8  f, 

when  g  is  the  girth  of  the  rope  in  inches,  and  N  and  M  are  in 
pounds.  The  wedging  of  the  ropes  in  grooves  causes  the 
coefficient  of  friction  to  be  virtually  of  its  usual  value  divided 
by  the  sine  of  half  the  angle  of  the  groove,  and  as  this  is 
usually  45°,  we  take  /z  to  be  about  0-6,  and  sometimes  more. 
Splices  are  about  10  feet  long.  The  speed  is  usually  about 
4,000  feet  per  minute.  The  bending  and  unbending  of  the 
rope  is  what  seems  to  shorten  its  life,  which  is  usually  one  of  a 
few  years.  There  is  probably  considerably  more  waste  of 
energy  by  friction  than  when  toothed  gearing  is  used. 

With  wire  ropes  (strands  of  wire  round  a  hempen  core)  the 
wedging  in  grooves  is  much  more  hurtful,  and  hence  reliance 
is  placed  upon  greater  velocities,  even  to  6,000  feet  per  minute. 
Under  usual  conditions  the  life  of  a  rope  is  one  of  less  than  a 
year.  In  calculations  of  the  tensions  due  to  the  weights  of 
ropes,  we  may  take  it  that  the  hanging  curve  is  practically 
parabolic  (see  Art.  133).  The  bottoms  of  the  grooves  on  the 
pulleys  are  usually  of  leather,  wood,  or  gutta-percha,  and  /K 
may  then  be  taken  as  0*25.  Each  relay  being  about  1,000 
yards,  we  may  take  its  efficiency  as  about  95  per  cent. 

Exercise. — If  there  are  nine  relays,  what  is  the  total 
efficiency  1  Answer. — 0*95  raised  to  the  9th  power,  or  0*63. 

At  Oberursel  a  rope  is  used  whose  diameter  is  0-6  inch,  o! 
36  wires  each  0'06  in  diameter;  pulleys,  12  feet  in  diameter, 
placed  400  feet  apart;  94  horse-power  is  transmitted  3,000 
feet,  the  rope  travelling  at  4,400  feet  per  minute. 

185.  Chain  Gearing. — The  use  of  tricycles  and  bicycles  has 


APPLIED    MECHANICS.  '        233 

caused  a  very  great  development  in  this  non-slipping,  certain, 
and  efficient  method  of  transmitting  power.  It  is  likely  to 
have  a  very  large  further  development  for  large  powers  in 
factories  on  account  of  its  success  for  small  power,  not  merely 
in  cycles,  but  in  electric- motor,  tramcar  and  locomotive  work. 
No  tension  is  needed  on  the  slack  side. 
EXEKCISES. 

1.  A  pulley  of  3  feet  diameter  receives  10  horse-power  at  150  revolu- 
tions per  minute ;  find  the  difference  of  pull  between  N  and  M.     If  the 
lapping  is  0'4  of  a  circumference,  and  p  =  0'25,  find  N/M.     Now  find  N. 

Am.,  233-4  Ibs. ;  N=501'3  Ibs. 

2.  Use  the  above  workshop  rule  to  find  the  breadth  of  a  single  belt 
(t  =  0-4  inch,  say)  to  transmit  20  horse-power  at  a  velocity  of  1,300  feet 
per  minute.  Am.,  7;7  inches. 

3.  A  rope  is  wrapped  three  times  round  a  post,  and  a  weight  of  12  Ibs. 
is  hung  from  one  end.     Find  the  least  pull  applied  at  the  other  end  neces- 
sary to  raise  this  weight  if  /*  =  '3.     What  weight  would  be  required  to 
just  prevent  the  12  Ibs.  from  slipping  down?     Ans.,  3,440  Ibs. ;  0-042  Ib. 

4.  The  pulley  on  an  engine  shaft  is  5  feet  in  diameter,  and  it  makes 
100  revolutions  per  minute.     The  motion  is  transmitted  from  this  pulley 
to  the  main  shaft  by  a  belt  running  on  a  pulley,  and  the  difference  in 
tension  between  the  tight  and  slack  sides  of  the  belt  is  115  Ibs.     What  is 
the  work  done  per  minute  in  overcoming  the  resistance  to  motion  of  Mie 
main  shaft  ?     What  is  the  horse-power  transmitted  ? 

Ans.,  180,714  ft.  Ibs.  ;  5-47. 

5.  50  horse-power  is  being  transmitted  by  a  belt  moving  at  a  speed  of 
70  feet  per  second.    What  width  of  belt  will  be  required  if  its  thickness  is 
0-6  inch,  assuming  the  maximum  working  stress  to  be  330  Ibs.  per  square 
inch,  and  the  tension  on  the  tight  side  being  double  that  on  the  slack  side  ? 

Ans.,  4  inches. 

6.  A  pulley  3  feet  6  inches  in  diameter,  and  making  150  revolutions  a 
minute,  drives,  by  means  of  a  belt,  a  machine  which  absorbs  7  horse- 
power.    What  must  be  the  width  of  the  belt  so  that  its  greatest  tension 
shall  be  70  Ibs.  per  inch  of  width,  it  being  assumed  that  the  tension  in  the 
driving  side  is  twice  that  on  the  slack  side  ?  Ans.,  4  inches. 

7.  A  rope  pulley,  carrying  20  ropes,  is  16  feet  in  diameter,  and  trans- 
mits 600   horse-power    when    running  at    90    revolutions    per  minute. 
Taking  p  =  0'7  and  the  angle  of  contact  180°,  find  the  tensions  on  the 
tight  and  slack  sides  of  the  rope.  Ans.,  246  Ibs. ;  27'2  Ibs. 

186.  Transmission  and  Absorption  Dynamometers.— I 
have  already  described  to  you  an  instrument  which  allows  us 
to  measure  the  horse-power  transmitted  by  a  shaft.  I  am  in 
the  habit  of  employing  a  somewhat  similar  arrangement  for 
measuring  the  power  transmitted  by  a  belt  to  any  machine.  It 
is  shown  in  Fig.  155,  and  is  easily  understood  from  the  descrip- 
tion of  Fig.  148.  I  can  take  it  near  any  machine,  and  drive  the 
machine  through  it,  using  two  belts  instead  of  one.  G  is  a 
loose  pulley.  A  belt  drives  H,  which  drives  the  plate  £  through 

f 


234 


APPLIED    MECHANICS. 


four  spiral  springs  B.  The  plate  E  is  keyed  to  a  shaft  carried 
on  the  frames  c  and  D,  and  the  pulley  F  is  keyed  on  the  shaft. 
A  belt  from  F,  therefore,  will  drive  any  machine.  When  much 
torque  is  acting,  the  springs  B  become  extended,  causing  a 


relative  motion  of  E  and  H,  and  this  motion  is  shown  by  the 
bright  bead  A,  at  the  end  of  the  lever  I  A,  approaching  the 
axis  of  rotation.  A  fixed  scale  attached  to  the  frame  C 
allows  the  motion  of  A  to  be  measured. 

In  Fig.  159,  showing  the  Fronde  or  Thorneycroft  Dynamo- 
meter, the  pulley  D  drives  the  pulley  F  by  a  belt  which  passes 


APPLIED    MECHANICS. 


235 


also  round  the  two  pulleys  A  and  B.  These  pulleys  are  on  a 
frame  A  B  L,  which  is  pivoted  at  E.  The  tension  N,  on  the  tight 
side,  is  greater  than  M,  and  a  measurable  force  p  must  be  ex- 
erted to  keep  the  frame  in  the  position 
shown.  Evidently  2  N.  a  =  2  M.  a  +  P.6 

P.b 

so  that  N  — M=-^ — 

This  transmission  dynamometer  is 
specially  useful  when  small  powers  are 
to.  be  measured  in  the  laboratory.  In 
the  Hefner-Alteneck  form,  the  long 


Fig._156. 


and  nearly  horizontal  belt  driving  a  pulley  passes  round  two 
guide  pulleys  D  and  c  on  a  suspended  and  balanced  frame. 
When  the  angle  made  by  the  slack  parts  MM  is  the  same 
as  that  made  by  the  tight  parts  N  N,  a  measurable  force  F  is 
required  to  maintain  the  frame  in  its  symmetrical  position. 

Exercise. — If  the  angle  between  M  and  M  is  180  —  0,  and 
there  is  the  same  angle  between  N  and  N,  and  the  pull  F  is 
symmetrical,  show  that  F  =  (N  -  M)  0  if  6  is  small.  In  truth, 
F  =  (N  —  M)  2  sin  J  0. 

187.  Absorption  Dynamometers  are  used  to  measure  the 
power  given  out  by  steam-, 
or  gas-,  or  oil-engines,  or 
electric  or  other  motors. 
They  measure  the  power, 
consuming  it  as  they  do  so. 
One  of  Professor  James 
Thomson's  forms,  as  ar- 
ranged for  measuring  the 
power  given  out  by  a  small 
electro  -  motor,  is  shown 
in  Fig.  157.  The  motor 
drives  the  grooved  pulley 
A,  and  the  pulley  B  turns 
along  with  A.  A  cord  hangs 
lapping  round  part  of  B, 
and  carries  at  its  one  end 
a  scale-pan  M,  containing  a 

weight.     The  other  end  N'  Pig.  157. 

is   pulled   by  means  of   a 

piece  of  metal  fastened  to  the  rim  of  a  loose  pulley  c,  which 
has  a  weight  N  always  acting  upon  it,  tending  to  turn  it  round. 


236 


APPLIED    MECHANICS. 


Evidently  the  cord  is  pulled  with  a  weight  M  at  one  end,  and 
a  weight  N  at  the  other.  If  now  there  is  slipping  between  the 
cord  and  B,  the  friction  is  measured  by  the  difference  of  the 
weights  N  and  M.  If  M  is  100  Ibs.  and  N  is  400  Ibs.  the 
friction  is  300  Ibs.  If  the  pulley  has  a  circumference  of 
2  feet,  and  makes  80  turns  per  minute,  the  amount  of  slipping 
is  80x2,  or  160  feet  per  minute,  and  the  work  done  against 
friction  is  160  x  300,  or  48,000  foot-pounds  per  minute — 
that  is,  1  '45  horse-power.  In  this  case  all  the  power  is  wasted 
in  friction,  and  this  is  called  an  Absorption  Dynamometer 
because  it  measures  the  power  but  absorbs  it  in  doing  so; 
whereas  the  coupling  of  Fig.  148  and  the  dynamometer  of 
Fig.  155  are  called  Transmission  Dynamometers,  because  they 
measure  the  power  transmitted  through  them  whilst  working 
any  machines. 

Should  we  try  to  use  merely  a  cord  or  belt  with  two 
different  weights  N  and  M  at  its  ends  (as  in  Fig.  158)  as  an 
absorption  dynamometer,  the  cord  or  belt  slipping  on  the 
pulley,  we  should  find  that,  after  adjusting  the  weights, 
when  we  leave  them  hanging  on,  N  will  gradually  overcome 
M  till  it  touches  the  floor,  and  after  that  we  are  ignorant 
of  N  —  M.  In  fact,  the  co-efficient  of  friction  gradually  alters. 

In  Fig.  157  this  is  automatically 
counteracted  by  a  change  in  the 
amount  of  lapping.  Sometimes, 
instead  of  the  lighter  weight  M,  we 
have  used  a  spring  balance  and  a 
dash-pot  to  still  the  jerky  motion 
which  occurs,  but  in  this  case  we 
must  take  readings  of  the  changing 
values  of  M ;  and  again,  it  is  usually 
very  important  during  a  test  to  keep 
N-M  constant.  We  have  met  with 
quite  wonderful  success  by  adopting 
the  following  very  simple  expedient : 
— A  B  is  a  grooved  pulley,  and  a  cord 
with  scale-pans  hangs  round  it  as  in 
Fig.  158.  At  P  there  is  a  knot  in 
the  cord,  or  an  excrescence  of  any 
kind  which,  when  it  is  drawn  up 
by  the  gradual  falling  of  N,  will  just  slightly  jam  itself 
in  the  groove.  When  measuring  the  power  from  a  three- 


Fig.  158. 


APPLIED    MECHANICS. 


237 


quarter  horse  electric  motor  we  bave  kept  the  same  weights  M 

and  N  in  the  scale-pans  for  an  hour  at  a  time,  a  slow  gentle  rise 

and  fall  taking  place  now  and  again.      Before  we  discovered 

the  value  of  the  knot  we  went  to  great 

trouble  in  having  a  thick  rough  cord  from 

p  to  Q  spliced  to  a  thinner  one,  so  that 

when   N  fell    a  greater  lapping  of  the 

rougher  cord  made  up  for  a  lessening  of 

the  friction  on  the  smoother  one,  but  as  a 

matter  of  fact  the  knot  is  excellent.    We 

have  used  the  same  expedient  with  a  can- 

vas belt  on  a  large  drum  when  measuring 

26  horse-power  ;  here  a  leather 

lace  passed  carelessly  a  number 

of  times  through  the  belt  in- 

troduced sufficient  resistance 

at  P  to  prevent  N  falling. 

In  all  these  cases  soapy 
water  is  kept  trickling  over 
the  rubbing  surfaces  to  carry 
off  the  heat,  for  all  the 
mechanical  energy  which  we 
measure  is  converted  into 
heat.  The  water  ought  to 
be  soapy,  because  when  pure 
water  acts  as  a  lubricant 
there  is  considerable,  and 
seemingly  spasmodic,  varia- 
tion in  the  friction.  A  brake 
on  the  principle  of  N,  a 
weight,  and  M,  the  pull  of  a 
spring  balance,  is  sometimes 
used  even  for  as  much  as 
50  horse-power,  N  being  at- 
tached to  two  ropes  going 
about  three-quarters  of  the 
way  round  an  ordinary 
pulley,  and  kept  apart  on  the 
smooth  rim  by  distance  pieces 
of  wood  which  fit  the  edges 


of  the  rim    one 


rope   com- 


pleting Ihe  round  so  as  to 


238 


APPLIED    MECHANICS. 


come  between  the  first  two  on  leaving,  M  is  then  an  upward 
pull,  and  N  downward. 

Fig.  160  shows  a  brake  which  maybe  used  for  large  powers, 
and  we  have  used  it  with  satisfactory  results.  It  consists  of 
blocks  of  wood  on  hoop  iron,  forming  a  brake  strap,  the  two 
ends  of  which  D  and  E  are  fastened  to  the  lever  D  H,  a  small 
angular  change  in  the  position  of  which  produces  great  increase 
in  tightness.  A  load  w  Ib.  is  hung  on  at  c  and  a  spring  balance 
acts  at  H.  The  pull  in  the  spring  balance,  w  Ib.,  is  usually 
small.  It  is  evident  that  w.  R— wr  is.  a  moment  in  pound 
feet  which,  multiplied  by  the  angular  velocity  in  radians 
per  minute,  and  divided  by  33,000,  gives  the  horse-power. 
Lastly,  Froude's  liquid  dynamometer  may  be  employed  up  to 

very  high  powers ; 
the  heat  is  devel- 
oped in  a  consid- 
erable mass  of 
water  which 
keeps  cool.  For 
the  latest  and 
best  form  of  this, 
students  are  re- 
ferred to  a  paper 
by  Prof.  Reynolds 
in  the  Transac- 
tions of  the  Royal 
Soc.,  May,  1897. 
If  any  student  has  trouble  in  seeing  why  we  assume  a 
virtual  surface  of  rubbing  coinciding  with  the  centre  of  the  rope 
in  Fig.  158,  let  him  return  to  the  consideration  of  Fig.  150 
Here  it  is  obviously  the  velocity  of  the  centre  of  the  rope  that 
is  taken,  and  we  calculate  this  from  angular  velocity  and  virtual 
diameter  of  the  pulley.  Now  suppose  the  pulley  of  Fig.  150 
to  be  at  rest  and  the  rope  slipping ;  the 
same  amount  of  power  is  now  wasted 
that  used  to  be  given  to  the  shaft ;  there 
will  be  exactly  the  same  forces  of  fric- 
tion and  waste  through  friction  if  the 
pulley  moves  and  the  weights  do  not. 

Generally,    let   A  B  be   a   revolving 
wheel   with  centre  p,  and  let  forces  F, 
w  and  w  keep  the  brake  block  p  Q  R  from  moving  with  the 


Fig.  160. 


Fig.  161. 


APPLIED    MECHANICS.  235 

wheel.  Let  there  be  no  other  forces  acting  on  the  brake 
block  which  exercises  moment  about  o. 

How  much  power  is  being  wasted  at  the  rubbing  surfaces  1 
It  is  the  sum  of  all  the  forces  of  friction  multiplied  by  their 
respective  rubbing  distances  (we  need  not  here  speak  of  pounds, 
feet,  etc.,  or  the  unit  in  which  power  is  measured).  But- this 
is  the  same  as  the  sum  of  all  the  moments  about  o  of  all  the 
forces  of  friction  multiplied  by  the  angular  velocity,  and  the 
sum  of  the  moments  of  the  forces  of  friction  is  measured  by 
w.  o  P  +  F.  o  R  -  w.  OQ.,  because  the  block  is  in  equilibrium 
under  the  action  of  the  forces  of  friction  and  w,  F  and  w  (see 
Art.  98). 

188.  Example. — In  an  Ayrton-Perry  Coupling,  the  bright 
bead  does  not  begin  to  move  inward  till  a  certain  horse- 
power is  reached.  If  a  radial  motion  of  10  inches  corre- 
sponds to  a  lengthening  O'l  inch  of  each  spring ;  if 
there  are  eight  springs,  whose  axes  are  10  inches  from 
the  centre,  and  the  shaft  runs  at  200  revolutions  per 
minute ;  if  the  maximum  horse-power  is  250,  and  if  it  is  to 
alter  from  200  to  250  for  an  18  inch  motion  of  the  bead,  how 
Dught  the  springs  to  be  ordered?  If  F  is  the  pull  in  each 
spring  at  H  horse-power,  8xFxlO-rl2in  pound  feet,  multi- 
plied by  200  x  2  TT  -4-  33,000  gives  H,  or  F  =  3-94  H.  Hence  the 
range  of  pull  in  each  spring  is  from  3'94  x  250,  or  985  Ibs.,  to 
3-94  x  200,  or  788  Ibs.,  and  for  this  the  bead  has  a  range  of  18 

inches;  and  therefore  each  spring  will  yield  by  18  x  y^,  or 

O'l 8  inch.  Thus  the  spring  yields  0-18  inch  for  a  difference 
of  pull  of  985-788,  or  197  Ibs.,  and  will  therefore  yield 
985x0-184-197,  or  0-9  inch  for  the  maximum  pull.  We 
therefore  order  eight  springs,  each  of  which  has  a  greatest  work- 
ing pull  of  985  Ibs.,  and  is  then  to  have  an  axial  lengthening 
0'9  inch.  It  will  also  be  necessary  to  tell  the  spring-maker 
what  the  exact  length  of  the  spring  is  to  be  between  the  end 
pins,  and  possibly  its  outside  diameter.  It  is  usual  to  arrange 
adjustable  stops  to  keep  the  springs  stretched,  so  that  they 
shall  not  elongate  for  any  force  less  than  788  Ibs.  When  the 
springs  are  long  we  use  diametral  wires  to  prevent  any  action 
due  to  centrifugal  force,  and  in  some  cases  of  very  fluctuating 
loads  we  have  used  dash-pots. 

Example. — The  middle  of  a  key  is  2  inches  from  the 
centre  of  a  shaft ;  the  key  is  the  only  fastening  of  a  wheel 


240  APPLIED   MECHANICS. 

which  makes  100  revolutions  per  minute,  and  gives  out  10 
horse-power.  What  is  the  force  transmitted  by  the  key? 
Answer. — The  force  ispifrx  2  -f-  12  (the  moment  in  pound 
feet),  multiplied  by  100  x  2  IT  -7-  33,000  =  10.  That  is, 
F  —  3,016  Ibs. 

Example. — A  6-feet  drum  at  150  revolutions  per  minute 
drives  through  a  Hefner  Alteneck  Dynamometer.  A  spring 
balance  registers  p  as  o  when  no  power  is  being  transmitted, 
and  registers  it  as  60  Ibs.  when  we  know  that  25  horse-power  is 
being  transmitted.  Find  6  if  (N  —  M)  6  =  F.  We  know  that  the 
velocity  of  the  belt  is  6  TT  x  150,  or  2,827  feet  per  minute ;  so 
that  N  -  M  is  25  x  33,000  -=-  2,827,  or  292  Ibs.  Hence,  0  is 
60  -h  292,  or  0'206  radians,  or  11-8  degrees. 

Example. — A  dynamo  machine  going  at  420  revolutions 
per  minute  is  supported  on  knife  edges.  The  driving-belt 
causes  the  machine  to  tilt,  and  the  tilting  tendency  is  prevented 
by  a  weight  w  of  15  Ibs.  suspended  from  an  arm  5  feet  from 
the  axis,  measured  horizontally.  Find  the  horse-power  given  to 
the  machine.  The  machine  gives  out  39 -2  amperes  at  98  volts. 
What  is  its  efficiency  1  Answer. — The  turning  moment  is 
15  x  5  pound  feet,  and  therefore  the  horse-power  is  15  x  5  x 
400  x  2  «•  -f  33,000,  or  571.  The  electrical  horse-power 
given  out  is  39 '2  x  98  -f  746  =  5-15.  The  efficiency  is  there- 
fore 5-15  ^  5-71,  or  0-902,  or  90'2  per  cent. 

Example. — A  brake  block  is  kept  at  rest  by  certain  forces, 
500  Ibs.  acting  at  2  feet  from  the  axis,  30  Ibs.  acting  at  8  feet 
from  the  axis,  50  Ibs.  acting  at  — 5  feet  from  the  axis  (we  mean 
here  that  the  moment  of  this  last  force  is  opposite  to  the 
others).  The  shaft  makes  120  revolutions  per  minute.  Find 
the  horse-power  consumed.  The  moment  is  500  x  2  +  30  x  8 
—  50  x  5,  or  990  pound -feet.  This,  multiplied  by  120  x  2  it 
-i-  33,000  gives  22-6  horse-power. 

Example. — The  following  readings  were  taken  when  testing 
an  electric  motor.     Find  the  electric  power  supplied  and  the 
mechanical  power  given  out  in  each  case.     The  knot  dynamo- , 
meter  was  used ;  each  string  was  found  to  be  almost  exactly 
6-1  inches  horizontally  from  the  centre.     If  n  is  the  number 

fvl 

of  revolutions  per  minute  (N  —  M)  x  —   X  2  IT  x  n  •  -  33,000 

=H,  or  (N  -  M)  n  -f-  10,000  =  H  nearly.  The  student  will 
remember  that  1  ampere  x  1  volt  =  1  watt,  and  746  watts  =  1 
horse-power. 


APPLIED    MECHANICS. 


241 


Electrical 

Brake 

Amperes. 

Volts. 

N 

• 

n 

Horse- 

Horse- 

Efficiency. 

power. 

power. 

8-4 

28-1 

4 

1-6 

1,504 

•318 

•162 

•51 

7'5 

30-4 

4 

2-4 

1,796 

•304 

•129 

•42 

13-1 

19-8 

18 

13 

572 

•349 

-128 

•37 

16-4 

31-0 

19' 

12 

1,008 

•681 

-316 

•46 

15-3 

30-7 

17 

10 

1,048 

•632 

•329 

•52 

EXERCISES    ON   BRAKES. 

1.  In  an  Ayrton  and  Perry  knot-brake  (Fig.  158)  the  centres  of  the 
cords  to  N  and  M  lie  respectively  6  and  6 -2  inches  horizontally  from  o  (it 
is  usual  to  take  these  distances  equal  and  equal  to  the  radius  of  the  circle 
of  the  centre  line  of  the  cord  on  the  pulley,  but  very  careful  measurement 
will  detect  slight  differences  due  to  stiffness  of  cord  and  effect  of  knot). 
The  weights  are  N  =  30  Ibs.,  M  =  5  Ibs.     The  pulley  makes  400  revolu- 
tions per  minute;    find  the  horse-power.      Here  we  had  better  take 
moments,  as  the  distances  are  unequal.   The  moment  is  (30  x  6  -i-  12)— (5 
+  62  -r-  12),  or  12-42  pound-feet.     The  speed  is  400  x  2  TT,  or  2,514  radiane 
per  minute,  and  12-42  x  2,514  -7-  33,000  =  0-946  horse-power. 

2.  In  a  rope-brake  on  a  flywheel  of  8  feet  diameter,  the  ropes  being  1 
inch  thick  (so  that  the  moment  due  to  friction  is  as  if  the  rope  had  no 
thickness,  but  the  wheel  is  8-083  feet  in  diameter),  the  load  is  500  Ibs. 
The  pull  in  the  spring  balance  varies  from  10  to  20  Ibs.  during  the  test. 
The  wheel  makes  105  revolutions  p'er  minute ;  find  the  horse-power.     The 
velocity  of  the  virtual  wheel  is  S-0837T  x  105  feet  per  minute,  and  this, 
multiplied  by  490  or  480,  divided  by  33,000,  gives  a  power  varying  from 
39-6  to  38-8  horse-power. 

3.  The  power  of  an  engine  is  tested  by  passing  a  belt  over  the  fly-wheel, 
which  is  5  feet  in  diameter.     One  end  of  the  belt  is  secured  to  a  spring 
balance,  and  a  weight  of  300  Ibs.  hangs  on  the  other  end.     What  is  the 
brake  horse-power  when  the  balance  registers  180  Ibs.  and  the  fly-wheel 
makes  150  revolutions  per  minute  ?  Am.,  8£. 

4.  In  Exercise  3  the  pull  in  the  spring  balance  gradually  alters  during 
the  test  to  200  Ibs.     This  is  due  to  a  change  in  //,  the  co- efficient  of 
friction.     Find  the  two  values  of  p.  Am.,  0-159  ;  0'129. 

If  the  extreme  tensions  300  Ibs.  and  180  Ibs.  were  maintained  con. 
stant,  as  in  the  James  Thomson  dynamometer  (Fig.  157),  and  the  lapping 
was  180°  to  begin  with,  what  would  it  be  at  the  end  ?  Ans.,  222°. 

5.  The  diameter  of  a  steam  cylinder  is  8  inches,  the  length  of  crank 
9  inches,  the  number  of  revolutions  per  minute  is  150,  and  the  mean 
effective  pressure  of  the  steam  is  35  Ibs.   Find  the  indicated  horse-power. 

The  same  engine  is  tested  by  a  brake  (like  Fig.  160)  on  the  crank 
shaft,  K  being  2£  feet,  w  being  310  Ibs.,  w  being  15  Ibs.,  and  r  being 
4£  feet.  Find  the  brake  horse-power  and  the  working  efficiency  of  the 
engine.  Ans.,  24  ;  20-2  ;  0.84. 

6.  Find   the   horse-power    given  out    by  a  steam-engine   driving  a 
Froude  dynamometer  which  discharges  700  gallons  of  water  per  hour 
with  a  rise  of  temperature  of  18*5  Centigrade  degrees.  Ans.,  92. 


242 


CHAPTER   XL 

KINETIC       ENEKGY. 

189.  WE  sometimes  assume  that  our  readers  know  quite 
well  the  fundamental  principles  of  mechanics,  and  then,  again, 
we  assume  that  they  do  not.     We  hope  that  they  agree  with 
us  that  we  are  right  in  proceeding  in  this  way. 

When  a  weight,  A  Fig.  22,  in  falling  lifts  a  weight  B  by 
the  use  of  a  machine  inside  the  box  c,  let  us  consider  the  store 
of  energy  at  any  instant.  The  store  of  energy  consists  in — 
first,  the  potential  energy  of  A — that  is,  the  weight  A  in  pounds, 
multiplied  by  the  distance  in  feet  through  which  it  is  possible 
to  let  it  fall  to  some  datum  level ;  second,  the  potential  energy 
of  B,  which  is  the  weight  of  B  multiplied  by  the  distance  through 
which  it  is  possible  to  let  B  fall ;  third,  the  energy  of  motion,  or 
kinetic  energy,  of  everything  which  is  moving — namely,  A,  B, 
and  the  parts  of  the  mechanism.  We  are  supposing  that  there 
are  no  other  weights  which  can  fall  or  rise,  and  that  there  are 
no  coiled  springs  or  other  stores  of  energy  in  the  mechanism. 
Now,  if  A  is  just  heavy  enough  to  maintain  a  steady  motion, 
the  kinetic  energy  remains  the  same ;  so  that,  whatever  energy 
is  given  out  by  A  in  falling  is  in  part  being  given  as  potential 
energy  to  B,  and  is  in  part  being  wasted  in  friction.  But 
suppose  A  to  be  heavier  than  this,  then  there  is  more  potential 
energy  being  lost  by  A  than  is  being  stored  by  B  or  wasted  in 
friction,  and  it  must  be  stored  up  in  some  other  form.  The 
surplus  stock  shows  itself  in  a  quicker  motion  of  everything ; 
it  is  being  stored  up  as  kinetic  energy. 

190.  We   have  now  to   consider  an  important   question. 
When  a  certain  amount  of  potential  energy  (measurable  in 
foot-pounds)    disappears,    and    becomes    kinetic   energy,    ho\v 
quickly  must  all  the  parts  of  the  machinery  move  to  store 
it  all  up?     This  problem  is  very  troublesome,  because  every- 
thing in  Fig.  22  is  in  motion  in  a  different  way;  some  parts 
of  the  mechanism  are  moving  slowly,  others  quickly.     It  is, 
however,  easy  to  find  out  how  much  kinetic  energy  a  small 
body  has  if  we  know  its  weight  and  its  speed.     Let  there  bo 
a  small  ball  hung  from  the  point  o,  Fig.  162,  by  a  silk  thread, 
BO  that  when  it  vibrates  we  can  call  it  a  simple  pendulum. 


APPLIED    MECHANICS. 


243 


Now,  you  know  that  when  it  reaches  the  end  of  its  swing  at  A 
it  is,  for  a  very  short  interval  of  time,  motionless,  and  has  no 
kinetic  energy.  It  falls  from  A  to  B  ;  and  as  there  is  almost 
no  friction,  we  may  suppose  that  the 
potential  energy  which  it  loses  in  falling 
through  the  vertical  height  from  A  to  B 
is  all  stored  up  as  kinetic  energy  when 
the  ball  reaches  B.  The  body  has  a 
certain  velocity  in  feet  per  second  when 
it  reaches  B,  and  it  is  on  account  of  its 
having  this  velocity  that  we  say  it  has  a 
store  of  kinetic  energy.  If  the  vertical 
height  from  B  to  A  is  h  feet  and  the 
weight  of  the  body  is  w  lb.,  we  have 
w  h  foot-pounds,  which  it  had  at  A,  con- 
verted into  kinetic  energy  at  B.  Now 
comes  the  question.  We  know  that 
this  kinetic  energy  is  proportional  to  w, 
but  how  does  it  depend  upon  the  velocity? 

It  evidently  does  not  depend  on  the 
direction  of  the  velocity  at  B,  and  some 
mathematicians    would    say    that    this 
shows  that  it  must  depend  upon   an  even 
this  is  metaphysics. 

Now,  experiment  shows  that  when  a  body  falls  freely  at 
London  without  friction,  its  velocity  when  it  has  fallen  freely 
without  friction  at  any  place  through  distances  proportional  to 
0,  1,  4,  9,  16,  25,  36,  &c.,  is  proportional  to  the  numbers  0,  1, 
2,  3,  4,  5,  6,  etc.  Experiment  shows  that  these  numbers  are 
quite  independent  of  whether  we  measure  in  feet  or  centi- 
metres or  yards,  in  minutes  or  seconds.  Hence  we  say  that 
the  velocity  is  proportional  to  the  square. root  of  the  height,  and 
we  assume  that  we  are  stating  our  result  when  we  say  v2  oc  h, 
or  v2  -f-  h  =  a  constant,  which  we  shall  call  b,  or  -y2  =  bh  . . .  (1). 

Notice  that  when  we  write  our  result  (1)  down  we  have 
written  down  something  which  is  beyond  mere  arithmetic ;  it 
is  on  the  very  much  higher  subject — physics,  which  in  its 
mathematical  form  is  called  algebra.  '  v,  b  and  h  are  not  mere 
numbers ;  they  are  quantities,  and  if  v  and  h  were  merely  to  be 
considered  numbers,  the  dimensions  being  feet  and  seconds,  the 

assumption  in  (1)  involves  vz  (  -?)    =  b  h  (feet),   or  b  =  j- 


Fig.  162. 

power  of   v,  but 


244  APPLIED    MECHANICS 


(second)2'  >  ^  *s  not  a  mere  num^er  >  ^ut  we  ma7 

it  to  be  the  mere  number  ^-  multiplied  by  .  —  —  —  „   In  truth, 

h  J  (seconds). 

we  find  out  in  various  ways  that  b  had  better  be  written  2  g 
where  g  is  an  acceleration  of  32-2  feet  per  second  per  second  in 
London. 

We  have,  then,  the  rule  for  bodies  falling  freely  from  rest, 
v=  ^/-2P....(1). 

Now,  h  is  v2/  2  <?,  and  hence,  when  a  body  has  fallen  h 
feet,  and  its  velocity  is  v,  the  potential  energy,  wh,  lost  by 

it,  may   be   written   w  -JL  or,  as  it  is  more  usually  written, 
3-JT 

1  M? 

x  —  v2  ...  (2).     w  is  the  gravitational  force  or  weight  of  the 

body  in.  pounds,  and  g  is  32-2  feet  per  second  per  second  in 
London. 

191.  It  was  by  experiments  on  falling  bodies  that  Galileo 
was  led  to  formulate  the  law  that  at  any  place  a  body's  w/g, 
which  we  call  the  mass  or  inertia  of  it,  is  constant  •;  and  we  have 
been  led  by  astronomical  observations  to  believe  that  the  mass  or 
inertia  of  a  body  is  a  constant  everywhere.     The  gravitational 
force  on  a  body,  w,  may  vary,  and  g,  the  acceleration  due  to 
gravitational  force,   may  vary;  but  the  ratio  between  them 
seems  everywhere  to  be  constant.     We  generally  denote  the 
mass  or  inertia  w/g  by  the  letter  m.     A  body  kept  in  London, 
defined  by  law  to  have  a  weight  of  1  Ib.  if  weighed  in  vacuo, 
has  a  mass  1  -f  32  '2  in  the  absolute  units  which  are  most  con- 
venient for  engineering  purposes.     Any  body  whose  weight  in 
London  is  w  Ib.  has  a  mass  numerically  expressed  by  w  -f-  32  '2. 
Any  body  whose  weight  in  any  kind  of  units  is  w  at  a  place 
where  the  gravitational  acceleration  in  any  units  is  g,  is  said  to 
have  a  mass  w/g. 

If  we  wish  to  be  very  exact,  we  notice  that  if  we  speak  of 
a  force  as  w  Ibs.,  the  w  is  a  mere  number  ;  if  we  speak  of  a 
force  as  w,  then  w  is  much  more  than  a  mere  number.  A 
body  whose  weight  in  London  is  32  -2  Ibs.  has  a  mass  1.  <  j  ** 

192.  We  have  now  seen  that  half  the  mass  of  a  body,  multi- 
plied by  the  square  of  its  speed  in  feet  per  second,  is  its  kinetic 
energy.     When  the  bob,  Fig.  162,  is  at  B,  let  us  say  that  its 
total  store  of  energy  is  kinetic.     When  it  is  at  A,  the  energy 


APPLIED    MECHANICS.  245 

is  all  potential.  When  it  is  anywhere  between,  its  total  amount 
of  energy  is  exactly  the  same  as  before,  but  part  is  potential 
and  part  is  kinetic.  During  the  swinging  of  a  pendulum  there 
is  a  constant  change  going  on,  potential  energy  changing  into 
kinetic  or  kinetic  into  potential,  and  the  sum  of  these  two 
would  always  remain  the  same  only  that  friction  is  constantly- 
reducing  this  sum  by  converting  part  of  it  into  energy  of 
another  order — namely,  heat. 

Exercise. — Imagine  no  frictional  resistance  to  the  motion 
of  a  projectile.  A  projectile  of  100  Ibs.,  with  a  muzzle  velocity 
of  1,000  feet  per  second.  What  is  its  kinetic  energy  when 
it  is  at  heights  of  10,  50,  and  100  feet?  When  its  whole 
velocity  is  707  feet  horizontally  per  second,  what  is  its  height  1 
This  being  the  horizontal  component  of  the  velocity  through- 
out, what  was  the  angle  of  elevation  of  the  gun  *? 
Ans.,  1,551,000,  1,547,000 and  1,542,000ft.  Ibs.;  8,241  ft.;  45°. 

It  is  to  be  noticed  that  we  assume  that  the  potential 
energy  of  a  body  is  w,  multiplied  by  the  vertical  height  through 
which  it  can  fall  (that  is,  to  some  fixed  datum  level),  so  that  at  a 
certain  height  we  know  its  potential  energy ;  and  if  we  know  its 
total  energy,  the  remainder  is  kinetic,  and  the  kinetic  being 

always  —  m  v*2,  the  velocity  is  known  when  the  height  is  known. 

193.  Test  of  our  Law. — We  now  have  our  rule  to  calculate 
the  energy  stored  up  in  a  moving  body,  every  part  of  which 
is  moving  with  the  same  velocity,  and  we  can  test  it  in  the 
following  way.  Get  a  pulley  (Fig.  163)  as  light  and  frictionless 
as  possible,  because  we  must,  at  the  beginning,  neglect  both  the 
energy  stored  up  in  the  pulley  itself  and  the  loss  by  friction. 
Fasten  the  pulley  at  a  considerable  height  above  the  floor. 
Let  two  equal  weights,  A  and  B,  balance  one  another  at  the 
ends  of  a  long  silk  cord,  passing  over  the  pulley ;  and  let  there 
be  a  wooden  scale,  close  alongside  which  A  passes  as  it  ascends 
and  descends.  Let  us  be  able  to  fix  to  this  scale,  at  any  place, 
a  plate  which  will  suddenly  stop  A,  and,  above  this,  a  ring 
which  will  just  allow  A  to  pass  through.  You  will  find  such 
an  arrangement  as  I  speak  of  in  almost  every  little  collection 
of  apparatus  in  the  kingdom,  and  it  is  called  an  Attwood's 
Machine.  Now  let  A  be  as  high  as  possible  at  the  beginning ; 
place  on  it  a  little  weight  w,  such  as  will  be  lifted  off  when  A 
passes  through  a  ring ;  and  place  a  ring  so  that  it  will  lift  the 
little  weight  off  A  when  A  has  fallen,  say,  3  feet.  You  know 


246 


APPLIED    MECHANICS. 


that,  so  long  as  the  little  weight  lies  on  A,  the  speed  of  A  down- 
wards and  B  upwards  must  become  greater  and  greater.  In 
fact,  the  potential  energy  lost  by  the 
little  weight  becomes  converted  into 
kinetic  energy  of  the  whole  arrangement. 
Now,  as  soon  as  the  little  weight  is 
stopped,  A  and  B  move  with  a  steady 
motion;  and  if  the  table  is  placed  by 
trial  so  that  one  second  after  A  passes  the 
ring  it  is  suddenly  stopped  by  the  table, 
the  distance  between  the  ring  and  table 
shows  the  velocity  which  A,  B,  and  the 
little  weight  had  when  the  little  weight 
was  removed.  In  one  experiment — A 
being  1  Ib.  and  B  the  same,  and  the  little 
weight  0*25  Ib.  —  the  velocity  was 
measured  after  A  had  fallen  3  feet,  and 
was  found  to  be  about  4*5  feet  per  second. 
Now,  the  potential  energy  lost  by  the  little 
weight  was  3  x  '25,  or  '75  foot-pound. 
The  kinetic  energy  was  stored  up  in  2 '25 
pounds,  moving  with  the  velocity  of  4 '5 
feet  per  second,  and,  according  to  the 
above  rule,  its  amount  is 

2-25  -f  64-4  x  4-5  x  4-5, 
or  *71  foot-pound,  or  -04  foot-pound  too 
little.     If   we   consider  that  there  was 
some  friction,  that   the  pulley  retained 
some   kinetic   energy,  and  that   it    was 

difficult  to  fix  the  table,  so  that  exactly  one  second  elapsed 
from  A'S  passing  through  the  ring  until  it  was  stopped,  we  see 
that  the  experiment  is  a  fairly  good  illustration  of  the  rule. 
You  ought,  with  your  own  hands,  to  make  a  number  of  such 
experiments.  In  exercise  work  we  shall  call  A  =  B  =  w  Ib., 
and  we  shall  call  the  little  extra  weight  w. 

In  Art.  50  we  found  the  sort  of  corrections  which  had  to 
be  introduced  by  friction  at  the  wheel  pivots  and  by  the  bend- 
ing of  the  cord.  Their  study  led  us  to  the  great  subject  of 
friction.  But  in  a  well-made  Att wood's  machine,  a  far  more 
important  matter  to  consider  is  the  energy  (so  far  neglected) 
in  the  pulley.  The  experimental  study  of  this  leads  us  to  a 
consideration  of  angular  motion. 


Fig.  163. 


APPLIED    MECHANICS.  247 

194.  Energy  in  a  Rotating  Body. — Suppose  now  that  the 
pulley  is  so  massive  that  its  kinetic  energy  is  considerable,  and 
may  not  be  neglected,  is  there  any  way  of  finding  from  its  speed 
how  much  energy  it  has  stored  up  in  it  ]    We  can  easily  calculate 
the  energy  in  any  little  portion  of  a  wheel  if  we  know  its 
velocity  and  mass,   but   those   portions   near  the  centre  are 
moving  more  slowly  than  portions  near  the  circumference,  so 
that  we  have  to  calculate  the  energy  in  each  little  portion 
separately,  and  add   all  the   results  together.     There  is  one 
thing  which  all  portions  of  a  wheel  have  in  common — they  all 
go  round  the  centre  the  same  number  of  times  per  minute. 
Suppose  now  that  the  number  of  revolutions  of  a  wheel  is 
doubled,  the  real  velocity  of  every  point  in  the  wheel  is  doubled, 
whether  that  point  be  near  the  axis  or  not,  so  that  the  kinetic 
energy  of  the  whole  wheel  is  quadrupled ;  in  fact,  then,  we  find 
that  the  kinetic  energy  stored  up  in  a  wheel  depends  on  the 
square  of  the  number  of  revolutions  which  it  makes  per  minute, 
so  that  the  energy  must  be  equal  to  a  constant  number  multiplied 
by  the  square  of  the  number  of  revolutions  per  minute. 

195.  To  find  experimentally  how  much  energy  is  possessed 
by  a  wheel  when  it  is  rotating,  let  the  wheel  be  mounted  on  an 
axle  supported  on  very  frictionless  bearings.     If  the  centre  of 
gravity  of  the  wheel  is  not  exactly  in  the  axis,  then  it  is  better 
to  place  the  wheel  as  in  Fig.  164.     Now  let  a  cord  with  a 
loop  from  the  pin  B  be  wound  round  the  axle  and  pulled  by  a 
weight  w.    Suppose  the  weight  to  be  1,000  Ibs.,  and  that  we  only 
allow  it  to  fall  8  feet  from  rest  before  the  cord  drops  off  the  pin, 
so  that  when  it  has  fallen  this  distance  it  no  longer  acts  on  the 
wheel,  which  will  then  rotate  with  a  constant  speed.     Roughly 
speaking,  the  wheel  possesses  1,000  x  8,  or  8,000  foot-pounds  of 
energy  stored  up  in  it.     This  is  not  quite  true,  because  the 
weight  itself  possessed  a  certain  amount  of  energy  of  motion 
which  must  be  subtracted.     Suppose  that  at  the  instant  before 
being  stopped  the  weight  was  moving  with  a  velocity  of  1  '5  foot 
per  second,  then  we  must  subtract 

L25?  x  1-5  X  1-5,  or  about  35  foot-pounds. 
64-4 

If  there  were  no  friction,  and  we  find  that  a  speed  of  10 
revolutions  per  minute  has  been  given  to  the  fly-wheel,  we 
know  that  we  have  to  find  a  constant  number,  M,  which,  when 
multiplied  by  the  square  of  10  or  100,  will  give  7,965  foot- 
pounds. Evidently  M  =  79-65,  and  hence,  if  ever  we  find  this 


248 


APPLIED    MECHANICS. 


fly-wheel  rotating,  we  know  that  it  has  stored  up  in  it  the 
amount  of  energy  in  foot-pounds  79*65  x  square  of  number  of 
revolutions  per  minute. 

196.  In  the  above  calculation  we  have  neglected  friction ; 
but,  as  a  matter  of  fact,  in  experiments  the  friction  never  is 
negligible.  As  for  the  friction  of  the  wheel  itself,  on  a  cord 
similar  to  that  which  you  have  already  used,  hang  a  small 
weight  such  as  will  merely  overcome  friction,  so  that  when  you 


w 


Fig.  164. 

give  the  wheel  a  jerk  for  the  purpose  of  starting  the  motion, 
this  weight  will  just  suffice  to  prevent  friction  reducing  the 
speed.  Suppose  this  weight  to  be  5  Ibs.,  then  it  is  quite  evident 
that  5  Ibs.  of  the  original  1,000  were  really  employed  in  over- 
coming friction  and  not  in  storage.  Hence  our  calculation  gives 

995  x  8  -  35,  or  7,925  foot-pounds  as  the  total  storage. 
This  is  at  ten  revolutions  per  minute.     When  it  makes  one 
revolution  per  minute  the  storage  is  79-25  foot-pounds,  and  at 
any  other  speed  we  multiply  79'25  by  the  square  of  the  number 
of  revolutions  per  minute ;  7 9  '25  is  called  the  M  of  the  wheel. 

As  for  the  friction  of  the  pulley  P  and  the  energy  wasted 
in  bending  the  rope,  the  pulley  must  be  tested  like  the  pulley 


APPLIED    MECHANICS.  249 

of  Art.  50,  and  a  correction  made  on  account  of  the  two 
parts  of  the  rope  being  at  right  angles  to  one  another.  Or 
two  pulleys  may  be  tested  at  the  same  time,  the  cord  going 
over  both  of  them  to  two  weights. 

197.  It  is  obvious  that  we  must  be  pretty  quick  in  count- 
ing the  number  of  revolutions  of  the  wheel  produced  by  the 
falling  of  the  weight.     Indeed,  we  ought  to  observe,  if  possible, 
the  time  taken  in  part  of  one  revolution,  using  some  special 
form  of  time-measurer,  because  the  speed  will  now  continually 
decrease  on  account  of  friction. 

But  there  is  another  way  in  which  it  is  easy  to  find  the 
speed  at  the  instant  when  the  weight  ceases  to  act.  Find  the 
total  number  of  revolutions  made  by  the  wheel  till  the 
cord  drops  off  its  pin,  and  let  someone  observe  this  time  in 
minutes.  Then,  as  we  know  that  the  speed  increases  uniformly 
during  this  interval  of  time,  the  mean  speed  is  just  half  the 
speed  at  the  end  of  the  interval ;  that  is,  divide  the  number  of 
revolutions  by  the  number  of  minutes  in  which  they  were  per- 
formed^ and  twice  the  quotient  will  give  the  number  of  revolutions 
per  minute  made  by  the  wheel  when  the  weight  jiist  ceases  to  act. 
You  can  test  your  result  by  counting  the  number  of  revolu- 
tions from  the  time  the  cord  drops  off  until  the  wheel  is 
stopped  by  its  own  friction  and  dividing  by  the  time  which 
elapses;  twice  this  quotient  ought  also  to  be  the  speed  you 
want  to  know. 

198.  It   is   not   necessary    even  to  measure  the   friction 
directly,  for  we  found  that  7,965  foot-pounds  were  given  out  by 
the  weight  in  falling;  now  if  we  count  the  total  number  ofrevolu 
tions  made  by  the  wheel  from  the  time  of  starting  until  stopped 
by  its  own  friction,  and  divide  7,965  by  the  total  number,  we 
shall  Jind  the  loss  of  energy  due  to  friction  during  one  revolution, 
since  there  is  just  as  much  energy  wasted  by  friction  in  any 
one  revolution  as  in  any  other  [a  statement  to  be  tested  by 
the  student].     Ten  times  this  must  be  the  same  amount  of 
energy  as  5  x  8,  or  40  foot-pounds,  for  we  measured  the  friction 
during  10  revolutions   of  the  wheel  as  equivalent   to  5  Ibs. 
falling  8  feet.     This,  then,  is  the  method  we  ought  to  employ. 

I  know  of  no  exercise  in  my  laboratory  which  is  so  useful 
as  this.  A  thoughtful  student  revives  his  knowledge  of 
nearly  all  the  important  dynamical  principles  in  making  the 
corrections,  which  enable  him  to  get  nearer  and  nearer  to  the 
correct  answer  for  M. 


250  APPLIED    MECHANICS. 

199.  You  see  that  M  is  a  number  which  ought  to  be  known 
for  every  fly-wheel ;  it  is  just  as  important  to  know  the  M  of  a 
fly-wheel  as  to  know  the  weight  of  an  ordinary  body.     We  have 
only  to  multiply  the  M  by  the  square  of  the  number  of  revolu- 
tions per  minute,  and  we  find  at  once  the  energy  in  foot-pounds 
stored  up  in  the  wheel.     I  have  shown  you  how  to  find  the  M 
of  a  fly-wheel  by  experiment ;  I  will  now  give  you  an  idea  of 
its  value  in  different  cases.      Imagine  a  grindstone  whose 
diameter   is   4*5  feet,   whose  breadth  is  1-4  foot,  the  weight 
of  its  material  per  cubic  foot  being   132   Ibs. ;   then  we  can 
calculate  its  M  by  first  finding 

132  x  1-4  x  4-5  x  4-5  x  4-5x4-5, 

and  dividing  this  answer  by  59,800.  For  any  rotating  object 
of  cylindrical  shape,  the  shape  of  a  grindstone,  this  rule  will 
always  find  M.  Multiply  the  weight  of  the  material  per  cubic 
foot  by  the  breadth  or  width  ;  multiply  this  by  the  fourth  power 
of  the  diameter,  and  divide  by  the  constant  number  59,800. 
Whether  the  material  is  wood  or  stone  or  metal,  this  will  give 
M,  and  this  multiplied  by  the  square  of  the  number  of  revolu- 
tions per  minute  will  give  the  energy  in  foot-pounds  stored  up 
in  the  rotating  body.  For  the  above  grindstone,  on  calculating 
out,  you  will  find  the  M  to  be  1-27.  So  that  when  it  makes 
1  revolution  per  minute,  there  is  stored  up  in  it  1'26  foot- 
pound of  energy ;  when  it  makes  2  revolutions  per  minute, 
there  is  stored  up  in  it  1-27  x  4,  or  5'04  foot-pounds. 

Foot-pounds. 

At       3  revolutions,  1'27  x  9,  or          11  "43 

„      20         „  1-27  x        400,    „        508 

„      50         „  1-27  x     2,500,   „     3,175 

„    100         „  1-27  x  10,000,   „  12,700 

200.  If  we  fix  a  small  weight  of  20  Ibs.  on  a  wheel,  at  12  feet 
from  the  axis,  this  adds  to  the  M  of  the  wheel  the  amount 

20  x  12  x  12  +  5,873,  or  0-49 ; 

or  the  weight  multiplied  by  the  square  of  its  distance  from  the 
axis,  divided  by  5,873. 

If  we  add  a  very  thin  rim  to  a  wheel,  the  addition  to  M  ia 
found  by  multiplying  the  weight  of  the  rim  by  the  square  of  its 
average  radius,  and  dividing  by  5,873. 

Note. — In  Table  II.  all  dimensions  are  supposed  to  be  in  feet  See 
Art.  203. 


APPLIED    MECHANICS. 
TABLE  II. 


251 


Nature  of  Rotating 
Body. 

I 

M 

k 

vjy 

Sphere  of  diara-^ 
eter  d,  rotating  f 
about    diameter  | 
as  axis.     .     .     .  / 

wd*  x  -001626 

wd5  ^  112,166 

•3162(Z 

0 

Spherical     shell,  \ 
whose      outside  / 
diameter     is     d  ' 
and  inside  is  d\,  ( 
rotating     about  I 
diameter  as  axis  J 

w  (d*  -  o^s)  ") 
x  '001626   J 

w  (d5  -  d\5) 
+  112,166 

^^i 

±- 

j 

Cylinder,     diam-  "S 
eter  d,  length  1,  ( 
rotating     about  j 
its  axis     .     .    .  / 

wld*  x  -00305 

wld4  -5-  59,814 

•3535d 

\ 

Hollow  cylinder,  > 

1     I     t 

outside  diameter  I 
d,   inside  diam-  j 

tol(d*-di*)\ 
x  -00305    } 

wZ  (d!4  -  di4) 
4-  59,814 

3535  v/dM1^ 

eter  c?i,  length  ^j 

I 

Thin    rim,   mean  "\ 
radius       r       of  > 
weight  W      .     .  j 

wr2  *  32-2 

wr2  *  5,873 

r 

: 

<i           > 
,      \            11 

Thin      rod,      of  \ 
length  Z,   rotat- 
ing   about    axis 
through    its  1 
middle       point,  / 
at   right   angles 
to    its     length. 
Weight  of  rod  wj 

WJ2-00258 

wJ2  4-  70,474 

•2887? 

<  i  > 

Thin  rectangular" 
plate,     rotating 
about      axis 
through  its  cen- 

i            OM 

tre    parallel    to  I 
side  6,  the  side 

wd2  -00258 

wd2  +  70,474 

•2882d 

j 

d  being  at  right 
angles    to   axis. 
Weight  of  plate 
w    .    .    . 

252  APPLIED    MECHANICS. 

It  will  be  found  that  if  a  fly-wheel  has  light  arms  and 
a  heavy  rim,  as  we  often  see  on  such  wheels,  a  fairly  good 
approximation  to  its  M  is  found  by  multiplying  the  weight  of 
the  wheel  by  the  square  of  the  mean  diameter  of  the  rim,  and 
dividing  by  23,000. 

Example. — The  rim  of  a  fly-wheel  weighs  15  tons;  its  mean 
diameter  is  20  feet.  Calculate  approximately  what  energy  is 
stored  up  in  it  when  it  makes  60  revolutions  per  minute.  Here 
you  will  find  the  M  of  the  fly-wheel  to  be  about  584,  and  hence 
the  stored  energy  is  584  x  60  x  60,  or  2,102,400  foot-pounds. 

The  mathematicians  do  not  use  our  M.  They  use  a 
quantity  called  I,  the  moment  of  inertia  of  a  rotating  body, 
which  is  numerically  equal  to  twice  the  energy  stored  in  the 
body  when  it  has  an  angular  velocity  of  o-ne  radian  per  second. 
See  Art.  203.  I  give  the  values  of  I  as  well  as  of  M  in  the 
above  table. 

In  the  case  of  any  rotating  body,  if  we  imagine  a  fly-wheel 
made  of  the  same  weight  and  same  M  or  I,  with  an  exceed- 
ingly thin  rim,  the  average  radius  of  this  rim  is  called  the 
radius  of  gyration  of  the  body.  In  fact,  the  mass  of  a  body 
multiplied  upon  the  square  of  its  radius  of  gyration  is  its 
moment  of  inertia.  The  radius  of  gyration  of  a  fly-wheel  rim 
is  usually  taken  to  be  the  average  radius  of  the  rim. 

201.  Steadiness  of  Machines. — A  fly-wheel  is  put  upon  a 
riveting-  or  shearing-machine,  or  other  machine,  because  the 
supply  of  energy  to  the  machine  is  not  given  regularly,  or  else 
because  the  demand  for  energy  from  the  machine  is  irregular. 
The  fly-wheel  enables  the  machine  to  maintain  a  more  constant 
speed.  In  calculating  the  proper  size  of  a  fly-wheel  for  any 
machine  we  must  know  two  things  :  first,  what  is  the  greatest 
alteration  of  speed  allowable  in  the  case ;  and  secondly,  the 
greatest  fluctuation  of  the  demand  and  supply  of  energy.  Thus, 
suppose  we  wish  never  to  have  the  speed  of  the  fly-wheel 
more  than  51  nor  less  than  49  revolutions  per  minute,  and 
that  during  some  interval  of  time  the  fly-wheel  has  to  give 
out  20,000  foot-pounds  more  than  it  receives  during  that  time; 
then,  although  the  fly-wheel  will  afterwards  have  this  deficiency 
made  up  to  it  by  some  steady  supply,  it  is  obvious  that 
its  speed  must  diminish.  We  wish  its  speed  to  diminish 
only  from  51  revolutions  to  49  revolutions  per  minute  in 
this  interval  of  time.  Now,  when  the  fly-wheel  runs  at  51 
revolutions,  it  has  stored  up  an  amount  of  energy  equal  to 


APPLIED    MECHANICS.  253 

its  M  x  51  x  51 ;  and  when  it  runs  at  49  revolutions,  its 
store  is  M  X  49  x  49,  and  the  difference  between  these  two 
ought  to  be  20,000.  Hence,  subtracting  49  x  49,  or  2,401, 
from  51  x  51,  or  2,601,  we  get  200;  and  dividing  20,000  by 
200,  we  find  100  as  the  required  value  for  M.  Subtract,  then, 
the  square  of  the  least  speed  from  the  square  of  the  greatest,  and 
divide  the  greatest  excess  of  demand  or  supply  by  this  remainder  ; 
the  quotient  is  the  M  of  the  fly-wheel.  Having  found  M,  the 
question  is,  how  can  you  tell  from  it  the  size  and  weight  of  the 
wheel1?  Find  the  M  of  any  wheel  of  the  same  shape  and 
material  as  that  which  you  want  to  use.  It  is  obvious  that  the 
diameters  of  the  wheels  are  as  the  fifth  roots  of  their  M's.*  We 
want  a  wheel  whose  M  is  100.  Suppose  I  find  a  wheel  of  the 
shape  I  wish  to  use  whose  outer  diameter  is  8  feet,  and  I 
calculate  its  M,  and  find  it  to  be  11;  then 

The  fifth  root  of  11  :  fifth  root  of  100  ::  8  :  answer. 

Log.  11  =  1-0414;  divided  by  5  it  is  0*2083,  which  is  the 
logarithm  of  1-615. 

Log.   100  =  2-0;   divided  by  5  it  is   0-4,  which  is  the 
logarithm  of  2*512.     Hence 

1-615  :  2-512  ::  8  :  answer. 

This  is  an  easy  exercise  in  simple  proportion.  I  find  my 
answer  to  be  12-44  feet,  or  12  feet  5J  inches,  the  diameter  of 
the  required  fly-wheel,  which  is  to  be  similar  in  form  to  the 
smaller  specimen  used  by  me  for  calculation. 

202.  The  total  kinetic  energy  stored  up  in  any  machine 
is  found  by  calculating  the  energy  in  every  wheel  and  in  every 

*  If  we  have  any  two  similar  wheels,  or  other  rotating  bodies  of  the  same 
material ;  if  we  consider  any  similar  small  portions  of  them ;  it  it  evident  that 
their  weights  are  proportional  to  their  cubic  contents,  or  to  the  cubes  of  any 
similar  linear  measurements.  Hence,  if  one  is,  say,  twice  the  diameter  of  the 
other,  as  every  dimension  of  the  one  is  twice  that  of  &ie  other,  the  weight  of 
one  must  be  2  X  2  X  2,  or  eight  times  that  of  the  other.  Now,  the  M  of  any 
rotating  body  depends  not  merely  on  the  weight  of  each  portion  of  the  body, 
but  on  the  square  of  its  distance  from  the  axis,  so  that  the  M  of  one  must  be 
8  X  2  X  2,  or  thirty-two  times  the  M  of  the  other.  Similarly,  if  the  linear 
dimensions  were  as  3  to  1,  the  values  of  M  would  be  as  243  to  1  for  a  pair  of 
similar  wheels. 

Example. — We  want  a  wheel  which  will  have  a  store  of  1,000  foot-pounds 
when  rotating  at  twenty  revolutions  per  minute,  and  it  is  to  be  of  the  same 
shape  as  that  of  an  already  existing  wheel,  which  is  4  feet  in  diameter,  and 
which  contains  a  store  of  1,350  foot-pounds  when  running  at  30  revolu- 
tions. Evidently  the  M  of  this  second  wheel  is  1,350  •*•  900,  or  1'5,  and  the 
M  of  the  first  wheel  is  to  be  2 '5.  Using  logarithms,  we  find  that  the  fifth 
root  of  1-5  is  to  the  fifth  root  of  2  '5  as  4  feet  is  to  4 '4  feet,  the  answer. 


254  APPLIED    MECHANICS. 

moving  part,  and  adding  all  together.  But  suppose  that  in  the 
machine  there  is  some  shaft  of  more  importance  than  any  other, 
it  is  usual  to  give  the  speed  of  this  shaft  only,  because  if  its 
speed  be  doubled,  the  speed  of  every  other  is  doubled.  Thus,  in 
a  steam-engine  we  state  the  number  of  revolutions  per  minute 
of  the  crank  shaft,  and  this  tells  us  the  speed  of  every  part  of 
the  engine.  Let,  then,  the  number  of  revolutions  of  some 
such  principal  axle  of  a  machine  be  found.  If  this  number 
of  revolutions  is  doubled,  the  kinetic  energy  stored  up  in  the 
machine  is  quadrupled ;  and,  in  fact,  the  kinetic  energy  stored 
up  is  equal  to  a  certain  number  which  can  be  found  for  the 
machine,  and  which  we  shall  call  its  M,  multiplied  by  the 
square  of  the  number  of  revolutions  of  this  particular  axle  per 
minute.  The  M  of  any  machine  may  be  experimentally  deter- 
mined in  exactly  the  same  way  as  we  have  shown  above. 

If  we  know  the  M  of  any  machine,  then  the  M  of  any  other 
machine  made  to  the  same  drawings,  and  of  the  same  materials, 
but  with  all  its  dimensions  twice  as  great,  is  thirty- two  times 
as  great,  because  the  M's  of  the  two  machines  are  proportional 
to  the  fifth  powers  of  their  corresponding  dimensions. 

203.  The  energy  stored  up  in  a  rotating  body  is  equal  to  ^ia2, 
where  I  is  moment  of  inertia  about  the  axis ;  that  is,  the  sum  of  all 
such  terms  as  mass  of  a  little  portion  multiplied  by  the  square  of  its 
distance  from  the  axis,  a  is  angular  velocity  in  radians  per  second. 

Hence,  as  o  =  -^- ,  if  n  is  number  of  revolutions  per  minute,  and 

rr  is  3-1416,  the  energy  is  i^f-  ;    so  that  our  M  is  il^.     In 

IjoOO  1,800 

Table  II.  we  give  values  both  of  M  and  of  I.  In  both  cases  w 
is  in  Ibs.  per  cubic  foot,  and  all  dimensions  are  supposed  to  be  in 
feet. 

We  ask  for  the  advice  of  students  in  an  important  matter.  Is 
it  good  to  use  the  idea  of  our  M,  the  energy  stored  in  a  body  when 
it  makes  one  revolution  per  minute  P  To  the  weaker  vessel,  the 
beginner,  or  the  man  who  dislikes  algebra,  we  know  it  to  Ite 
useful.  Indeed,  to  any  practical  engineer,  however  mathematical 
he  may  be,  it  is  convenient  to  have  such  a  quantity.  But  is  its 
convenience  overpowered  by  the  inconvenience  of  having  a  new 
quantity  to  think  of  ?  We  do  not  know,  and  we  should  like  to 
receive  advice. 

As  for  the  idea  of  moment  of  inertia,  it  comes  in  in  this  way. 
If  we  have  a  small  mass  in  moving  round  an  axis  at  the  distance 
T  feet  with  angular  velocity  a,  its  linear  velocity  is  ra  feet  per 
second ;  its  kinetic  energy  is  ^w?>2a2  foot-pounds.  Now,  if  we  have 
many  small  masses  and  make  this  calculation,  we  see  that  every 
little  torm  has  £aa  in  it,  common  to  them  all.  Hence  we  multiply 


APPLIED    MECHANICS.  255 

every  little  w  by  its  r2  and  add  up,  calling  the  stun  I,  or,  as  we 
say  mathematically,  £wr2  =  i,  and  evidently  ^ia2  is  the  kinetic 
energy.  In  any  book  on  the  calculus,  exercises  will  be  found  on 
the  calculation  of  the  various  values  of  I  given  in  Table  II. 
Every  student  who  knows  how  to  do  so  ought  to  calculate  all 
these,  and  also  the  value  of  I  for  various  other  bodies  in  such 
shapes  as  may  be  defined  mathematically. 

EXERCISES. 

1.  A  fly-wheel  of  cast  iron,  whose  rim  is  10  feet  mean  diameter  and 
section  8"  x  10",  has  a  volume  8xlOxl07rXl2  cubic  inches,  and  as 
a  cubic  inch  weighs  0'26  Ib.  the  weight  is  7,840  Ibs.      Its  moment  of 
inertia  is,  approximately,  7,840  x  52  -j-  32-2,  or  6,087.     Find  its  M   also. 

The  M  of  a  fly-wheel  is  its  kinetic  energy  when  going  at  one  revo- 
lution per  minute.  This  is  an  angular  velocity  o  =  2ir  •£•  60,  or  0'1047 
radians  per  second;  so  that  M  is  |  x  6,087  X  (-1047)2  =  33 -48. 

2.  What  energy  will  this  fly-wheel  store  in  changing  from  98  to  102 
revolutions  per  minute  ?      Ans.,  M  (1022  —  982),  or  26,782*8  foot-pounds. 

3.  A  gas  engine  has  6  indicated  horse-power  at  150  revolutions  per 
minute  ;  what  is  the  indicated  work  of  one  cycle  (or  two  revolutions)  ? 

Ans.,  6  x  33,000  -r-  75,  or  2,640  foot-pounds. 

4.  If  one-half  of  this  is  stored  in  changing  speed  from  149  to  151 
revolutions  per  minute,  what  is  the  M  of  the  fly-wheel  ?     What  is  its 
moment  of  inertia  ?    If  it  is  made  to  the  same  drawings  as  the  fly-wheel 
of  Exercise  1,  what  is  the  average  radius  of  its  rim  ?    What  is  its  weight  ? 

Ans.,  2-2  ;  400  ;  2-9  ft.  1,534  Ibs. 

5.  If  the  gas-engine  of  Exercise  3  goes  at  200  revolutions  per  minute, 
and  if  there  is  a  complete  cycle  in  every  revolution  and  half  of  its  energy 
is  stored,  what  is  the  M  of  the  fly-wheel  if  fluctuation  of  speed  is  to  be 
the  same  as  in  (4)  ?  Ans.,  0-46. 

6.  In  Attwood's  machine  each  w  =  1'5  Ib.,  w  =  0-3  Ib.      If  w  is 
applied  for  a  height  of  2  feet,  what  ought  to  be  the  velocity  ? 

Ans.,  Total  mass  is  3-3  4-  32-2,  or  0-1025,  and  its  kinetic  energy  is 
£  X  '1025  x  v3.  This  is  equal  to  0'3  X  2,  or  0'6  foot-pound.  Hence 
v9  =  1-2  -f-  1,025,  or  v  =  3 -42  feet  per  second. 

7.  If  in  Exercise  6  the  wheel  weighs  0'17  Ib.,  with  a  radius  of  gyration 
which  happens  to  be  equal  to  the  radius  of  the  circle  of  the  centre  of  the 
cord  as  it  goes  round,  we  have  simply  to  imagine  the  moving  mass  to 
have  0-17  Ib.,  or  rather  0'17  -r-  32-2,  or  -0053,  added  to  it.     What  is  the 
corrected  value  of  v  at  the  time  when  w  is  lifted  off  ?v 

Ans.,  The  whole  mass  is  now  '1078,  and  ^  x  -1078  v2  =  0'6 ;  so 
that  v  =  3-34  feet  per  second. 

8.  If  the  radius  of  the  cord  circle  on  the  pulley  is  0'24  foot  and  the 
radius  of  gyration  of  the  pulley  is   0'21   foot,  imagine  the  pulley  of 
0-17  Ib.  and  radius  of  gyration  0-21  replaced  by  another  of  as  Ibs.  and 
radius    of    gyration    0-24 ;    so    that    x  x   (0-24)2  =  0-17  x  (0-21)2,    or 
x  =  0-13.  ^  We  therefore  imagine  a  mass  0-13  -^  32-2,  or  -0040,  added  on 
to  the  original  weight.     In  this  case  find  v. 

Ans.,  v  =  3-36  feet  per  second. 

It  is  this  sort  of  calculation,  when  experimenting  with  an 
Attwood's  machine,  which  gives  a  man  a  practical  knowledge  of 
mechanics. 

9.  A  homogeneous  cylinder  of  mass  m  and  radius  r  rolls  down  an 


256  APPLIED    MECHANICS. 

inclined  plane.     If  the  linear  speed  of  -its  centre  Is  v,  show  that  its 
angular  velocity  a  about  the  centre  is  vjr.     What  is  its  kinetic  energy  ? 

Ans.,  |m-2  +  |io2. 

Table  I.  shows  that  its  radius  of  gyration,  k,  is  -3535^  ;  and 
as  i  =  mk2,  we  have  the  kinetic  energy  fyn  (v2  +  £2a2),  or 

\m  (v*  +  £2  ^V   or    \rntf  (\  +  ^Y     "Notice  that  in  a  cylinder 

&2/r2  =  £  ;  so  that  the  total  kinetic  energy  is  3my2/4.  If  the 
cylinder  is  cast  iron,  r  =  0-17  foot,  its  length  0-3  foot  ;  then  its 
weight  is  12-257  Ibs.,  its  mass  is  0-3808,  and  its  kinetic  energy 
is  0-2855v2. 

If  this  cylinder  has  rolled  without  friction  along  a  switchback 
path  through  a  vertical  height  of  1*5  foot,  show  that  v  =  8-02. 

If  the  path  was  an  inclined  plane,  such  that  for  a  vertical  fall 
of  1'5  feet  the  distance  traversed  is  20  feet,  and  if  the  average 
velocity  was  half  the  final  velocity,  find  the  time  taken  to  reach 
this  point.  Ans.,  4  -99  seconds. 

On  a  plane  at  the  angle  a  with  the  horizontal,  show  that 
the  time  taken  for  a  sloping  distance  I  feet,  being  I  -f-  %v  ,  is 


10.  A  riveting-machine  needs  3  horse-power;    a  fly-wheel  upon  it 
fluctuates  in  speed  between   80  and   120   revolutions  per  minute;   an 
operation  occurs  every  two  seconds,  and  this  requires  fths   of  all  the 
energy  supply  for  two  seconds.     Find  the  M  and  I  of  the  fly-wheel. 

The  energy  supply  for  one  minute  is  3  x  33,000,  and  fths  of  the 
supply  for  two  seconds  is  3  x  33,000  x  £  x  -fa  x  2,  or  2,875  foot- 
pounds. This  is  equal  to  M  (1202  —  802)  ;  so  that  M  is  0*36,  I  is 
•36  x  1,800  4-  7T2,  or  65;6. 

11.  A  machine  (consisting  of  a  grindstone  and  the  shafting  of  a  shop) 
has  an  M  =  12  '13.     Suppose  the  loss   of    energy  by  friction  in  one 
revolution  to  be  always  1,550  foot-pounds,  whatever  the  speed.     If  the 
machine  (the  grindstone  spindle)  is  making  140  revolutions  per  minute 
and  ceases  to  receive  energy,  in  how  many  revolutions  will  it  come 
to  rest? 

Let  the  answer  be  x  ;  then  x  x  1,550  =  12-13  x  HO2,  so  that  x  =  154 
revolutions.  As  the  average  speed  will  be  70  revolutions  per  minute, 
the  stoppage  will  take  2-2  minutes. 

12.  Find  the  M  and  I  of  a  fly-wheel  which  stores  25,000  foot-pounds 
of  energy  when  changing  in  speed  from  50  to  70  revolutions  per  minute. 
It  is  to  be  similar,  and  of  the  same  material,  to  an  existing  fly-wheel 
whose  M  is  0-047.     What  is  the  ratio  of  their  sizes  ? 

Ans.,  10-4;  189-7;  2'93  :  1. 

13.  If  the  earth  described  a  circular  path  of  92  x  106  miles  radius  in 
365J  days,  if  it  were  a  sphere  of  8,000  miles  diameter,  of  uniform  density 
6'6  times  that  of  water,  what  is  its  mass  in  engineer's  units  ?    What  is 
its  kinetic  energy  because  of  the  motion  of  its   centre?    What  is  its 
kinetic  energy  because  it  rotates  on  its  axis  once  in  23  hours  56  minutes 
4  seconds  ? 

14.  An  engine  is  running  at  200  revolutions  per  minute.     Suppose 
that  after  the  steam  is  shut  off  and  the  load,  removed  the  engine  made 
250   revolutions  before  coming  to  rest.     Assume  a  constant  moment  of 
resistance,  and  calculate  its  amount  if  the  moving  parts  are  capable  of 


APPLIED    MECHANICS.  257 

storing  as  much  energy  as  a  fly-wheel  weighing  1  ton  and  having  a  radius 
of  gyration  of  2£  feet.  Ans.,  7 3  6  Ibs. 

15.  The  rim  of  a  fly-wheel  is  9  inches  broad  and  4^  inches  deep,  the 
external  diameter  being  9  feet.  Find  the  moment  of  inertia  of  the  rim.  If 
an  engine  be  supposed  to  drive  this  wheel  (neglecting  other  resistances), 
how  many  revolutions  will  have  been  made  before  the  speed  acquired  from 
rest  by  the  wheel  is  120  revolutions  per  minute,  the  diameter  of  the 
cylinder  being  18  inches,  the  stroke  2  feet  6  inches,  and  the  mean  effective 
pressure  15  Ibs.  per  square  inch?  What  time  would  be  required  ? 

Ans.,  1977;  8-2  revs.;  t=8-8. 

1G.  The  fly-wheel  of  an  engine  of  4  horse-power,  running  at  75  revolu- 
tions per  minute,  is  equivalent  to  a  heavy  rim,  2  feet  9  inches  mean 
diameter,  weighing  500  Ibs.  Estimate  the  ratio  of  the  kinetic  energy  in 
the  fly-wheel  to  the  energy  developed  in  a  revolution,  and  determine  the 
maximum  and  minimum  speeds  of  rotation  when  the  fluctuation  of  energy 
is  one-fourth  the  energy  of  a  revolution.  Ans.,  1 : 1-94  ;  84'17,  65*83. 

17.  A  fly-wheel  is  made  to  rotate  by  means  of  a  weight  of  5  Ibs. 
attached  to  the  end  of  a  cord  passing  round  the  shaft,  the  diameter  of 
which  is,  together  with  thickness  of  cord,  1£  inches.  Find  the  moment 
of  inertia  of  the  wheel  if  the  weight  falls  a  distance  of  8  feet  from  rest 
in  1  minute,  friction  being  neglected.  Ans.,  4-35. 

204.  Exercise. — In  the  Attwood's  machine,  if  the  pulley  of  moment  of 
inertia  I  is  supported  on  four  friction  wheels  eacn  of  moment  of  inertia  Ix 
about  its  own  axis,  and  if  the  spindle  of  the  pulley  has  a  diameter  one- 
twentieth  of  the  diameter  of  the  rim  of  each  friction  wheel,  show  that  the 
pulley  has  a  virtual  moment  of  inertia  I  .+  '01  Ij. 

In  a  good  Attwood's  machine  calculate  carefully  the  virtual  moment 
of  inertia  of  the  pulley.  Find  experimentally  for  any  values  of  A  and  B 
the  lessening  of  the  preponderating  force  which  represents  the  friction 
and  stiffness  of  the  cord.  If  you  can  work  with  a  discarded  old  machine 
with  massive  wheels  and  much  friction  and  hemp-cords,  you  will  find  it 
far  more  instructive  than  an  expensive,  nearly  frictionless  contrivance. 
In  measuring  time,  exercise  a  little  ingenuity  of  your  own  in  getting 
accuracy  ;  but  if  your  teacher  has  made  everything  very  easy  for  accurate 
measurement,  you  will  learn  nothing.  If  you  have  read  too  much  about 
other  people's  methods,  you  will  learn  very  little.  You  will  learn  most 
when  you  try  to  get  accuracy  of  measurement  with  very  rough  apparatus. 
Teachers  who  do  not  think  will  introduce  roughness  and  inaccuracy  in 
the  wrong  places ;  they  will  give  you  a  bad  timekeeper,  perhaps,  or  an 
inaccurate  scale  for  measuring  distances. 

After  having  worked  with  the  Attwood's  machine,  in  which  the  velocity 
ratio  is  1,  you  will  find  it  very  interesting  (if  you  have  time  enough)  to 
experiment  with  the  apparatus  shown  in  Fig.  22,  in  which  the  velocity 
ratio  may  have  any  value.  In  the  experiments  described  in  Art.  51  you 
aimed  at  keeping  speed  constant.  Now  let  there  be  an  excess  weight  B, 
and  measure  the  kinetic  energy  as  with  the  Attwood  machine.  We  need 
not  here  describe  more  fully  the  obviously  interesting  and  instructive 
experiments  which  may  be  made. 

205.  In  a  Bull  engine  (see  Fig.  16 6)  steam-pressure  in  the 
space  B,  p  Ib.  per  unit  area  (in  excess  of  the  pressure  from  above 
the  piston) acts  on  a  piston  of  area  A  with  a  total  upward  force/? A. 


258 


APPLIED    MECHANICS. 


This  causes  a  total  weight  w  to  be  lifted  (consisting  of  steam- 
piston,  rod,  pump-plunger,  etc.).  If  p  varies  when  the  piston 
rises  through  a  height  7i,  the  total  work  E  done  by  the  steam  must 
be  calculated  for  small  changes  of  level,  and  the  results  added 
up.  Thus  in  Table  HA.  we  give  the  pressure  of  the  steam  as 
the  piston  rises,  the  pressure  being  measured  by  an  indicator. 
The  area  of  the  piston  is  900  square  inches,  so  that  900  p  is 
the  lifting  force.  It  is  therefore  evident  that  the  fourth  column 
shows  at  each  place  (approximately)  the  work  already  done 
upon  the  piston  when  w  (w  =  20  tons,  or  44,800  Ibs.)  has 
risen  h  feet.  If  it  is  rising  with  the  velocity  v  feet  per 

•«V 

second,  it  possesses  w/i  foot-pounds  of  potential  and  J— v2  foot- 
pounds of  kinetic  energy.  But  we  know  the  total  E  ;  and  if 
we  subtract  w  h,  we  know  the  kinetic  energy,  and  we  can 
therefore  calculate  v. 

TABLE   UA. 


fcFeet. 

p  Pounds 
aer  Square 
Inch. 

A  p  Pounds. 

E,  the  total 
Work  done  in 
Foot-pounds. 

wfc. 

& 

~v. 

0 

100 

90,000 

0 

0 

0 

0 

0-5 

100 

90,000 

1-0 

100 

90,000 

90,000 

44,800 

45,200 

8-06 

1-5 

100 

90,000 

2-0 

100 

90,000 

180,000 

89,600 

90,400 

11-4  f 

2-5 

80 

72,000 

3-0 

66-7 

60,030 

252,000 

134,400 

117,600 

13-0 

3-5 

57-1 

51,390 

4-0 

50-0 

45,000 

303,390 

179,200 

124,190 

13-36 

4-5 

44-4 

39,960 

5-0 

40-0 

36,000 

343,350 

224,000 

119,350 

13-1 

5-5 

36-4 

32,760 

6-0 

33-3 

29,970 

376,110 

268,800 

107,310 

12-41 

6-5 

30-8 

27,720 

7-0 

28-6 

25,740 

403,830 

313,600 

90,230 

11-38 

7-5 

26-7 

24,030 

8-0 

25 

22,500 

427,860 

358,400 

69,460 

9-98 

8-5 

23-5 

21,150 

9-0 

22-2 

19,980 

449,010 

403,200 

45,810 

8-10 

9-5 

21 

18,900 

10-0 

20 

18,000 

467,910 

448,000 

19,910 

5-33 

10-25 

19-5 

17,550 

10-5 

19 

17,100 

10-75 

18-6 

16,740 

481,600 

481,600 

0 

0 

I 

APPLIED    MECHANICS. 


259 


206.  The  earnest  student  will  work  out  the  numbers  in 
the  above  table,  and  draw -curves  showing  h  andp  and  h  and  v. 
In  Fig.  165  OGIIIJ  shows  p  the  pressure,  o  F  being  the  distance 
travelled  by  the  piston  and  weight.  The  ordinates  of  o  P  N  repre- 
sent the  velocity  at  every  instant.  Notice  that  if  we  divide  the 


Pig.  165. 


Fig.  166. 


force  p  A  and  the  weight  w  by  A,  we  get  p  and.  w/A  to  compare. 
The  straight  line  K  M  represents  w/A.  The  area  of  o  K  M  N  ia 
equal  to  the  area  o  G  H  u  N ;  that  is,  w/A  is  the  average  value 
of  the  pressure  until  h  =  ON.  The  student  must  see  clearly 
that  the  force  increasing  the  velocity  is  represented  to  scale  by 
the  ordinate  of  KGHLIJMK,  and  this  force  becomes  negative 
after  L,  so  that  the  point  L  tells  us  where  v  reaches  a  maxi- 
mum and  begins  to  diminish.  The  positive  area  K.G  H  L  K  is 
equal  to  the  negative  area  L  u  M  L.  It  is  worth  while  spending 
a  good  deal  of  time  over  such  curves  and  such  an  investigation 
as  this. 

The  student  ought  now  to  assume  that  besides  the  force 
w  there  is  a  constant  force  of  friction  (say  2,000  Ibs.)  to  be 
overcome.  He  will  therefore  subtract  2,000  h  foot-pounds 

w 

from  E,  as  well  as  w  h,  to  find  the  kinetic  energy,  J  -  v2,  and 

he  will  calculate  v,  again  obtaining  a  new  curve.  In  this 
engine,  when  the  steam  is  allowed  to  escape,  the  weight  w 
falls,  and  performs  the  actual  pumping  operation. 

207.  Should  the  advanced  sti^ent  care  to  take  up  a  problem  that 
more  nearly  approaches  the  actual  case  (namely,  assume  that  the 
force  of  friction  F  is  not  constant,  hut  is  some  function  of  the 
velocity),  he  must  find  out  for  himself  some  graphical  method  of 
working.  Our  own  plan  is  this : — We  calculate  v  at  the  end  of  an 
interval  from  a  rough  notion  of  the  velocity,  and  therefore  of  p,  the 


260  APPLIED    MECHANICS. 

force  of  friction  during  the  interval.  We  can  now  approximate 
more  closely  to  the  actual  friction  and'  calculate  more  exactly. 
We  advise  the  student  to  proceed  in  this  way  in  the  above  case, 
taking  p  as  following  the  law 

F  =  500  +  100  v. 

When  a  torque  M  turns  a  body  through  the  angle  9  radians,  it 
does  work  of  the  amount  M  0.  Thus,  if  a  turning  moment  of  M 
pound-inches  is  acting  on  a  shaft  which  revolves  at  n  revolutions 
per  minute,  or  2irn  radians  per  minute,  it  does  work  M  (pound  x 
inch)  lirn  per  minute,  or  irn  M  -f-  6  foot-pounds  per  minute. 

If  the  twisting  moment  on  a  shaft  M  is  proportional  to  fl 
radians,  the  angle  through  which  a  certain  length  of  shaft  is 
twisted,  then  |M  9  is  the  strain  energy. 

If  bending  moment  M  acting  on  a  certain  length  of  a  beam 
between  two  cross-sections  causes  an  angular  change  0,  and  6  is 
proportional  to  M,  then  ^M  9  is  the  strain  energy  produced  by  the 
action  of  M.  These  two  propositions  enable  us  to  calculate  the 
resilience  of  many  springs  ;  that  is,  the  energy  which  it  is  possible 
to  store  up  in  them. 

The  force  p,  acting  through  a  little  distance  Has,  does  work 
p  .  5a?.  If  F  varies,  a  curve  must  be  drawn  showing  its  value  for 
each  value  of  x,  and  the  work  done  in  any  distance  is  shown  as  an 
area  or  an  integral.  Thus,  if  p  =  ax,  it  is  easy  to  show  that  the 
work  done  from  x  =  0  to  any  other  value  is  ^ax2  ;  or  from  x  =  xl 
to  x  =  a?2  the  work  done  is  \a  (x.22  —  x-?}.  Or  we  may  put  it  that 
the  work  done  is  x  multiplied  by  the  average  value  of  F,  the  average 
value  being  the  half  sum  of  the  two  extreme  values. 

Thus  if  the  gradually  applied  load  w  on  a  beam  produces  the 
deflection  y  where  y  is  proportional  to  w,  the  energy  stored  in  the 
loaded  beam  is  \wi 

But  if  there  is  no  loss  of  the  energy  due  to  loading,  any  method 
of  loading  which  during  the  deflection  y  is  calculated  to  do  the 
same  work  will  cause  the  same  deflection.  One  such  method 
is  to  suddenly  apply  half  the  final  load  or  \w.  Here,  as  in. 
structures  generally,  we  have  assumed  that  deformation  is  pro- 
portional to  load.  If  the  law  of  yielding  is  force  =f(y]  where 
there  is  any  curious  law,  and  if  the  integral  of  f(y]  with  regard  to 
y  is  r(y)  ;  or  if  in  case  the  deformation  is  through  an  angle  9  the 
moment  M  =  /(0),  then  the  energy  stored  in  any  configuration  is 
•p(y)  or  p(0).  If  a  force  or  moment  which,  gradually  applied, 
would  only  produce  a  yielding  y  or  9  is  suddenly  applied  of  its 
full  amount,  the  yielding  is  y1  or  01  where  ylf(y]  =  ?(yl)  or 


Thus  if  the  righting  moment  M  of  a  ship  is  a  known  function  of 
the  heeling  angle  6  (say  M  =  A  sin.  ad],  if  the  moment  M  due  to 
the  wind  produces  the  steady  heeling  angle  9,  and  the  energy 

ri  . 

A-  cos.  a9.d9  =  —  (1—  cos.  ad),  if  this 
a  a 

moment  were  suddenly  applied,  and  the  vessel  heeled  over  under  a 
gust  of  wind  from  o  to  019  the  work  done  is  0jM  or  0tA  sin.  a6  ;  and 


APPLIED    MECHANIC'S. 


261 


as  this  is  all  stored  as  -  (l  ~  cos.  «0j),  we  can  calculate  015  knowing 


The  work  is  easily  done 


6  from  61  A  sin.  a6  = 

graphically.  If  the 
curve  o  P  R  v  repre- 
sents the  righting 
moment  (o,  P)  corre- 
sponding to  the  in- 
clination 0  (or  TP), 
then  a  steady  mo- 
ment P  Q  produces  the 
heel  o  s  if  the  rectan- 
gular area  o  T  P  u  s  o 
is  equal  to  the  area 
of  OPRSO.  If  the 
law  is  M  =  A  sin.  40, 
the  vessel  cannot 

right  herself  if  6  is  more  than  45  degrees.  But  if  the  student 
studies  the  matter,  he  will  see  that  if  a  steady  moment  gets  the 
vessel  over  to  anything  approaching  45  degrees,  she  must  heel 
further  than  45  degrees.  We  are  neglecting  friction. 

There  is  the  same  moment  at  02  as  at  0  when  sin.  «02  =  sin.  aQ  ; 

and  if  02  sin.  aQ  =  -  (1  -  cos.  «02),  the  steady  wind  which  would 

keep  the  ship  at  0  will,  as  a  steady  gust  acting  from  0  =  0,  heel 
the  vessel  to  02  and  beyond.  Thus,  taking  the  above  example 
a  =  4, 


Fig.  107. 


sin. 


=  sin.  40,   or  402  =  n   —   40,  or  02  =  _  — 


cos.  40 


02  sin.  40  =  |  (1  -  cos.  402) 


cos.  (TT  —  40)  =  —  cos.  40, 


(j  -  0)  sin.  40  =  I  (1  +  cos.  40),         cos.  40  =  2  cos.8  20  -  1, 


(7T  -  49)  = 


sin.  40 
cos.  20 


sin.  20 
v  _  40  —  cot.  20  =  0. 
Calling  this  $  (0)=0, 1  find  that  0  =^11| 
degrees.  Hence,  if  a  wind  is  such  ihat 
it  would  maintain  a  steady  heel  of  llf 
degrees  ;  if  it  caught  the  vessel  sud- 
denly and  acted  steadily,  and  we  ne- 
glect friction,  the  heel  would  be  as 
much  as  33£  degrees;  and  if  the 
wind  still  continued  to  act  with  the 
same  moment,  the  righting  moment 
being  now  less  than  the  moment  due 
to  the  wind,  the  vessel  must  go  on  her 
beam- ends.  She  will  recover  from 
a  gust  of  wind  less  strcng  than  this. 


cot.  20, 


0deg. 

0 

0(0) 

18° 
16° 

•3142 

•2793 

0-5088 
0-4241 

12° 
11° 

•2095 
•1920 

•0576 
-  -1015 

262  APPLIED   MECHANICS. 

208.  So  long  as  a  constant  force  acts  it  produces  a  uniform 
acceleration  in  the  direction  in  which  it  acts.  We  may  experi- 
ment with  Attwood's  machine,  or  simply  use,  as  the  body  acted 
on,  a  small  carriage  running  very  freely  on  a  veiy  smooth  level 
table;  and  the  force  acting,  the  pull  in  a  string  passing  over  a 
pulley  on  the  edge  of  the  table,  and  having  weights  in  a 
scale-pan  at  its  end.  Fig.  168.  Here,  however,  friction  will 


be  greater  than  in  Attwood's  machine, 
We  can  illustrate  the  following  rule.  If 
a  force  of  2  Ibs.  acts  on  a  body  whose 
weight,  is  50  Ibs.  at  London  (the  50  Ibs.  includes  the 
weight  of  everything  which  is  set  in  motion,  so  that  if  we 
use  a  little  weight  of  2  Ibs.  for  the  purpose  of  exerting  the 
force,  we  must  remember  that  this  little  weight  of  2  Ibs.  is 
included  in  the  50  Ibs.  ;  we  may  take  48  Ibs.  as  the  body 
acted  upon,  and  the  pull  in  the  string,  which  is  less  than 
2  Ibs.,  as  the  force),  then  the  acceleration  or  increase  of  the 
velocity  every  second  is  equal  to  the  force  divided  by  the  whole 
mass  moved.  In  this  case  the  mass  is  50 -f  32'2,  or  1-553,  so 
that  we  have  2-s-l  -553,  or  1-223  feet  per  second  per  second  as 
the  acceleration.  Thus,  if  the  body  started  from  rest,  the 
velocity  would  be  1-223  x  5,  or  6-115  feet  per  second  at  the  end 
of  5  seconds.  And  now  comes  the  question,  how  far  will  the 
body  move  from  rest  in  five  seconds  ?  Evidently  its  average 
velocity  during  this  time  is  half  its  terminal  velocity,  or  3-058, 
so  that  5  x  3-058,  or  15-388  feet  is  the  distance.  It  is  evident 
that  to  get  the  space  passed  over  we  have  multiplied  half  the 
acceleration  by  the  square  of  the  time. 

I  do  not  suggest  that  the  apparatus  of  Fig.  168  is  the  most 
suitable  apparatus  for  use  in  the  laboratory ;  there  is  too 
much  friction,  and  it  seems  difficult  to  measure  the  velocity. 
Attwood's  machine  is  used  for  this  illustration,  and  is  described 


ALLIED    MECHANICS.  263 

in  Art.  193.  In  both  pieces  of  apparatus  corrections  must  be 
made  for  the  motion  of  wheels. 

When  a  body  falls  freely,  its  own  weight  is  acting  on  its 
own  mass.  For  instance,  say  the  weight  is  2  Ibs.,  then  the 
mass  is  2  -r  32-2,  and  weight  divided  by  mass  is  acceleration, 
which  we  find  to  be  in  every  case  32-2  feet  per  second  per 
second  at  London.  Anywhere  else  than  in  London  its  weight 

is  w  Ibs.,  and  its  mass  is,  as  before,  2  -*-  32-2  ;  and  consequently 

o 
its  acceleration  is  w  -*- But  we  call  this  acceleration  g, 

Oa  a 

and  hence  w  =  2  x     £         We  see  that  —   anywhere   is   the 

same  as  anywhere  else.  Keeping  to  London,  where  g  =  3  2 -2 
feet  per  second  per  second,  the  velocity  at  the  end  of  any 
number  of  seconds  is  32*2  multiplied  by  this  number ;  and  the 
space  fallen  through  in  any  number  of  seconds  is  half  32*2,  or 
16-1  multiplied  by  the  square  of  the  number  of  seconds.  You 
can  check  these  rules  by  the  rules  given  you  for  potential  and 
kinetic  energy,  and  you  will  find  them  quite  consistent  with 
one  another. 

209.  Momentum. — If  a  body's  weight  is  2  Ibs.,  its  mass  is 
2  -v-  3 2 '2,  or  -0621.  Now,  if  the  body  is  moving  with  a  velocity 
of  20  feet  per  second,  its  momentum  is  '0621  x  20,  or  1*242. 
If  this  momentum  is  created  or  destroyed  by  a  force  acting  for 
only  one  second,  the  force  must  be  1-242  Ibs. ;  if  it  is  created  or 
destroyed  by  a  force  acting  for  5  seconds,  the  force  must  be 
1-242  -r-  5,  or  -2485  Ib.  The  mass  of  a  body  multiplied  by  its 
velocity  represents  its  momentum.  Momentum  is  sometimes 
denned  as  the  quantity  of  motion  possessed  by  a  body.  It 
represents  the  constant  force  which,  acting  for  one  second, 
would  stop  the  body.  Suppose  a  certain  amount  of  momentum 
is  produced  by  a  force  of  1  Ib.  acting  on  a  body  for  one 
second,  the  same  amount  of  momentum  would  be  produced  by 
a  force  of  2  Ibs.  acting  for  half  a  second,  or  by  1,000  Ibs.  acting 
for  the  thousandth  of  a  second,  or  by  '001  Ib.  acting  for  1,000 
seconds. 

Example. — A  bullet  of  2  ounces,  or  -125  Ib.,  at  500  feet  per 
second,  directed  towards  the  centre  of  mass  of  a  body  of  200  Ibs. 
at  rest,  in  which  it  lodges  and  which  is  free  to  move ;  what  is 
the  velocity  after  the  impact  1  The  momentum  before  impact  is 

.1  OK  900-1  9>ri 

4±:  x  500.  The  momentum  after  impact  is  ---_-  x  the 
32*2  32*  2 


264  APPLIED    MECHANICS. 

•1 

*          1C!  _._ 

200-125 


-    j       i    •*         TT  .,  !     '125   x  500 

required    velocity.       Hence,    the    answer   is    — onn.io-   >    OI 


0'312  feet  per  second 

Exercise. — Chipping  hammer,  1|  Ibs.  ;  a  velocity  of  30  feet 
per  second  is  destroyed  in  the  one  live  hundredth  of  a  second. 
What  is  the  average  force  of  the  blow  ? 

Ans.,  700  Ibs.  (nearly). 

Example. — Onehundred  and  twenty  cubic  feet  of  water  leave 
the  rim  of  the  wheel  of  a  centrifugal  pump  every  minute;  its  com- 
ponent velocity  in  the  direction  of  motion  of  the  rim  is  25  feet 
per  second.  What  is  the  retarding  force  on  the  vanes  at  the 
rim  of  the  wheel? — Ans.,  Two  cubic  feet  of  water  per 

second  have  the  mass  -    *  9  2     ;  this,  multiplied  by  25,  gives 

the  momentum  lost  by  the  wheel  per  second,  which  is 
force,  and  amounts  to  97  Ibs.  If  the  velocity  of  the  rim 
is  30  feet  per  second,  the  work  done  per  second  is  30  x  97 
foot-pounds;  the  work  done  per  minute  is  30  x  97  x  GO; 
dividing  by  33,000,  we  find  5-56  horse-power.  Assuming  that 
there  is  no  force  at  the  inside  of  the  wheel,  the  water  entering 
radially  and  without  shock,  this  is  the  power  given  to  the 
water.  If  we  neglect  friction  inside  the  wheel  and  also  out- 
side it  retarding  its  motion,  this  is  the  total  power  given 
to  the  wheel  itself. 

What  is  the  work  done  per  pound  of  water  1  There  are 
2  x  62-4  Ibs.  of  water  per  second,  and  the  work  done  per  second 
is  30  x  97  foot-pounds,  so  that  the  work  done  per  pound  of  water 

30  x  97 
is pijT7>  or  23*3  foot-pounds,  or  energy  sufficient  to  lift  the 

water  to  a  height  of  23-3  feet. 

210.  If  the  velocity  of  a  body  has  been  produced  or  destroyed 
by  various  forces,  each  acting  for  a  certain  time,  multiply  each 
force  by  the  time  during  which  it  acted  (each  of  these  products 
is   called   an    impulse),    and   the  sum  must  be  equal  to  the 
whole  momentum  generated  or  destroyed.      When  we  know 
the  time  during  which  a  certain  force  has  acted  on  a  body 
giving  to  it   motion,  we  generally  determine  the  motion  by 
calculating  the  momentum  of  the  body.     When  we  know  the 
distance  through  which  a  force  has  acted  on  a  body  giving  to  it 
motion,  we  generally  first  find  the  kinetic  energy  of  the  body. 

211.  Knowing  the  force  F  acting  at  any  instant  on  the  mass 


APPLIED    MECHANICS. 


265 


if,  the  acceleration  a  is  P  -f-  M.  Thus,  suppose  the  following 
values  of  P  to  be  given ;  the  varying  force  acting  on  the  mass 
of  a  body  whose  weight  is  64-4  Ibs.  in  London.  In  engineers' 
units  its  mass  is  64-4-^32-2,  or  2.  Suppose  that  at  time  0  the 
body  has  a  velocity  v=30  feet  per  second. 


a, 

Time, 
in  seconds. 

in  pounds. 

in  feet  per 
second 
per  second. 

in  feet  per 
second. 

in  feet. 

0 

20 

10 

30 

0 

•1 

20 

10 

31 

305 

•2 

19 

9-5 

31-975 

6-199 

•3 

18 

9 

32-900 

9-443 

•4 

16 

8 

33-750 

12775 

•5 

14 

7 

34-500 

16-188 

•6 

11 

5-5 

35-125 

19-669 

•7 

8 

4 

35-600 

23-205 

•8 

5 

2-5 

35-925 

26-781 

•9 

2 

1-0 

36-100 

30-382 

1-0 

-1 

-0-5 

36-125 

33993 

1-1 

-3 

-1-5 

36-025 

37601 

1-2 

-4 

-2-0 

35-040 

41-163 

To  obtain  the  numbers  in  column  .4,  take  an  example. 
Suppose  we  know  that  when  t  =  f4  second  from  the  beginning 
v  =  33-750  feet  per  second.  Now,  in  the  next  Ofl  second,  the 
average  acceleration  is  approximately  J  (8  +  7),  or  7-5 ;  and  in 
the  time  O'l  the  actual  increase  of  velocity  is  7-5  x  0*1,  or  -75  ; 
and  this  is  what  we  add  to  33*750  to  get  34 '500  the  velocity  at 
the  end  of  the  little  interval. 

We  warn  beginners  that  there  is  no  easier  way  than  this, 
of  getting  several  very  important,  essential  ideas,  and  every 
number  of  such  a  table  ought  to  be  calculated.  Now,  notice 
that  s,  the  space  passed  over,  is  made  up  by  multiplying  an 
interval  of  time  O'l  by  the  average  velocity  during  that 
interval.  Thus,  at  t  =  -4,  s  =  12-775  feet  is  the  distance 
passed  through  since  the  time  t  =  0.  During  the  interval  from 
t  =  -4  to  t  —  -5  the  average  velocity  is  \  (33-75  +  34-50),  or 
34-125  feet  per  second,  and  the  space  passed  through  in  this 
0-1  second  is  34-125  x  (H,  or  3-4125  ;  and  this,  added  to  12-775, 
gives  s=16-1875,  or,  rejecting  the  last  figure,  s=  16-188  at  the 


266  APPLIED    MECHANICS. 

time£=0'5  seconds  from  the  beginning.  Note  that  in  approximate 
calculations  of  this  kind  we  cannot  be  certain  of  our  last  figures 
in  each  number.  It  will  assist  the  student  now  to  illustrate 
this  work  by  curves.  Plot  a  and  t  so  that  the  ordinate  of  B  c  D  is 

a  and  the  abscissa  is  t. 
Anyone  who  thinks  a 
little  must  see  that 
the  area  of  B  c  E  o  re- 
presents (to  scale)  the 
total  increase  of  velo- 
city at  the  time  repre- 
sented by  o  E.  Add 
this  on  (taking  care 
about  the  scale  of 
measurement)  to  30,  the  velocity  at  o,  and  we  have  the  true 
velocity  at  the  time  o  E.  Let  the  velocities  be  found  and 
plotted  as  o  G  H  u.  In  the  same  way  the  area  of  the  v  curve 
must  give  the  space  s  curve ;  that  is,  the  area  of  o  G  H  E  o 
represents  to  some  scale  the  space  s  passed  over  since  the  time 
o,  and  we  can  now  show  on  a  curve  s  at  every  instant.  This  is 
given  as  o  K  L. 

To  repeat :  E  c  represents  to  scale  the  acceleration  at  the 
time  o  E  j  E  H  represents  to  scale  the  velocity  at  the  time  o  E  ; 
it  is  the  area  of  o  B  c  E  o  added  to  the  velocity  at  the  time  o, 
care  being  taken  as  to- the  scale  of  the  diagrams ;  E  K  represents 
s  ;  it  is  measured  as  the  area  of  o  G  H  E  o. 

A  student  who  works  such  an.  exercise  as  this  carefully  is 
getting  all  sorts  of  valuable  notions,  not  merely  of  mechanics 
but  of  practical  mathematics.  Unfortunately,  twenty  academic 
exercises  can  be  worked  out  without  thought  or  trouble  to 
teacher  or  student,  and  by  the  rules  of  the  game  this  is 
sufficient  for  the  passing  of  examinations.  For  the  present, 
therefore,  my  advice  will  be  followed  by  a  few  earnest  students 
only — the  men  who  want  to  know,  the  men  who  are  not  merely 
in  search  of  examination  tips,  the  men  ~tyho  find  academic 
exercises  difficult  because  they  think  abofct  what  they  do. 
If  such  exercises  as  the  above  ever  become  obligatory  on  all 
examination  candidates,  of  course  their  academic  friends  will 
discover  ways  of  doing  such  exercises  without  the  trouble 
of  thinking  about  them. 

212.  When  the  resultant  force  F  acting  on  a  body  of  mass  M  is 
Qonstant,  the  acceleration  «=F  f  M  is  constant.  There  is  no  sue}) 


APPLIED    MECHANICS.  267 

case  in  Nature,  but  it  is  commonly  studied.  When  a  body  falls  in 
a  vacuum  in  an  ordinary  laboratory,  and  there  are  no  magnetic 
or  electric  or  frictional  effects,  we  may  for  all  practical  purposes 
assume  that  the  force,  the  weight  of  the  body  w,  is  constant. 
The  mass  is  w  -^  g,  and  the  constant  acceleration  is  g,  or  32-2 
feet  per  second  per  second  in  London.  If  the  student  treats  this 
case  in  a  table  like  that  of  p.  265,  or  by  means  of  curves  like 
Fig.  169,  he  will  see  that  v  =  g  t,  s  =  £  g  fi,  and  hence  that 
#3  —  2  g  s.  Or  if  v0  is  the  velocity  downwards  at  the  time 
t  -.—  o  and  s  is  the  vertical  height  through  which  the  body 
falls  from  t  =  o  to  any  other  time  t,  then  v  =  v0  +gt,  s=.v^t  + 


If  a  body  w  Ibs.  falls  without  friction  down  an  inclined 
plane,   making  an  angle  a  with  the  horizontal,  the  constant 

w 

force  is  w  sin.  a,  the  constant  acceleration  is  w  sin.  a  -=-  -  ,  or 

g  sin.  a. 

In  any  case  when  the  accelerations  is  constant,  v  =  v^ 


213.  I  have  described  the  units  of  measurement  employed 
practically,  not  merely  in  calculation  but  in  thought,  by 
English  speaking  people.  In  some  parts  of  our  work  we  find 
it  necessary  to  calculate  in  a  system  based  upon  other  units  — 
the  centimetre  as  the  unit  of  length,  the  inertia  or  mass  of  the 
amount  of  stuff  called  one  gramme  as  the  unit  of  mass  or 
inertia,  and  the  second  as  the  unit  of  time.  This  is  called  the 
C.G.S.  system.  Its  advantages  lie  in  its  being  used  by  scientific 
men  of  all  countries.  One  of  its  disadvantages  lies  in  this, 
that  all  answers  to  problems  must  be  multiplied  by  coefficients 
to  bring  them  into  practical  language  (see  p.  655). 

EXERCISES. 

1.  A  bullet  takes  2J  seconds  to  fall  to  the  bottom  of  a  well.     What 
are  the  depth  and  the  velocity  at  the  bottom  ?    Assume  no  resistance  of 
the  atmosphere. 

Ans.,  Depth  9  =  \g  (2*)2  ;  and  taking  g  =  32%  *  =  100-6  feet.     The 
velocity  is  2£y,  or  79  5  feet  per  second. 

2.  The  "bullet  of  (1)  leaves  the  hand  with  a  velocity  of  20  feet  per 
second  downwards.     What  are  the  two  answers  ? 

*  =  20  x  2£  +  \g  (2i)2  =  150-6   feet  ;     also    v  =  1\g  +  20  =  99'5 
feet  per  second. 

3.  The  bullet  of  (1)  leaves  the  hand  with  a  velocity  of  20  feet  per 
second  upwards.     What  are  the  two  answers  ? 

s  =  -  20  x  2£  +  \g  (2±)2  =  50  6  feet  ;    v  =  1\g  -  20  =  59'5  feet 
per  second, 


268  APPLIED   MECHANICS. 

4.  How  high  did  the  bullet  reach  in  (3)  ? 

Ans.,v*  ==  2ffh,  or  202  4-  2g  =-  h  =  6-21  feet. 
In  the  above  exercises  time  and  space  are  measured  from  leaving 
the  hand. 

5.  A  bullet  leaves  o,  a  point  on  the  ground,  with  an  upward  velocity 
of  300  feet  per  second.     Find  y,  the  vertical  height  of  it,  at  the  times 
t  =  0,  t  =  -1,  t  —  -2,  etc.,  seconds. 

6.  A  bullet  leaves  o  with  a  horizontal  velocity  of  400  feet  per  second, 
and  no  force  acts  upon  it  to  alter  its  horizontal  velocity.     Find  x,  its 
horizontal  distance  from  o,  at  the  times  0,  •!,  -2,  "3,  etc.,  seconds. 

7.  If  a  bullet  has  both  the  velocities  of  (5)  and  (6)  when  leaving  o, 
plot  its  positions  on  a  sheet  of  squared  paper  at  the  times  0,  •!,  '2,  etc., 
and  show  that  the  path  is  parabolic. 

8.  "With  different  horizontal  and  vertical  components,  but  the  same 
total  velocity  (500  feet  per  second),  let  the  bullet  of  (7)  leave  o  and  again 
plot  the  path.     Do  this  in  several  cases.     If  you  knew  a  little  mathe- 
matics, you  could  prove  that  an  angle  of  elevation  of  45  degrees  will 
give  the  greatest  range  on  a  horizontal  plane. 

9.  A  force  acts  on  a  body  of  8  ozs.  for  6-9125  minutes,  and  produces  a 
velocity  of   10  feet  per  second.    Find  the  force.     Express  it  in  dynes 
(see  p.  655). 

Ans.,  0-000372  lb.,  or  165'6  dynes. 

10.  How  far  will  a  lateral  force  of  1  oz.  move  100  Ibs.  on  a  smooth 
horizontal  plane  in  5  minutes  ?  Ans.,  900  feet. 

11.  In  Attwood's  machine,  where  the  weights  are  17  ozs.  and  16  ozs., 
find  the  acceleration  and  the  tension  of  the  cord. 

Ans.,  0'976  foot  per  second  per  second  ;  1  lb. 

12.  How  long  must  a  force  of  14  Ibs.  act  on  a  body  of  1,000  tons  to 
give  it  a  velocity  of  1  foot  per  second  ?  Ans.,  5,000  seconds. 

13.  A  rifle  5  feet  above  a  lake  discharges  a  bullet  horizontally,  which 
strikes  the  water  400  feet  away.     What  was  the  velocity  of  the  bullet  ? 

Ans.,  720  feet  per  second. 

14.  A  man  weighing  168  Ibs.  is  standing  on  the  floor  of  a  lift.     What 
force  does  he  exert  on  it  (1)  when  the  lift  is  stationary?  (2)  when  it  is 
falling  freely — that  is,  with  an  acceleration  of  32-2  feet  per  second  per 
second  ?  (3)  when  it  is  descending  with  an  acceleration  of  12  feet  per 
second  per  second  ?  (4)  if  it  is  ascending  with  the  latter  acceleration  ?    In 
each  case  indicate  by  a  figure  what  are  the  forces  which  act  on  the  man, 
and  give  the  resultant  force  in  the  direction  of  motion. 

Ans.,  (3)  105  Ibs.,  (4)  231  Ibs. 

15.  A  jet  of  water  having  a  sectional  area  of  12  square  inches,  and  a 
velocity  of  16  feet  per  second,  impinges  normally  on  a  fixed  plane  surface. 
What  is  the  mass  of  the  water  which  comes  on  the  plane  per  second  ?  What 
is  the  momentum  of   this  quantity  before  impact  ?    What  is  the  force 
on  the  plane  ?    If  the  jet  impinges  normally  on  a  plane  surface  which  has 
a  velocity  of  6  feet  per  second  in  the  direction  of  the  jet,  what  is  the 
velocity  of  the  water  relatively  to  the  surface  ?    And  what  is  the  force 
exerted  on  the  surface  ?     Find  the  amount  of  work  per  second  necessary 
to  maintain  the  jet,  and  the  work  done  by  it  per  second ;  and  find  the 
efficiency.      Ans.,  2-58  ;  41-28  ;  41-28  Ibs. ;  10  feet  per  second  ;  16-1  Ibs. ; 
332  ft.  Ibs. ;  96-7  ft.  Ibs. ;  -29. 

16.  The  head  of  a  steam-hammer  weighs  50  cwts. ;  steam  is  admitted 


APPLIED   MECHANICS. 


269 


on  the  under  side  for  lifting  only,  and  there  is  a  drop  of  5  feet.  What 
will  be  the  velocity  and  momentum  of  the  head  the  instant  before  the 
blow  is  given  if  the  fall  is  without  resistance  ?  If  the  time  during  which 
the  compression  of  the  iron  takes  place  be  -fo  second,  find  the  average 
force  of  the  blow.  Ans.y  17*95  feet  per  second  ;  3,121 ;  111-5  tons. 

17.  A  body  has  its  velocity  diminished  by  one-third.  By  how  much 
are  its  kinetic  energy  and  momentum  diminished  ?  If  this  diminution 
was  brought  about  by  a  certain  constant  force  acting  on  the  body  through 
a  distance  of  5  feet,  through  what  further  distance  would  this  force  have 
to  act  in  order  to  bring  the  body  to  rest?  If,  on  the  other  hand,  the 
diminution  of  velocity  had  taken  place  in  five  seconds,  what  additional 
time  would  be  required  to  bring  the  body  to  rest,  the  same  constant  force 
still  acting  ?  Ans.,  4  feet ;  10  seconds. 

213.  Force  may  be  defined  as  the  space-rate  at  which  work  is  done 
or  any  form  of  energy  converted  into  another,  or  it  may  be  defined 
as  the  time -rate  of  transference  of  momentum. 

We  would  advise  students  to  make  two  sets  of  curves  from 
Table  HA.,  p.  265.  The  first  is  given  in  Fig.  169.  The  second  set 


Distance. 

8 

Pressure. 
P 

Acceleration. 

a 

Area  = 
V2 

2 

V 

„ 

t 

0 

100 

32-5 

0 

•5 

100 

32-5 

•2481 

1 

100 

32-5 

32-5 

8-  6 

•2481 

1-5 

100 

32-5 

•1027 

2 

100 

32-5 

65-0 

11-4 

•3508 

2-5 

80 

19-5 

•082 

3 

66-7 

10-9 

84-5 

13-0 

•4328 

3-5 

57-1 

4-7 

•0758 

4 

50 

•1 

89-2 

13-36 

•5086 

4-5 

44-4 

-   3-5 

•0756 

5 

40 

-  6-3 

85-7 

13-1 

•5842 

5-5 

36-4 

-  8-7 

•0784 

6 

33-3 

-10-7 

77-0 

12-41 

•6626 

6-5 

30-8 

-12-3 

•0841 

7 

28-6 

-13-7 

64-7 

11:38 

•7467 

7-5 

26-7 

-14-9 

•0936 

8 

25 

-16-0 

49-8 

9-98 

•3403 

8-5 

23-5 

-17-0 

•1106 

9 

22-2 

-17-8 

32-8 

8-1 

•9509 

9-5 

21 

-18-6 

•1490 

10 
10-25 

20 
19-5 

-19-3) 
-19-6T-9-8 

14-2 

5-33 

•1217 

1-0999 

10-5 

19 

-19-9{ 

4-4 

2-97 

1-2216 

10-75 

18'6 

-20    j  ~4'9 

-0-5 

0 

.•1700 

1-3916 

1-3916 

270  APPLIED    MECHANICS. 

ought  to  have  *  for  the  horizontal  co-ordinates  or  abscissas.  "We  hav  3 
seen  that  it  was  a  most  instructive  problem,  when  given  the  o,  t 
curve,  to  find  the  v,  t  and  s,  t  curves.  Now,  suppose  we  have  the 
a,  s  curve  and  we  wish  to  find  the  v,  s  and  the  t,  s  curves.  As  a 
matter  of  fact,  we  have  already  worked  out  a  v,  s  curve  when  given 
an  a,  *  curve  in  Art.  211.  But  let  us  look  at  it  from  our  new  point 
of  view.  In  the  Bull  engine  of  Art.  205  the  force  causing  upward 
motion  of  w  is  A.p  —  w  ;  the  mass  is  \\/g,  and  so  the  acceleration 

is  a=  (A,p  —  w)  •*-  ^,  or  a  =  (p  —  -  j  -*-  — .      This  accelera- 

g  \          A/        Ay 

tion  is  given  in  Column  3  of  the  table  on  p.  269  ;  Column  1  shows  s, 
pjid  from  s  and  a  we  can  draw  our  s,  a  curve,  which  is  really  shown  as 
QHLIJMK  of  Fig.  165,  a  being  the  ordinate  measured  upwards  from 

K  M.    Now.  a  =  —  =  —  .  —  =  v  -,- ,  and  hence  o  .  ds  =  v  .  dv,  or 
dt       ds    dt          ds 

a  .  ds  =  *  v2,  or  twice  the  area  of  the  a ,  *  diagram  up  to  any 

place  is  the  square  of  the  velocity. 

We  have,  in  our  usual  way,  worked  the  integral  of  o  .  da 
numerically  in  the  table,  and  we  give  the  values  of  v. 

Note  that  as  v  =  -^,  or  tit  =  8s/v,  we  get  the  intervals  of  time 

by  dividing  the  intervals  of  space  by  the  average  velocity  during 
the  interval.  Thus  the  interval  of  time  from  *  =  4  to  *  =  5  feet 
is  1  foot  -T-  £  (13-36  +  13-1),  or  0-0756  second;  and  if  we  had 
already  determined  that  t  =  '5086  for  *  =  4,  we  know  that 
t  =  -5086  +  -0756  for  *  =  5.  In  this  way  the  numbers  of 
column  6  were  obtained. 

If  the  v,  s  diagram  is  given,  and  we  are  asked  to  find  the  a,  t 
diagram,  we  notice  that 

_dv dv     ds dv 

Therefore  o  is  represented  by  the  sub-normal  to  the  v  diagram.  We 
advise  no  student  to  use  the  measurement  of  the  normal  of  a  curve 
for  any  useful  purpose.  It  is  practically  too  inaccurate.  But 
from  a  table  of  the  values  of  v  and  s,  taken  from  a  curve  which 
will  correct  errors  of  observation,  values  of  8v/5s,  and  therefore  of  v 
8v/$s  or  a,  may  be  calculated  with  no  great  error.  The  values  of  t 
may  be  obtained  as  in  the  last  example,  and  the  whole  motion  is 
known. 

•  Of  course  if  of  the  variables  t,  *,  v,  *  any  one  is  known  as  an 
algebraic  function  of  the  other,  it  is  an  exercise  in  the  calculus  to 
find  an^  or  all  the  others  in  terms  of  any  one ;  as  also  the  work 
done,  or  the  kinetic  energy,  and  other  things. 

Example. — A  body  has  reached  the  earth  at  London  from  space, 
no  other  force  than  the  earth's  gravitation  having  acted  upon  it ; 
what  is  its  velocity  ?  If  r0  feet  is  the  distance  of  London  from 
the  earth's  centre,  and  at  any  other  place  reached  by  the  body  if 

the  distance  is  r,  then  g,  the  acceleration  at  r,  is  a0  ^.  Let  2^0r,s 
be  called  P,  then  H 


APPLIED    MECHANICS.  271 

,-»£«.*£—,--*£.  ...ID. 

etr    <Pr          If    dr  ,  /dry  ,  V 

2 


Let  —  =  0  when  r  =  GO.  and  we  need  to  add  no  constant.     Hence 
dt 

^  =  _JL?  and  therefore  r*dr  =  —  b  .  dt.     We  take  the  -  sign 

as   we  imagine  the  body  falling.     Integrating  again,  we  have 

§  r  i  =  —  bt  +  c.     Let  t  =  0  when  r  =  r0,  so  that  all  our  times 
before  reaching  the  earth  will  he  negative. 


___  ,  orr  =    r08-—     J  ----  (4). 


It  is  (2)  that  we  asked  for. 

Note  that  a  body  of  w  Ibs.,  when  it  reaches  r  from  space,  has 

w  J2 
a  kinetic  energy  ^  ~  — ,  or  w  P2/P0.    We  are  therefore  prompted  to 

study  the  problem  from  the  energy  rather  than  the  momentum 
point  of  view.  Imagine  a  body  of  1  Ib.  leaving  the  earth  (say  at 
London). 

At  the  distance  r  feet  from  the  earth's  centre  the  weight  of  the 

r  2 
body  is  ~  if  r0  is  the  distance  of  the  earth's  surface  from  the 

centre.      The  potential  energy  here  being  v,  and  being  v  +  Sv 

r  2                                          r  " 
at  r  +  Sr,   Sv  =  weight  x  Sr  =  4  .  Sr.      Hence  v  = ^-  + 

const. 

If  v0  is  the  potential  energy  of  1  Ib.  (we  are  really  in  all  this 
neglecting  the  fact  that  the  earth  moves  relatively  to  the  object)  at 

T   " 

the  earth's  surface,  v0  = -f-  const.,  so  that  our  constant  is 

r2     ^ 
v0  +  r0.     Hence  v  =  -  ^-  +  v0  +  r0. 

The  potential  energy  when  r  =  oo  is  V0  -f-  »V  Hence  we  have 
the  easily  remembered  fact : — A  body  of  w  Ibs.  lifted  infinitely  high 
from  the  earth's  surface  would  receive  a  store  of  w  r0  foot-pounds  of 
energy  if  r0  is  the  radius  of  the  earth  in  feet ;  and  if  we  imagine 
all  tliis  converted  into  kinetic  energy,  we  see  that  the  velocity  of 
the  body  coming  from  rest  in  space  would  be  V  2gnr(t  when  gQ  =* 
82'2.  If  g  is  the  gravitational  acceleration  at  the  place  r,  the 


272  APPLIED    MECHANICS. 

body's  speed  on  reaching  this  place  would  be  A/  Igr  ;  and  if  we 
remember  that  g  =  0V0'Y;'2>  calculation  is  easy,     v  =  V  fytf-ffr. 
If  we  write  v  =  —  -57  where  t  is  time,  and  if  V  2y0r02  is  called 

b,  then  —  r\  .  dr  =  b  .  dty  or  —  §  r  i  =  bt  -f  const.  Thus,  if  we 
count  time  from  the  time  of  reaching  the  earth's  surface,  so 
that  it  is  always  negative,  let  t  =  0  when  r  r0.  The  constant 

is  -  §r0  J,  and  so  t  =  ^  (r0i  -  ri). 

Also,  as  vr\  =  b,  ri  =  T3,  and  hence  t  =  —^  (r05  --  ),  or 

o  o   \  v  i 


125.  Force  F  is  rate  of  change  of  momentum  M.  If  force  F  acts 
for  time  5t,  it  increases  the  momentum  of  a  body  by  the  amount  8  M  ; 

(?M        r*T 

so  we  can  say  either  that  F  =  -T-,  or  that  I    p  .  dt  =  whole  gain 

J  o 

of  momentum  '  if  T  is  the  time  during  which  the  gain  occurs. 
Of  course  if  F  is  constant,  we  have  F  T  =  gain  in  momentum  ; 
or  the  whole  momentum  gained  is  the  force  multiplied  by  the 
time.  But  if  F  varies,  we  can  only  say  that  the  whole  gain  of 
momentum,  divided  by  the  time,  is  the  average  force  during  the 
time.  Here  we  have  a  time  average. 

.  work  done 

Space  average  value  of  force  =  w^ole  distance  '  tune  avera£e 

:        „  „  momentum  gained 

value  of  force  =  -  =—  ^  —  ^  -  . 
whole  time 

Continually  in  dynamics  we  are  considering  the  two  great  ideas 
of  energy  and  momentum.  On  any  system  if  a  force  acts  from 
outside  bodies  it  gives  energy,  and  it  gives  momentum.  If  no 
force  acts  from  the  outside,  the  momentum  and  moment  of 
momentum  remain  unaltered  ;  and  the  total  energy  would  remain 
unaltered  were  it  not  that  other  forms  of  energy  become  changed 
to  heat,  and  a  system  loses  heat  by  radiation.  If  the  earth-moon 
system  were  alone  in  space,  we  have  to  consider  that  its  moment  of 
momentum  remains  constant,  whereas  its  total  store  of  mechanical 
energy  is  diminishing.  Professor  Purser  pointed  out  that  this  idea 
gives  us  the  past  and  future  history  of  the  earth-moon  system,  and 
Professor  Darwin  has  worked  it  out  for  us. 

One  of  the  most  instructive  of  laboratory  experiments  is  that 
in  which  two  bodies,  A  and  B,  are  suspended  so  that  they  may 
collide,  their  motions  before  and  after  collision  being  measurable. 
The  whole  momentum  after  impact  is  the  same  as  before  impact  ; 
and  it  is  very  interesting  to  notice  that  when  A  strikes  B  at  rest 
the  extent  of  swing  of  the  two  in  combination  after  impact  is  not 
affected  by  the  initial  rebounding  and  chattering  and  the  many 
little  circumstances  which  cause  the  energy  after  impact  to  be  quite 
different  from  what  it  was  before.  Men  who  do  not  make 
experiments  of  this  kind  have  no  clear  notions  of  dynamical 


APPLIED   MECHANICS. 


273 


phenomena,  unless  they  are  very  exceptional  (that  is,  men  of 
genius). 

When  the  momentum  M,  as  of  a  pile-driver,  is  destroyed  in  the 
time  T,  M/T  is  the  time  average  force  of  the  blow ;  we  have  a 
perfectly  definite  idea  of  what  is  meant.  But  when  we  are  told 
that  the  whole  energy  of  the  falling  pile-driver,  divided  by  the 
distance  through  which  the  pile  is  forced  into  the  ground,  gives  us 
the  resistance  of  the  ground  to  the  pile,  we  get  a  misleading 
academic  statement.  The  blow  is  a  complicated  phenomenon ;  and 
even  in  the  much  simpler  case  of  the  history  of  the  collision  of  two 
billiard-balls  we  are  only  now  beginning  to  see  how  and  where  the 
total  energy  is  converted  into  heat  (see  Art.  404). 

Indicator  diagrams  of  engines  are  space  diagrams  of  force. 
Their  average  heights  enable  the  work  done  to  "be  calculated. 


t 

p 

—   a 

-  a&*  = 
-  8v 

V 

8s 
=  vSt 

8 

F.S* 

0 

0 

0 

0 

8-8 

•1 

49 

•141 

•0014 

8-756 

1-760 

86 

•2 

97-9 

•281 

•0211 

8-772 

1-76 

•4 

195-8 

•562 

8-687 

3-475 

680 

•G 

293-8 

•844 

•337 

5'24 

•8 

391-7 

1-125 

8-350 

3-334 

1,310 

1-0 

489-6 

1-406 

•562 

8-53 

1-2 

587-5 

1-687 

7'787. 

3-115 

1,830 

1-4 

678-4 

1-968 

•787 

11-69 

1-6 

783-4 

2-250 

7-000 

2-800 

2,194 

1-8 

881-3 

2-540 

1-016 

14-49 

2-0 

979-2 

2-812 

5-984 

2-393 

2,350 

2-2 

1077-1 

3-193 

1-277 

16-93 

2-4 

1175-0 

3-374 

4-707 

1-883 

2,220 

2-5 

1224-0 

3-519 

2-6 

1175-0 

3-374 

1-350 

18-85 

2-8 

1077-1 

3-193 

3-357 

1-353 

1,460 

3-0 

979-2 

2-812 

1-125 

20-22 

3-2 

881-3 

2-540 

2-233 

$93 

790 

3'4 

783-4 

2-250 

•900 

21-16 

3-6 

678-4 

1-968 

1-333 

•533 

362 

3-8 

587-5 

1-687 

•675 

21-74 

4-0 

489-6 

1-406 

•658 

•263 

129 

4-2 

391-7 

1-125 

•450 

22-04 

4-4 

293-8 

•844 

•208 

•083 

24  -.5 

4-6 

195-8 

•562 

•225 

-•017 

22-16 

4-8 

97-9 

•281 

•028 

•055 

-•007 

-  '7 

5-0 

0 

0 

0 

-•045 

22-19 

8-844 

21-892 

13,635 

APPLIED    MECHANICS. 


When  a  mass  is  vibrating  at  the  end  of  a  spiral  spring,  the 
space  diagram  of  the  force  exerted  by  the  spring  upon  the  body  in 
«  straight  line.  The  space  average  of  the  force  (neglecting  the 
weight  of  the  body)  between  the  end  of  a  swing  and  the  mid 
position  is  half  what  the  force  was  at  the  end  of  the  swing; 
whereas  the  time  average  of  the  force  in  this  interval  is  the 

2 
fraction  -  of  the  force  at  the  end. 

7T 

216.  Example. — A  body  of  5  tons  moving  at  6  miles  per  hour ; 
what  are  its  momentum  and  kinetic  energy?  Find  the  time 
average  of  the  force  which  will  stop  it  in  5  seconds. — Ans.,  The 
mass  is  347'8;  momentum,  T97  x  105;  the  kinetic  energy  is 
13466-8  ;  time  average  of  force  =  612  Ibs. 

If  the  force  increases  for  2^  seconds,  and  then  diminishes  again, 
in  both  cases  uniformly  with  time,  draw  curves  showing  the 
velocity  with  time,  and  also  with  distance ;  also  of  force  with 
distance.  What  is  the  space  average  of  the  stopping  force  ? 

Draw  a  curve  showing  acceleration  a  and  t  (a  is  negative) ;  the 
integral  of  this  shows  v  and  t,  the  ordinate  at  t  =  o  being  6  miles 
per  hour,  or  8 -8  feet  per  second.  The  integral  of  the  t>,  t  curve 
shows  s  and  t  where  *  is  distance  from  where  the  force  began  to 
act.  Now  a  x  mass  represents  force.  Hence  the  values  of  a 
represent  force  to  some  scale.  Now  plot  a  new  curve  showing 
force  and  s.  The  whole  area  of  it  is  the  kinetic  energy.  The 
average  height  of  it  is  the  space  average  of  the  force. — Ans.,  621  Ibs. 

We  have  performed  the  integrations  numerically,  and  have 
shown  the  results  in  the  table  on  p.  273,  and  we  have  shown  the 
various  curves  in  Fig.  169 A. 


Fig.  100A. 

O  A  A  A  if  shows  a  and  t,  t  parallel  to  o  D. 
o  c  B  E  K  shows  v  and  t,  t  parallel  to  o  D. 
o  i  i  J  B  shows  *  and  t,  t  parallel  to  o  D. 
O  H  H  H  o  shows  F  and  s,  s  parallel  to  o  D. 
o  c  v  v  o  shows  v  and  s,  s  parallel  to  o  D. 


275 
CHAPTER  XII. 

MATERIALS    USED  IN   CONSTRUCTION. 

217.  Mere  reading  will  give  to  no  student  a  knowledge  of 
the  properties  of  materials.     I  insist  on  the  necessity  for  work 
in  a  pattern-shop  and  a  fitting-shop  and  forge ;  and  setting  work 
in  machine  tools.     Workshops  at  a  college  or  school  are  not 
intended  for  the  teaching  of  trades,  which  can  only  be  done  in 
real  shops  beside  real  workmen.     A  student  learns  facts  about 
materials  which  are  necessary  for  his  study  of  mechanics  ;  it  is 
a  secondary  matter  that  he  also   acquires  some  skill  which 
enables  him  to  learn  his  trade  quickly  in  a  real  shop  afterwards. 
Town  boys  buy  their  toys  and  never  come  in  contact  with 
nature ;  country  boys  are  always  making  things  and  learning 
much  besides  the  properties  of  materials. 

218.  Stone. — The  rocks  which  have  once  been  melted,  and 
have  cooled  slowly,  are  usually  hard,  compact,  strong,  and  dur- 
able.    They  are  most  easily  worked  when  regard  is  paid  to  the 
fact  that  they  naturally  divide  up  into  certain  regular  shapes. 
They  are  all  more  or  less  crystalline  in  texture.     Stratified 
rocks  are  those  which  have  been  deposited  at  the  bottom  of  a 
sea  or  river;    they  are  often   easily  divided   in   a  direction 
parallel  to  the  layers  of  which  they  are  built  up,  but  sometimes 
there  are  lines  of  easy  cleavage  in  other  directions.     These 
rocks  vary  very  much  in  appearance,  according  to  the  method 
of  their  formation,  and  to  the  heat  and  pressure  to  which  they 
have  been  subjected,  sometimes  being  very  Crystalline,  strong, 
and  durable,  like  marble.    Slaty  rocks  may  be  hard  and  durable, 
or  soft  and  perishable ;  they  are  not  much  used  in  construction, 
except  as  roof  covering.     Sandstones  are  hardened  sand   of 
very  different  degrees  of  compactness,  porosity,  strength,  and 
durability.     There  are  limestones  whose  particles  seem  to  form 
one  continuous  mass,  and  which,  when  they  have  been  subjected 
to  great  heat  and  pressure,  become  marbles/  there  are  also 
limestones  which   are  composed  of    distinct  grains  cemented 
together,    and   which  may  vary  very  much    in  compactness, 
strength,  and  durability.    Besides  these  there  are  conglomerates, 
in  which  fragments  of   older  rocks  are  imbedded.     A  little 
knowledge  of  geology  is  necessary  in  order  to  understand  the 
properties  of  rocks.     Stones  are  preserved  by  coating  them 


276  APPLIED  MECHANICS. 

with  some  material  such  as  coal-tar,  various  kinds  of  oil  and 
paint,  or  soluble  glass,  which  fills  their  pores  and  prevents  the 
entrance  of  moisture.  An  artificial  stone,  which  can  be  made 
in  blocks  of  any  required  size  and  shape,  is  obtained  by  turning 
out  of  moulds  and  afterwards  saturating  with  a  solution  of 
chloride  of  calcium,  a  mixture  of  clean  sharp  sand  and  silicate 
of  soda.  The  chloride  of  calcium  and  silicate  of  soda  produce 
silicate  of  lime,  which  cements  the  sand  together,  and  thus 
gradually  consolidates  the  whole  mass. 

219.  Bricks. — Bricks  are  made  of  tempered  (that  is,  freed 
from  pebbles,  saturated  with  water,  and  well  ground  and  mixed) 
clay,  moulded,  dried  gently,  then  raised  to  and  kept  at  a  white 
heat  in  a  kiln  for  some  days,  and  cooled  gradually.     Sand  in 
the  clay  prevents  too  much  contraction  and  helps  vitrification. 
Bricks  should  have  plane  parallel  surfaces  and  sharp  right- 
angled  edges,  should  give  a  clear  ringing  sound  when  struck, 
should  be  compact,  uniform,  and  somewhat  glassy  when  broken, 
free  from  cracks,  and  able  to  absorb  not  more  than  one-fifteenth 
of  their  weight  of  water.     They  ought  to  require  at  least  half 
a  ton  per  square  inch  to  crush  them.     The  published  tests  are 
sometimes  much  more  than  a  ton  per  square  inch.     Probably 
half  the  published  strengths  are  the  true  strengths  of  bricks  or 
of  brickwork.     The  standard  brick  is  8J  x  4$  x  2|  inches. 
The  average  weight  of  brickwork  is  116  Ibs.  per  cubic  foot. 
A  bricklayer  lays  from  100  to  200  bricks  per  hour. 

220.  Limestone,  when  burnt  in  kilns,  gives  off  carbonic 
acid.    If  pure  it  forms  quick-lime,  which  combines  readily  with 
water,  becoming  larger  in  volume.     Mixed  with  clean  sand  this 
forms  mortar,  which,  in  the  course  of  time,  hardens  by  losing  its 
water  and  combining  with  carbonic  acid  from  the  air.     If  the 
burnt  limestone  were  not  pure,  but  contained  certain  kinds  of 
clayey  materials,  iron,  &c.,  it  would  not  combine  with  much 
water,  but  when  ground  up  fine,  water  enables  its  particles  to 
combine  chemically  with  one  another,  forming  compound  sili- 
cates with  greater  or  less  rapidity,  depending  on  its  composition. 
Such  cement  first  sets,  acquiring  a  large  degree  of  firmness,  and 
then  more  slowly  and  without  much  expansion  becomes  as  hard 
as  many  limestones.     These  natural  hydraulic  limestones  are 
not  much  used  now.     Nearly  pure  limestone  or  chalk  is  mixed 
with  about  one-third  of  its  volume  of  blue  clay  to  produce — 
when  ground  and  mixed  in  plenty  of  water,  then  drained  and 
dried,  then  burnt  to  incipient  vitrification  and  ground  up  again 


APPLIED    MECHANICS.  277 

very  finely  indeed — an  artificial  cement,  which  is  equal,  if  not 
superior,  to  the  natural  cement.  This  is  the  Portland  cement 
now  in  use.  Fineness  in  the  particles  is  exceedingly  important. 
Sand  in  mortar  saves  expense,  and  prevents  the  cracking  ot 
the  mortar  in  drying;  coarse  sand  seems  better  than  fine. 
Two  measures  of  sand  to  1  of  slaked  lime  or  3  to  7  of  sand  to 
1  of  cement  are  the  average  allowances,  but  every  person  who 
uses  mortar  ought  to  test  a  particular  lime  or  cement  to  see 
how  much  sand  it  will  bear  to  have  mixed  with  it.  Concrete 
is  a  mixture  of  6  of  gravel  or  broken  stones  and  1  of  cement. 

From  the  time  of  setting  the  tensile  strength  of  cement 
increases  at  first  rapidly  and  gradually  more  slowly.  The 
French  standard  is  280  Ibs.  per  square  inch  at  the  end  of  seven 
days,  500  Ibs.  in  28  days,  640  Ibs.  in  84  days. 

The  initial  strengths  of  neat  cement,  1  of  cement  to  2  of 
sand,  1  of  cement  to  5  of  sand  being  about  1  :  l  :  f- ;  the 
gains  of  strength  in  a  year  are  about  as  5  :  4  :  3 .  Using  too 
much  water  weakens  cement ;  water  about  £  to  J  of  the  weight 
of  the  cement  is  found  to  give  the  best  results  in  testing.  The 
true  crushing  stress  of  cement  is  probably  about  six  times  the 
tensile  stress, 

221.  Timber. — A  tree  is  made  up  of  a  great  number  of  little 
tubes  and  cells  arranged  roughly  in  concentric  circles,  one  circle 
for  each  year  of  growth,  because  the  sap  which  circulates  out- 
side is  checked  every  winter.  The  process  of  seasoning  consists 
in  uniformly  drying  the  timber.  As  each  little  portion  dries, 
it  contracts  and  becomes  more  rigid,  and  it  contracts  much 
more  readily  in  the  direction  of  the  circular  arrangement  of  the 
tubes  than  it  does  towards  the  centre  of  the  tree,  and  least 
easily  in  the  direction  along  the  tree.  It  is  obvious,  then,  that 
if  the  tree  is  dried  whole,  there  will  be  a  tendency  to  splitting 
radially.  If  the  tree  is  cut  up  before  drying  we  can  tell  the 
way  in  which  the  planks  will  warp  if  we 'remember  the  above 
facts. 

Firwoods  are  easily  wrought,  and  possess  straightness  in 
fibre  and  great  resistance  to  direct  pull  and  transverse  load, 
and  are  largely  used  because  of  their  cheapness.  They  differ 
greatly  in  strength,  but  their  weak  point  is  their  inability  to 
resist  shearing.  The  best  of  these  is  the  red  pine  or  Memel 
timber  from  Russia,  which  can  be  had  in  large  scantlings,  and 
thus  used  without  trussing.  The  white  fir  or  Norway  spruce 
is  suitable  for  planking  and  light  framing,  and  is  imported 


278  APPLIED    MECHANICS. 

from  Christiania  in  "  deals,"  "  battens,"  and  "  planks."  Larch 
is  a  very  strong  timber,  hard  to  work,  and  has  a  tendency  to 
warp  in  drying,  and  is  therefore  not  suitable  for  framing,  but 
is  largely  used  for  railway-sleepers  and  fences,  because  of  its 
durability  when  exposed  to  the  weather.  Cedar  lasts  long  in 
roofs,  but  is  deficient  in  strength. 

The  English  oak  is  the  strongest  and  most  durable  of  all 
woods  grown  in  temperate  climates,  but  is  very  slow-growing 
and  expensive.  Its  great  durability  when  exposed  to  the 
weather  seems  to  be  due  to  the  presence  of  gallic  acid,  which, 
however,  in  any  wood  corrodes  iron  fastenings  ;  trenails  or 
wooden  spikes  should  be  used  instead.  Teak,  which  is  grown 
in  the  East,  is  the  finest  of  all  woods  for  the  engineer.  It  is 
very  uniform  and  compact  in  texture,  and  contains  an  oily 
matter  which  contributes  greatly  to  its  durability.  It  is  used 
specially  in  ship-building  and  railway  carriages.  Mahogany  is 
unsuitable  for  exposure  to  the  weather,  but  it  has  a  fine  appear- 
ance and  is  not  likely  to  warp  much  in  drying.  It  is  chiefly 
used  for  furniture  and  ornamental  purposes,  and  to  some 
extent  in  pattern-making.  Ash  is  noted  for  its  toughness  and 
flexibility,  and  a  capability  of  resisting  sudden  stresses  of  all 
kinds,  which  make  it  specially  adapted  for  handles  of  tools  and 
shafts  of  carriages.  It  is  very  durable  when  kept  dry.  It  is 
not  obtainable  in  large  scantlings,  and  is  sometimes  very 
difficult  to  work.  Elm  is  valuable  for  its  durability  when 
constantly  wet,  which  makes  it  useful  for  piles  or  foundations 
under  water.  It  is  noted  for  its  toughness,  though  inferior  to 
oak  in  this  respect,  as  also  in  its  strength  and  stiffness.  It  is 
very  liable  to  warp.  Beech  is  smooth  and  close  in  its  grain. 
It  is  nearly  as  strong  as  oak,  but  is  durable  only  when  kept 
either  very  dry  or  constantly  wet.  It  is  very  tough,  but  not 
so  stiff  as  oak.  (See  also  Table  VII.) 

The  best  time  for  felling  timber  is  when  the  tree  has 
reached  its  maturity,  and  in  autumn  when  the  sap  is  not 
circulating.  We  want  to  have  as  little  sap  in  the  timber  as 
possible,  and  in  order  to  harden  the  sapwood,  some  foresters 
are  of  opinion  that  the  bark  should  be  taken  off  in  the  spring 
before  felling.  After  timber  is  felled,  it  is  well  to  square  it  by 
taking  off  the  outer  slabs. 

Timber  is  for  the  most  part  dried  by  putting  it  into  hot- 
air  chambers,  from  ono  to  ten  weeks  according  to  the  thickness. 
Even  when  kept  quite  dry,  ventilation  is  necessary  to  prevent 


APPLIED    MECHANICS.  279 

dry  rot.  The  circumstances  least  favourable  to  the  durability 
of  timber  are  alternate  wetting  and  drying,  as  in  the  case  of 
timber  between  high  and  low  water  mark,  whereas  good 
seasoning  and  ventilation  are  most  favourable  conditions.  The 
most  effective  means  adopted  for  preserving  timber  is  by 
saturating  it  with  a  black  oily  liquid  called  creosote.  The 
timber  is  placed  in  an  air-tight  vessel,  and  the  air  and  moisture 
extracted  from  its  pores  as  far  as  possible.  The  warm  creosote 
is  then  forced  into  these  pores  at  a  pressure  of  170  Ibs.  per 
square  inch.  In  this  way  timber  may  be  made  to  absorb  from 
a  tenth  to  a  twelfth  of  its  weight  of  creosote. 

I  shall  give  few  numbers  here  for  the  strength  of  timber. 
Tests  of  small  specimens  are  not  to  be  relied  upon.  The  time 
of  felling,  the  duration  of  drying,  the  part  of  the  tree  from 
which  the  specimen  is  cut,  and  many  other  circumstances  affect 
the  strength.  A  beam  will  sometimes  break  with  a  long  con- 
tinued load  only  about  half  of  what  will  fracture  it  in  the 
ordinary  way.  The  ultimate  shearing  stress  along  the  grain  of 
ash  is  about  600,  oak  850,  pine  and  spruce  300  ibs.  per  square 
inch.  For  bending,  the  average  value  of  f  in  (1)  of  Art.  341 
seems  to  be  for  spruce  5,000,  yellow  pine  7,300,  oak  6,000,  and 
white  pine  5,000  Ibs.  per  square  inch ;  their  Young's  moduli 
being  respectively  14  x  105,  17  x  105,  13  x  105,  and  11  x  105  Ibs. 
per  square  inch.  The  crushing  strength  of  timber  may  roughly 
be  taken  as  from  4,000  to  3,000  Ibs.  per  square  inch,  and  its 
tensile  strength  as  about  10,000  Ibs.  per  square  inch.* 

222.  Glass. — Glass  is  a  combined  silicate  of  potassium  or 
sodium,  or  both,  with  silicates  of  calcium,  aluminium,  iron,  lead, 
and  other  chemical  substances.  Certain  mixtures  of  flint  and 
chemicals  are  melted  in  crucibles,  formed  when  hot  into  the 
required  shapes,  and  cooled  as  slowly  as  possible.  This  may  be 
called  the  devitrification  of  glass  by  slow  cooling,  giving  rise 
to  crystallisation.  The  more  slowly  and  more  uniformly  the 
cooling  is  effected,  the  more  likely  is  it  that  the  glass  will  be 
without  internal  strains.  When  glass  is  suddenly  cooled,  as 
when  a  melted  drop  falls  into  water,  the  outside  is  suddenly 
contracted,  becomes  hard  and  brittle,  and  there  are  such 
internal  strains  that  if  the  tapering  part  be  broken  or  scratched 
at  the  point  the  whole  drop  crumbles  into  a  state  of  dust.  A 
blow  or  scratch  on  the  thick  part  produces  no  such  effect. 
Heating  and  gradual  cooling  destroy  this  property.  Many 
peculiarities  in  the  behaviour  of  metals  when  heated  and  copied 

*  S|ee  Appemlig. 


280  APPLIED   MECHANICS. 

seem  to  be  caricatured  in  glass,  possibly  because  they  are  due 
to  the  fact  that  all  the  portions  of  matter  which  are  about  to 
form  one  crystal  must  be  at  the  same  temperature,  and  when 
the  substance  is  a  bad  conductor  of  heat  there  is  great  variation 
in  temperature.  Pure  metals  are  good  conductors,  but  the 
admixture  of  small  quantities  of  carbon  and  of  gases  hurts 
their  conductivity.  Toughened  glass  is  the  name  wrongly 
given  to  the  hardened  glass  produced  by  plunging  glass,  in  a 
nearly  melting  state,  into  a  rather  hot  oily  bath.  This  glass 
is  somewhat  in  the  condition  of  the  glass  in  a  Rupert's  drop. 
It  is  so  hard  that  it  is  difficult  to  cut  it  with  a  diamond,  but  if 
the  diamond  cuts  too  deep  the  whole  mass  breaks  up  into  little 
pieces.  Objects  made  of  it  may  be  thrown  violently  on  the 
floor  without  breaking. 

223.  Cast  Iron. — Certain  chemical  changes  occur  when  tho 
ores  of  iron,  generally  oxides,  are  smelted  with  coke ;  the  iron 
ceases  to  be  in  combination  with  the  oxygen,  and  appears  in 
metallic  form,  associated,  however,  with  carbon  derived  from 
the  fuel.  There  are  usually  other  impurities  besides,  derived 
from  the  same  source  or  from  the  ores.  When  the  carbon  is 
all  combined  with  the  iron  the  cast  iron  is  "white,"  and  is 
very  hard  and  brittle.  When  only  a  little  is  combined,  and 
most  of  its  particles  crystallise  separately,  the  cast  iron  is  grey 
in  colour ;  it  is  weaker  and  more  fusible.  Using  the  common 
names  for  the  different  varieties,  No.  1  is  darkest  in  colour,  and 
from  No.  4  to  No.  1  there  is  a  gradual  darkening  in  colour.  In 
the  cupola  of  the  foundry  a  little  purification  is  effected,  and  it  is 
found  that  the  composition  of  a  casting  is  from  97  to  95  per 
cent,  of  iron,  the  remainder  being  nearly  pure  carbon,  often 
very  largely  in  the  combined  form  owing  to  the  elimination  of 
one  of  the  impurities — namely,  silicon.  Nos.  2,  3,  4  are  com- 
monly used  in  the  foundry,  mixtures  being  made  of  them  in 
various  proportions  according  to  circumstances.  A  greater 
proportion  of  No.  3  or  No.  4  gives  greater  strength,  whereas  a 
greater  proportion  of  No.  1  gives  greater  fluidity  and  a  better 
power  of  expanding  at  the  moment  when  the  metal  solidifies, 
so  that  the  sharp  corners  of  the  mould  are  better  filled. 
Higher  numbers  than  4,  as  8,  7,  6,  and  5 — the  white  varieties 
— are  seldom  used  in  the  foundry,  but  they  may  be  converted 
into  grey  varieties  by  cooling  from  a  very  high  temperature 
at  a  slow  rate,  but  much  more  easily  and  immediately  by 
the  addition  of  certain  brands  of  cast  iron  containing  specia] 


APPLIED    MECHANICS. 


281 


impurities  such  as  "  siliconeisen  "  or  "glazed  pig."  A  special 
degree  of  fluidity  and  resistance  to  action  of  acids  is  conferred 
upon  cast  iron  by  the  presence  of  a  little  phosphorus,  but  this 
impurity  renders  iron  fragile  at  low  temperatures,  just  as  the 
presence  of  much  silicon  will  render  it  weak  and  liable  to 
fracture  from  shock.  To  soften  a  hard  casting,  it  is  heated  in 
a  mixture  of  bone-ash  and  coal-dust  or  sand,  and  allowed  to 
cool  there  slowly. 

The  density  of  oast  iron  varies  from  6 '8  in  dark  grey 
foundry  iron  to  7 '6  in  white  iron.  Of  late  years  cast  iron  has 
greatly  improved  in  strength ;  this  is  due  probably  to  our 
better  knowledge.  Contracts  are  sometimes  undertaken  to 
deliver  iron  of  nearly  50  per  cent,  greater  strength  than  the 
average  number  of  our  table,  p.  411.  The  crushing  fracture 
usually  makes  an  angle  of  56  degrees  with  the  axis  of  a  test 
column.  The  strengths  of  little  round  columns  of  lengths  equal 
to  from  one  to  three  diameters  are  much  the  same,  but  shorter 
columns  are  very  much  stronger,  and  longer  columns  are  very 
much  weaker.  The  reason  for  this  is  given  in  Art.  256.  The 
specified  test  for  cast  iron  is  often  this — that  a  bar  3  feet 
between  supports,  section  2  inches  deep  and  1  inch  broad,  should 
carry  a  middle  load  of  25  to  35  cwt.,  and  will  deflect  before 
fracture  0*2  to  0-5  inch.  The  average  ultimate  shearing  stress 
is  15  tons  per  square  inch  as  tested  by  torsion.  Remelting  im- 
proves the  strength,  but  not  beyond  a  certain  number  of  times, 
a  tenacity  of  6  tons  per  square  inch  in  the  pig  becoming  9  tons 
after  the  second  melting,  and  12  tons  after  the  fifth.  This 
seems  connected  with  the  change  of  the  iron  from  grey  to 
white  by  increase  of  the  combined  carbon  and  decrease  of 
silicon. 

Mr.  Turner  has  arrived  at  the  following  percentages  as 
giving  the  following  qualities  in  the  highest  degrees  : — 


Combined 
Carbon. 

Graphitic 
Carbon. 

Silicon. 

Softness  

0-15 

3-1 

2'S 

Hardness 





Under  0-8 

General  strength 

0-50 

2-8 

1'42 

Stiffness 

— 

•    — 

1-0 

Tensile  strength 

— 



1-8 

Crushing  strength 

.   Over  1-0 

Under  2'6 

About  0-8 

282  APPLIED  MECHANICS. 

224.  Patterns  of  objects  are  usually  made  in  yellow  pine 
(sometimes  of  metal  if  many  castings  are  wanted),  about  one- 
eighth  of  an  inch  per  foot  in  every  direction  larger  than  the  object 
is  to  be,  because  the  iron  object  contracts  to  this  extent  in  cooling. 
Gun-metal  contracts  about  one-eleventh  of  an  inch  per  foot, 
and  brass  about  twice  as  much.     A  thoughtful  pattern-maker 
can  often  greatly  diminish  the  labour  of  moulding.     Prints  are 
excrescences  made  on  the  patterns  to  show  in  the  mould  where 
certain  cores  are  to  be  placed.     These  cores  are  made  of  loam 
or  core-sand  in  core-boxes,  which  the  pattern-maker  supplies ; 
they  represent  the  spaces  in  the  object  where  the  melted  metal 
is  not  to  flow.     They  are  coated  with  a  wash  of  charcoal  dust 
and  clay.     Common  casting  is  green-sand ;  there  is  the  more 
elaborate  dry-sand  for  such  objects  as  pipes,  and  there  is  the 
most  expensive  loam  moulding,  in  which  the  mould  is  built  up 
without  a  pattern.     You  must  see  for  yourself  in  a  foundry 
what  are  the  usual  methods  of  preparing  a  mould  :  How  the 
pattern  is  made  so  as  to  draw  out  easily ;  how  the  surface  of 
the  sand  is  blackened ;  how  the  moulder  arranges  his  vents  to 
let  gases  escape  from  the  more  compact  parts  of  the  sand ; 
how  he  places  his  gates  to  let  the  metal  run  into  the  mould 
with  just  enough  rapidity  and  yet  without  hurt  to  the  mould. 
You  must  also  see  for  yourself,  taking  sketches  in  your  note- 
book and  making  a  drawing  of  the  cupola,  how  the  pig  iron  is 
melted  and  poured  into  the  moulds ;  how  the  moulder  stands 
moving  an  iron  rod  up  and  down  in  one  of  the  gates,  producing 
just  so  much  circulation  and  eddying  motion  in  the  melted  iron 
as  is  likely  to  remove  bubbles  of  gas  which  may  otherwise  be 
unable  to  escape  from  the  sides  and  corners  of  the  mould,  as 
well  as  to  prevent  the  formation  of  cavities  by  shrinkage  or 
"piping";  how  in  some  castings  he  exposes  to  the  air  certain 
parts  wjiich  would  otherwise  cool  too  slowly  for  the  rest  of  the 
object;  how  next  morning  he  screens  his  sand  and  wets  it. 
You  ought  to  observe  the  appearance  of  the  castings  before 
and  after  they  are  cleaned  up  next  morning. 

225.  The  Cooling  of  Castings. — The  most  important  matter 
in  connection  with  moulding  is  that  there  shall  be  the  same 
amount  of  contraction  at  the  same  time  in  every  portion  of  the 
mass  of  metal  as  it  cools;  otherwise,  when  finished,  there  may  be 
internal  strains,  which  very  much  weaken  the  object,  and  often 
produce  fracture.     In  designing  the  shape  of  an  object  which  is 
to  be  cast,  care  is  taken  that  when  a  thin  portion  joins  a  thick 


APPLIED   MECHANICS.  283 

one  it  shall  do  so  by  getting  gradually  thicker,  and  not  by  an 
abrupt  change  of  size.  The  thin  piece  exposes  more  surface, 
and  cooling  is  effected  through  the  surface.  The  thin  rim  of  a 
pulley  cools  sooner  than  the  arms,  and  becomes  rigid  sooner ; 
when  the  arms  cool  they  contract  so  much  as  sometimes  to 
produce  fracture  near  the  junction.  In  a  thick  cylindric  object 
the  outer  portion  becomes  rigid  first;  now  when  the  inner 
portion  contracts  it  tends  to  make  the  outer  portion  contract 
too  much,  and  the  outer  portion  prevents  the  inner  from 
contracting  so  much  as  it  ought  to,  so  that  the  outer  portion 
retains  a  compressive  strain,  and  the  inner  a  tensile  strain. 
When  a  hollow  cylinder  is  cast,  and  is  required  to  withstand  a 
great  bursting  pressure — that  is,  all  the  metal  is  required  to 
withstand  tensile  tresses — it  is  usual  to  cool  it  from  the  inside 
by  means  of  a  metal  core,  in  which  cold  water  circulates.  The 
inside  now  becomes  rigid  sooner,  the  outer  portions  as  they 
solidify  contract,  and  tend  to  make  the  inner  portion  con- 
tract more  than  it  naturally  would,  and  there  is  a  per- 
manent state  of  compressive  strain  inside  and  tensile  strain 
outside  in  the  object,  which  materially  helps  it  to  resist 
a  bursting  pressure.  This  inequality  of  contraction  and 
production  of  internal  strains  in  objects  cause  them  to  vary  in 
their  total  bulk  as  compared  with  that  of  their  patterns,  but  it 
is  probable  that  some  of  this  variation  is  due  to  the  fact  that 
the  contraction  of  grey  oast  iron  is  only  1  per  cent,  of  its 
linear  dimension,  whereas  white  cast  iron  contracts  2  to 
2-J  per  cent.  The  fractional  difference  between  size  of  pattern 
and  the  finished  object  varies  from  one-twenty-fifth  of  an 
inch  per  foot  in  small,  thin  objects  to  one-eighth  of  an  inch 
per  foot  in  heavy  pipe  castings  and  girders.  Small  castings 
seem  to  be  considerably  stronger  than  large  ones.  As  there  is 
always  great  inequality  in  the  rate  of  cooling  of  a  casting  near 
a  sharp  corner,  internal  strains  may  be  expected  here,  and  also 
an  inequality  in  the  nature  of  the  cast  iron,  since  the  grey 
variety  gets  whiter  the  more  rapidly  it  is  cooled.  In  nearly  all 
bodies  a  re-entrant  corner  is  a  place  of  weakness  (see  Art.  303), 
and  is  specially  to  be  guarded  against  in  castings.  Crystals  of 
cast  iron  and  other  metals  group  themselves  along  lines  of  flow 
of  heat.  When  a  plate  or  wire  of  iron  or  steel  is  rolled  or 
pulled,  the  crystals  become  more  longitudinal,  and  the  wire  or 
plate  becomes  stronger,  whereas  annealing  allows  the  crystals 
to  arrange  themselves  laterally,  and  the  material  is  weakened. 


284  APPLIED    MECHANICS. 

It  is  said  that  time  gradually  reduces  some  internal  strains. 
Castings  which  have  been  rapidly  cooled  by  being  cast  in  an 
iron  mould  (painted  on  its  inside  with  loam)  are  white,  and 
very  hard  in  those  parts  which  lie  nearest  the  mould, 
whereas  they  are  grey  and  strong  inside.  These,  like  the 
hollow  cylinder  above  mentioned,  are  called  chilled  casting's. 
When  a  cleaned  casting,  preferably  of  white  iron,  is  put  in  a 
box,  surrounded  with  oxide  of  iron  (hematite  iron  ore  or  rolling 
mill  scale),  and  kept  at  a  white  heat  for  a  length  of  time  (say 
a  week),  its  surface,  to  a  depth  dependent  on  the  time,  loses 
its  carbon  and  becomes  pure  or  wrought  iron,  which  is  much 
tougher  than  cast  iron.  The  teeth  of  wheels  are  sometimes 
heated  in  this  way.  Such  are  malleable  castings.  Malleable 
cast  iron  seems  to  be  of  50  per  cent,  greater  tensile  strength  than 
cast  iron,  with  a  contraction  of  8  per  cent,  before  fracture ;  it 
stands  about  60  tons  per  square  inch  crushing  stress.  Melted  cast 
iron  possesses  the  property  of  dissolving  pieces  of  wrought  iron, 
and  is  then  said  to  be  toughened  cast  iron.  When  sufficient 
iron  is  so  added  it  becomes  an  inferior  variety  of  cast  steel. 

226.  Wrought  Iron. — Cast  iron  is  exposed  to  the  air  in  a 
melted  state  for  a  long  time,  and  the  carbon  is  burnt  out  of  it. 
The  pig-iron  really  undergoes  two  processes,  one  called  refining, 
the  other  puddling.  It  is  then  hammered  and  rolled,  when  hot, 
into  bars  of  various  shapes.  The  quality  of  wrought-iron  bars 
as  bought  in  the  market  varies  greatly.  We  have  common  iron, 
used  for  rails,  ships,  and  bridges  ;  best,  double  best,  and  treble 
best  Staffordshire  iron,  used  for  boilers  and  f orgings  generally ; 
Lowmoor,  Bowling,  and  other  good  irons  for  the  most  difficult 
forgings  ;  and,  lastly,  charcoal  iron,  which  is  nearly  pure.  Iron 
is  softer  and  more  ductile  the  purer  it  is.  Traces  of  sulphur 
make  it  red-short,  difficult  to  work  hot.  Phosphorus  has 
effects  like  carbon,  but  also  makes  the  iron  cold-short,  or 
treacherous  cold.  Mansrarese  and  silicon  seem  abo  to  act  like 
carbon,  but  we  are  still  afraid  of  them.  There  is  usually  J  to 
^  of  1  per  cent,  of  manganese  present,  and  one-third  of  these 
amounts  of  silicon  if  castings  are  wanted.  Up  to  the  tempera- 
tures of  ordinary  boilers,  the  tensile  strength  of  iron  is  not  much 
diminished  by  heating,  but  at  a  red  heat  it  is  very  much  less 
than  in  the  cold  state.  Repeated  forging  increases  the  strength 
of  wrought  iron  up  to  a  certain  number  of  times,  after  which 
it  diminishes  the  strength.  This  is  why  small  rods  and  small 
forgings  and  the  outside  of  large  forgings  are  respectively  of 


APPLIED    MECHANICS.  285 

stronger  material  than  large  rods  and  large  forgings  and  the 
inside  of  large  forgings.  Cuts  of  metal  in  certain  directions 
from  heavy  forgings  seem  surprisingly  weak.  By  rolling  and 
hammering  when  hot,  iron  gets  a  fibrous  texture,  and  becomes 
more  tenacious.  By  hammering  when  cold,  or  by  long  con- 
tinued strains  of  a  vibratory  kind,  wrought  iron  changes  its 
fibrous  and  tough  for  a  crystallised  and  more  brittle  condition. 
This  brittle  condition  may  be  removed  by  heating  and  slowly 
cooling  (annealing).  Iron  wire  is  stronger  the  thinner  it  is. 
Bar  iron  is  generally  stronger  than  angle  or  T-iron,  and  this 
again  than  plate  iron.  The  toughness  of  an  iron  bar  is  best 
shown  by  the  contraction  it  undergoes  before  it  breaks.  The 
section  of  a  very  tough  bar  may  contract  as  much  as  45  per 
cent,  in  area.  Case  hardening"  of  a  wrought-iron  object  is 
effected  by  heating  it  in  a  box  with  bone  dust  and  horn 
shavings.  The  iron  absorbs  carbon,  and  is  partially  converted 
into  steel. 

227.  Steel. — Steel  contains  less  carbon  and  impurities  than 
cast  iron,  and  thus  lies  intermediate  bet  ween  cast  iron  and  wrought 
iron.  Expensive  steel  is  produced  by  giving  carbon  to  wrought 
iron  (the  best  Swedish  charcoal  iron),  keeping  the  iron  heated 
for  some  days  in  contact  with  powdered  charcoal,  and  then 
hammering  it,  whilst  hot,  till  it  is  homogeneous  (shear  steel),  or 
else  (and  this  is  the  most  usual  practice)  casting  it  when  melted 
into  ingots. 

Cheap  steel  is  produced  by  taking  only  a  portion  of  the 
carbon  from  very  pure  varieties  of  cast  iron  by  a  puddling 
process  such  as  is  employed  in  the  production  of  wrought  iron, 
or  by  the  Bessemer  process.  In  the  Bessemer  process,  air  is 
forced  into  the  melted  cast  iron  for  a  time,  and  very  pure  white 
cast  iron  is  then  added  to  help  in  removing  bubbles  of  gas.  In 
Art.  235  I  shall  speak  about  the  tempering  of  steel.  Many 
varieties  of  soft  steel  do  not  harden  when  suddenly  cooled. 
Some  of  it  is  like  pure  iron  almost,  and  some  of  it  has  half  as 
much  carbon  as  the  hardest  cast  steel.  It  differs  from  wrought 
iron  in  having  been  melted,  and  so  there  are  no  minute  streaks 
of  slag  giving  heterogeneousness ;  and  so  a  plate  is  nearly  as 
strong  one  way  as  another.  The  sudden  hardening  of  steel  when 
rapidly  cooled  seems  greater  the  more  carbon  there  is,  up  to  a 
certain  limit.  It  is  more  fusible  than  wrought  iron,  and  much 
success  has  been  met  with  in  the  production  of  steel  castings  in 
spite  of  the  fact  that  unless  certain  precautions  are  taken,  and  the 


286 


APPLIED    MECHANICS. 


addition  of  silicon,  aluminium,  etc.,  there  is  a  tendency  to  con- 
tain cavities.  This  steel  has  about  twice  the  strength  of  cast 
iron.  Annealing  is  necessary  after  casting.  The  strength  of 
crucible  cast  steel  is  greater  than  that  of  any  other  material,  and 
is  greater  as  it  contains  more  carbon  and  is  harder.  The  pro- 
perties of  steel  depend  so  much  on  so  many  seemingly  small 
things — small  impurities,  a  little  too  much  heating  or  variation 
in  the  rate  of  cooling  at  different  places  — that  great  care  must 
be  taken  in  working  it.  By  the  Bessemer,  Basic,  and  Siemens 
processes  great  quantities  of  steel  are  produced  cheaply,  contain- 
ing small  percentages  of  carbon.  This  steel  has  largely  replaced 
wrought  iron  in  its  use  in  locomotive  rails,  bridges,  and  ships. 
Taking  it  that  crucible  steel  has  elastic  limits  of  stress  26 
to  20  tons  per  square  inch,  and  ultimate  stress  52  to  34  tons, 
and  contraction  of  section  5  to  20  per  cent,  before  fracture, 
these  are  for  Bessemer  steel  17  tons,  34  tons,  and  20  to  45  per 
cent.  Steels  to  be  easily  welded  together  ought  to  be  of  the 
same  kind. 

As  steel  cools  from  a  welding  heat  it  passes  through  the 
"  blue  heat "  stage  (about  300°  0.),  where  it  is  brittle.  If 
hammered  or  bent  in  this  state  it  is  permanently  injured,  and 
if  the  work  was  local  in  a  large  plate  the  plate  becomes 
treacherous.  Soft  steel  and  iron  seem  to  get  a  little  stronger 
at  very  low  temperatures. 

The  amount  of  carbon  present  in  steel  does  not  seem  to 
affect  much  the  Young's  modulus,  which  in  all  kinds  of  steel 
and  wrought  iron  seems  to  be  about  3  x  107,  as  the  modulus 
of  rigidity  is  not  very  different  from  12  x  106.  Various 
numbers  are  given  in  Table  III.,  in  deference  to  custom.  The 
carbon  produces  other  effects,  shown  in  the  following  table  : — 


%  Carbon. 

Elastic  limit 
stress. 

Breaking 
stress. 

Contraction  of 
area. 

Ultimate 
shearing 
stress 

•14 

18 

28 

•50 

22 

•51 

22 

36 

•25 

26 

•96 

31 

63 

•10 

37 

Stresses  are  in  tons  per  square  inch.  A  formula  has  been 
given  -.—Strength^  19-5  +  11-4  C2  +  302  +  11-4  Mn  +  9'5  P, 
and  elongation  per  cent.  =  42  -  36  0.  —  5 -5  Mn  -  6  Si,  where 


APPLIED   MECHANICS.  287 

C,  Mn,  P,  and  Si  represent  the  fractional  amounts  of  carbon, 
manganese,  phosphorus,  and  silicon  in  the  steel. 

The  oxide  and  silicate  skin  on  cast  iron  is  less  liable  to 
corrosion  than  clean  iron,  but  it  is  advisable  to  paint  or  varnish 
all  iron  exposed  to  oxidation.  Sometimes,  as  for  water-pipes, 
the  iron  is  heated  to  150°  C.,  and  placed  in  pitch  with  some  oil 
in  it  at  100°  C.  Mr.  Barff  keeps  the  iron  surface  exposed  to 
superheated  steam  at  a  high  temperature,  and  this  coats  it  with 
a  film  of  protecting  oxide.  Iron  is  "  galvanised  "  by  putting 
it  in  a  bath  of  melted  zinc.  Boilers  are  sometimes  protected 
inside  by  the  contact  of  blocks  of  zinc. 

Some  alloys  of  iron  and  manganese  are  very  strong,  and  so 
hard  as  to  prevent  their  being  readily  tooled.  They  can,  how- 
ever, be  both  cast  and  forged.  They  are  specially  important  to 
electrical  engineers,  as  they  are  not  magnetic.  The  power  of  a 
small  trace  of  manganese  to  destroy  all  the  magnetic  properties 
of  iron  is  remarkable.  Certain  alloys  of  iron  and  nickel  are  not 
quite  so  hard,  but  they  are  extremely  tough  and  strong.  Steels 
containing  a  little  tungsten  or  chromium  have  the  special 
properties  that  fit  them  for  self-hardening  tools  or  projectiles. 

Mitis  castings  are  of  wrought  iron,  to  which  0'5  to  1  per 
cent,  of  aluminium  has  been  added  to  lower  the  fusing  point. 
The  tenacity  seems  to  be  20  per  cent,  greater  than  that  of 
wrought  iron,  and  the  ductility  about  equal  to  that  of  wrought 
iron. 

228.  Copper  is  noted  for  its  malleability  and  ductility  when 
both  hot  and  cold,  so  that  it  is  readily  hammered  into  any  shape, 
rolled  into  plates,  and  drawn  into  wires.  When  cast  it  usually 
contains  much  oxide  and  many  cavities,  but  when  pure  it  may 
be  worked  up  by  hammering  into  a  state  of  great  strength  and 
toughness,  whereas  slight  traces  of  carbon,  sulphur,  and  other 
impurities  necessitate  its  being  refined  to  do  away  with  its 
brittleness.  The  brittleness  produced  by  hammering  when  cold 
is  very  different,  as  it  is  removable  by  annealing.  Phosphorus 
is  sometimes  used  to  assist  casting,  and  the  strength  is  greater 
with  more  phosphorus,  whose  function  seems  that  of  reducing 
the  oxides.  Copper  is  an  expensive  metal,  and  is  only  used 
now  for  pipes  which  require  to  be  bent  cold,  for  bolts  and 
plates  in  places  where  iron  would  be  more  readily  corroded, 
and  for  electrical  purposes.  Its  tensile  strength  is  more 
reduced  by  heating  than  that  of  iron. 

Boron  seems  to  affect  copper  as  carbon  does  iron,  the  wire 


288  APPLIED   MECHANICS. 

alloy  standing  22  to  27  tons  per  square  inch  without  loss  of 
conductivity. 

229.  Brass  consists  of  about  two  parts  by  weight  of  copper  to 
one  of  zinc,  with  a  little  tin  and  lead.     The  copper  is  first  melted 
and  the  zinc  added,  not  long  before  casting.     It  is  used  chiefly 
on  account  of  its  fine  appearance  and  the  ease  with  which  it 
can  be  worked.     Cheap  brass  things  have  more  zinc  usually. 
The  small  amount  of  lead  added  in  melting  makes  it  much 
softer.    Muntz  metal  contains  more  zinc  than  ordinary  brass — 
3  copper  to  2  zinc,  or  2  to  1  with  a  little  lead.     Like  copper,  it 
is  not  much  weakened  by  heating  up  to  260°  C. ;  it  can  be 
rolled  hot.     It  is  used  for  sheathing  ships  and  for  the  tubes 
of  uoilers. 

230.  Sterro  and  Delta  metal  are  brasses  to  which  iron  has 
been  added,  the  latter  name  being  given  to  the  metal  after  it 
has  also  received  special  mechanical  treatment,  the  rods  being 
made  by  being  forced  to  flow  under  great  pressure  through 
dies.     Delta  metal  can  be  worked  hot  or  cold,   and  may  be 
brazed.     Bronze  and  gun-metal  are  alloys  of  copper  and  tin  in 
varying  proportions,  more  tin  giving  greater  hardness.     Twelve 
of  copper  to  1  of  tin  seems  best  for  guns.     Five  of  copper  to 
1  of  tin  is  the   hardest  alloy  used  by  the  engineer  in  bear- 
ings, but  bell  metal  is  3 '3  to  1      A  slight  addition  of  zinc  helps 
in  casting,   and   increases    the   malleability.     A  great  many 
experiments  have  been  made  on  bronze.     Its  strength  depends 
very  much  upon  the  care  taken  in  mixing  and  melting  the 
metals,  for  it  is  easily  injured  by  oxidation.     Gun-metal  is  a 
good  material  for  castings,  which,  however,  should  be  quickly 
cooled  to  give  more  uniformity,  density,  strength,  and  toughness ; 
and  hence  they  are  sometimes  made  in  cast-iron  moulds.    Hard 
bronze  is  much  used  for  the  bearings  of  shafts.     There  are  also 
various  soft  alloys  of  copper  with  lead,  zinc,  tin,  and  antimony, 
which  are  used  for  this  purpose.     Phosphor  bronze  is  an  alloy 
of  copper  and  tin  to  which  some  phosphorus  has  been  added. 
It  bears  re-melting  better  than  gun-metal,  and  its  properties 
may  be  varied  at  will.     It  may  be  either  strong  and  hard,  or 
weaker  but  very  tough.     The  phosphorus  appears  to  act  by 
removal  of  the  oxide  of  tin,  and  to  tend  to  prevent  segregation 
in  cooling ;  and,  indeed,  we  may  say  that  when  care  is  taken 
against  oxidation  the  difficulties  in  the  way  of  re-melting  gun- 
metal  vanish.     Hard  wire  has  broken  with  from   100  to  150 
tons  per  square  inch,  and  after  annealing  has  stood  half  these 


APPLIED    MECHANICS.  289 

amounts.  It  has  been  used  successfully  in  railway  axle  .and 
crank  shaft  bearings.  It  is  good  in  resisting  shocks,  and  has 
been  used  instead  of  steel  for  chisels  in  powder  factories. 
Gun-metal  slowly  decreases  in  strength  as  it  is  heated,  until  at 
a  certain  temperature  its  strength  is  suddenly  halved,  and  there 
is  almost  no  ductility.  Phosphor  bronze  is  less  affected. 

231.  Manganese  bronze  and  Silicon  bronze  are  special  alloys 
of  copper  with  manganese  and  with  silicon^  the  former  made  by 
adding  ferro-manganese  to  bronze  or  brass,  and  used  where 
unusual  strength  and  power  (as  for  screw  propeller  blades)  of 
resisting  sea-water  are    required.     It  may  be  cast  and   also 
forged.      The   manganese   seems   to   act   like    phosphorus   in 
clearing  off  the  oxide.     Some  containing  zinc  can  be  forged 
and  rolled  hot.     Silicon  bronze  has  fair  electric  conductivity, 
and  resists  atmospheric  corrosion  when  used  as  telephone  wire, 
and  is  of  great  strength.     As  tested  for  tension  in  the  condi- 
tion of  wire  for  telephonic  purposes,  it  seems  to  stand  from  30 
to  50  tons  per  square  inch;  the  drawn  copper  wire  for  the 
same  purpose  standing  about    29   tons    to  the  square  inch ; 
phosphor  bronze,   44  to  70;  brass,   25;   German  silver,    30; 
iron,  57 ;  Martin  steel,  86  ;  and  crucible  steel,   102  tons  per 
square  inch.     Pianoforte  wire  sometimes  stands  150  tons  pei 
square  inch. 

232.  Aluminium  bronze  is  formed  of  9  parts  of  copper  and 
1  of  aluminium,  and  has  a  tenacity  of  43  tons  per  square  inch. 
The  5  per  cent,  alloy  stands  30  tons  per  square  inch.     Copper 
with  only  2  to  3  per  cent,  of  aluminium  is  stronger  than  brass. 
The  usual  alloy  is  of  the  colour  of  gold,  but,  like  all  aluminium 
alloys,  must  be  prepared  from  materials  free  from  iron.    When 
ad\antage  is  to  be  taken  of  the  lightness  of  aluminium,  there 
is  an  alloy  of  32  parts  of  aluminium  and  1  of  nickel  that  can 
be  employed,  and  to  which  1  part  of  copper  may  also  be  added. 
Specimens  have  broken  with  45  and  50  tons*  per  square  inch 
tensile  stress,  the  first  with  no  elongation  and  the  second  with 
33  per  cent,  elongation.     The  aluminium  and  silicon  bronzes 
and  alloys  of  silver  and  other  metals  with  aluminium  are  pro- 
duced in  an  electric  furnace,  the  ores  being  mixed  with  retort 
carbon  and  an  electric  current  passed. 

About  7  per  cent,  each  of  copper  and  antimony  being 
added  to  tin,  we  have  white  metal  or  Babbitt's  metal,  which 
fuse  easily  and  may  be  cast  inside  a  bracket,  round  a  journal, 
as  the  step  of  a  bearing  if  not  too  large 


290 


CHAPTER   XIII. 

TENSION   AND    COMPRESSION. 

233.  THE  following  chapter  on  the  behaviour  of  material 
when  subjected  to  tension  and  to  compression  has  so  much  to 
do  with  the  physical  properties  of  matter,  that  students  ought, 
before  reading  it,  to  refresh  their  memories  in  regard  to  the  sim- 

Eler  principles  of  chemistry  and  physics  and  the  notions  which 
iboratory  work  in  these  subjects  gives  to  us  in  regard  to  the 
probable  molecular  condition  of  matter,  and  also  of  that  very 
much  larger  coarse-grainedness  which  we  sometimes  call  hetero- 
geneity. Besides  giving  us  general  notions,  chemistry  gives  us 
useful  facts  as  to  the  changes  which  occur  in  the  manufacture 
of  metals,  the  cause  of  the  rusting  of  metal,  the  burning  of 
fuel,  etc.  A  little  knowledge  of  electricity  enables  us  to  have 
clear  ideas  as  to  the  action  by  which  when  two  metals  touch, 
and  are  also  connected  by  liquid,  one  of  them  rapidly  corrodes 
and  the  other  does  riot,  and  how  it  is  that  oil  preserves  a 
polished  metal  surface.  A  little  knowledge  of  heat  tells  us 
how  friction  wastes  mechanical  energy ;  how  heat  energy  is 
measured ;  when  a  body  is  heated  how  much  it  expands  ;  the 
laws  of  expansion  of  gases  ;  the  properties  of  steam  ;  the  laws 
of  flow  of  heat  in  conduction  and  radiation,  and  other  phano- 
mena  which  continually  influence  our  mere  mechanical  work. 

234.  It  will  be  found  by  students  who  read  what  is  in  the 
smaller  printing  in  this  chapter  that  the  usual  statements  on 
which  we  base  our  mathematical  calculations  are  very  incom- 
plete.    The  manufacturer  depends  on  many  curious  properties 
of  materials,   many  of  which  are  familiar  to  and  helpful  to 
inarticulate  workmen  and  unknown  in  the  laboratory  as  yet. 
Some  rude  processes  and  shop  beliefs,  which  sometimes  seem 
to  be  no  more   worthy  of  attention  than  superstitions,  have 
suggested   scientific  experiments   and  new  industries.     Some 
phenomena  that  seem  curious  at  first  sight  are  really  easily  ex- 
plained— the  behaviour  of  James  Thomson's  overtwisted  shaft, 
for  example,  explains  many  curious  things  ;  and  our  knowledge 
of  such  things  as  how  initial  strains  are  induced  in  castings  by 
unequal  cooling  has  helped  us  greatly  in  systematising  our  notions. 

235.  There  is  one  phenomenon  which  has  been  known  almost 


APPLIED    MECHANICS.  291 

since  iron  was  discovered  by  man — namely,  that  steel  hardens 
with  sudden  cooling,  and  we  seem  to  be  almost  as  far  from 
understanding  it  as  our  ancestors  were.  There  has  been  some 
advance,  for  we  have  no  such  notions  about  incantations  and 
the  virtues  of  particular  kinds  of  water  for  quenching  the  heat 
as  were  held  in  the  Middle  Ages.  Mild  steel  is  almost  like  pure 
iron,  and  does  not  harden,  but  with  more  carbon  (over  0'4  per 
cent.),  such  as  there  is  in  cast  steel,  the  more  rapidly  the  steel 
is  cooled  the  harder  it  gets.  So  that,  for  example,  thin  pieces 
of  steel  are  apt  to  be  harder  than  thick  pieces.  The  more 
carbon  a  steel  contains,  the  less  need  it  be  Ideated  before  being 
suddenly  cooled  to  acquire  a  particular  hardness.  Re-heating 
diminishes  the  hardness,  or,  as  it  is  called,  tempers  the  steel ; 
and  the  higher  the  temperature  of  re-heating,  the  softer  does 
the  steel  become.  It  may  roughly  be  taken  that  sudden  cool- 
ing in  oil  doubles  the  proof  tensile  strength  of  the  material — • 
that  is,  the  stress  which  would  permanently  hurt  the  material. 

In  every  workshop  the  common  method  adopted  for 
tempering  a  fitter's  chisel  is  as  follows : — Heat  the  chisel  to 
a  dull  red  colour,  put  the  edge  in  water  to  a  distance  of  say 
half  an  inch,  so  that  it  may  become  very  hard;  quickly  brighten 
the  edge  with  pumice  or  a  file ;  watch  it  till,  as  it  heats  by  con- 
duction from  the  thicker  portion,  you  know  that  a  certain 
temperature  has  been  reached  by  seeing  a  certain  colour 
(purplish-yellow  for  a  chisel)  of  oxide  of  iron  making  its 
appearance.  When  this  colour  appears,  plunge  the  whole  chisel 
into  water.  Thus  the  steel  is  first  made  extremely  hard  at  its 
edge,  and  is  then  brought  back  to  the  required  degree  of 
hardness  by  re-heating  up  to  a  certain  temperature  and  then 
cooling.  This  simple  process  is  in  common  use.  In  tempering 
other  objects  sometimes  much  greater  care  must  be  taken,  since 
it  is  often  necessary  that  every  portion  of  the  object  shall  be  of 
the  same  hardness,  and  in  such  cases  the  whole  may  be  cooled 
at  first  and  then  re-heated  in  a  bath  of  oil,  mercury,  or  other 
melted  metal  whose  temperature  is  definitely  known.  The 
effect  is  of  the  same  kind,  however,  whether  the  process  is  the 
rough  one  which  I  have  described  or  a  more  careful  one. 

There  must  be  no  attempt  to  make  large  objects  glass-hard : 
they  would  cool  very  unequally  and  might  fly  to  pieces  or 
develop  flaws ;  a  less  rapid  cooling  in  hot  oil  or  melted  lead 
tempers  such  objects  in  one  process.  It  is  interesting  to  see 
that  one  or  other  of  the  above  two  principles  is  carried  out  in 


292  APPLIED    MECHANICS. 

all  sorts  of  industries,  but  in  a  great  number  of  different  ways. 
A  certain  size  of  watchmaker's  drill  is  stuck  when  at  a  red 
heat  into  sealing-wax.  This  gives  the  right  temper.  Another 
smaller  drill  is  merely  waved  about  in  the  atmosphere  to  cool 
it.  The  tempering  colours  of  steel,  beginning  with  the  hardest, 
are : — Straw  colour  to  yellow,  (this  is  about  220°C)  for  light  turn- 
ing tools,  milling  cutters,  screw-cutting  dies  and  taps,  punches, 
chasers ;  straw  to  purplish-yellow,  rimers,  wood-chisels,  plane- 
irons,  twist-drills  ;  light  purple  to  dark  blue,  augers,  chisels  for 
steel,  axes,  chisels  for  cast  iron,  chisels  for  wrought  iron,  saws 
for  metal;  less  dayk  blue  (this  is  about  320°C),  screw-drivers 
and  springs.  These  colours  are  supposed  to  indicate  fairly 
exactly  the  temperature,  irrespective  of  time ;  but  we  cannot 
say  that  there  is  conclusive  evidence  yet  that  time  produces  no 
effect  on  the  thickness  of  the  film  of  oxide.  If  true,  it  is  a  very 
curious  phenomenon.  But  surely  it  cannot  be  true  ! 

236.  The  volume  gets  greater  in  hardening.  Curiously  enough, 
I  have  a  note  saying  that  steel  becomes  denser  by  hardening,  but  its 
authority  is  unknown.  Eepeated  hardening  and  annealing  seem 
to  strengthen  the  steel.  It  is  usual  to  explain  the  hardening  by 
saying  that  in  sudden  cooling  the  particles  of  iron  and  carbon  have 
not  had  time  to  get  into  their  natural  positions  when  cold,  and  that 
they  jam  one  another  somehow,  getting  into  positions  of  instability. 
As  regards  the  influence  of  impurities,  of  gases  from  the  atmosphere 
which  may  possibly  be  suddenly  imprisoned  among  the  particles, 
very  little  is  known.  It  may  help  towards  an  explanation  to  say 
that  Abel  found  that  in  annealed  steel  the  carbon  is  in  the  form  of 
a  chemical  carbide  Fe3C  mixed  in  the  mass.  In  hardening,  the 
formation  of  the  carbide'  is  prevented  (just  as  suddenly-cooled 
gases  remain  dissociated).  At  various  tempers  we  have  various  pro- 
portions of  the  carbide,  but  it  is  always  the  same  kind  of  carbide. 

Speculation  as  to  the  molecular  constitution  of  iron  does  not 
yet  seem  to  have  sufficient  facts  to  go  upon.  It  is  sometimes 
assumed  that  there  are  two  kinds  of  iron  mixed  together,  the  soft 
a  particles  and  the  harder  ft  particles.  With  slow  cooling  from  a 
high  temperature,  when  the  mass  is  soft,  although  the  particles 
may  be  hard,  the  ft  particles  (practically  all  are  of  the  ft  kind  at  a 
high  temperature)  change  to  a.  That  there  is  some  great  molecular 
change  even  in  the  purest  iron  is  evidenced  by  recalescence  and 
other  allied  phenomena.  In  sudden  cooling  most  of  the  ft  particles 
have  no  time  to  change.  Any  effect  due  to  the  carbon  is  produced 
at  a  much  lower  temperature  than  that  at  which  the  change  from 
ft  to  a  occurs  in  slow  cooling ;  and  although  the  presence  of  carbon 
seems  necessary  to  the  hardening  of  steel,  changes  in  its  mode  of 
existence  are  not  of  much  importance.  The  a  particles  are  changed 
to  /3  in  the  plastic  condition  of  iron  in  the  ordinary  testing  opera- 
tions at  low  temperatures. 


APPLIED    MECHANICS. 


293 


237.  How  a  pull  is  exerted. — How  is  it  that  a  cord  transmits 
force  from  my  hand  to  an  object  when  I  pull  the  object  by 
means  of  a  string  ?    If  you  study  this  matter,  you  will  see  that 
every  particle  of  the  string  coheres  to  the  next ;  and  although 
the  refusal  of  one  particle  to  come  away  from  its  neighbour 
might   easily   be    overcome,   there   are  so 

many  of  them  to  be  separated  at  any 
particular  section  of  the  string  that  it 
requires  a  considerable  pull  to  perform 
this  operation.  When  a  string  is  pulled 
it  really  lengthens  a  little,  and  it  lengthens 
more  the  more  force  is  applied,  although 
it  may  not  break.  A  string  is  not  so  easy 
to  experiment  with  as  a  wire  of  metal, 
because  we  find  that  it  differs  more  in  its 
quality  at  different  sections,  and  it  is 
affected  by  dampness  arid  many  other 
circumstances.  No  doubt  it  is  also  dif- 
ficult to  obtain  a  metal  wire  which  shall 
just  be  as  willing  to  break  at  one  place 
as  another — that  is,  which  shall  be  exactly 
of  the  same  material  everywhere ;  but 
metal  wire  is  certainly  more  uniform 
than  string. 

238.  Strain. — Take,,  then,  a  steel  wire, 
A  B  (Fig.  170),  fastened  near  the  ceiling  at 
A,  between  two  pieces  of  wood,    screwed 
together  firmly  so  that  there  may  be  no 
tendency  for  the  wire  to  break  just  at  the 
fastening.     Similarly  fasten  at  B  a  scale- 
pan  arrangement,   and  first  place  just  so 
much  weight  in  the  pan  as  keeps  the  wire 
taut.       Let    there    be    two    light    little 
pointers  stuck  or  tied  on  at  a  and  6,  and 
let  there  be  a  vertical  scale  on  the  wall. 

Now  read  off  the  distance  between  a  and  b  on  the  scale,  and 
note  the  weight.  Add  more  weight,  and  again  read  the 
distance,  and  continue  doing  this  until  the  wire  breaks.  You 
will  prove  by  means  of  squared  paper  that  the  amount  of  the 
extension  of  a  wire  is  nearly  proportional  to  the  weight  which 
produces  the  extension. 

239.  My  students  are  in  the  habit  cf  using  a  more  exact  method 


Fig.  170. 


294 


APPLIED    MECHANICS. 


of  measurement,  a  scale  hanging  from  A  and  two  verniers  being 
attached  to  the  wire  at  a  and  b.  They  also  use  a  method  in  which 
a  vernier  on  a  is  brought  close  to  a  scale  on  J,  the  wire  being 
passed  over  pulleys.  In  some  cases  they  use  micrometer  methods 
of  measurement  for  greater  accuracy,  and  they  also  experiment 
with  larger  specimens,  loading  them  by  means  of  levers  or  wheels 
and  screws;  and  advanced  students  may  "be  allowed  to  use 
machines  in  which  loads  up  to  100  tons  are  applied,  and  arrange- 
ments may  be  employed  for  making  automatic  records  of  the  load 
and  the  extension.  If,  however,  the  apparatus  is  elaborate  and 
imposing  as  compared  with  the  specimen,  a  beginner  cannot 
readily  pick  up  the  essential  idea  of  an  experiment,  and  hence  he 
had  better  begin  with  visible  specimens  loaded  with  visible  weights. 
He  may  proceed  to  use  such  a  machine  as  Bailey's  wire-testing 

machine,  and  after- 
wards make  a  few 
tests  with  large 
commercial  testing 
machines. 

Bailey's  latest 
form  of  machine  is 
shown  in  Fig.  171. 
The  specimen,  say 
of  |-inch  wire,  is 
shown  at  D,  being 
gripped  at  c  and  B. 
By  turni»g  the 
handle,  A,  we  turn 
a  worm  driving  a 
worm-wheel,  turn- 
ing a  screw  whose 
nut  is  part  of  the 
frame,  and  so  the 
gripping  piece  B  pulls  on  the  specimen  D.  The  other  gripping 
piece,  c,  tilts  the  weight,  F,  and  the  amount  of  tilting  which  measures 
the  pulling  force  is  indicated  by  a  pointer  on  the  dial,  E. 

In  some  English  engineering  laboratories  experimenting  with  a 
steam-engine  and  testing  specimens  in  tension  by  means  of  a  large 
testing-machine  are  supposed  to  be  the  only  experimental  exercises 
in  which  students  ought  to  engage.  Tests  of  specimens  in  com- 
pression, bending,  and  twisting  are,  however,  sometimes  made. 
Consequently  a  description  of  all  the  kinds  of  large  testing-machines 
which  have  ever  been  constructed  forms  a  large  part  of  the  college 
instruction  in  mechanical  engineering.  It  seems  to  me,  however, 
somewhat  out  of  place  in  a  text-book,  as  almost  every  student  has 
opportunities  of  examining  some  such  machine  for  himself.  Complete 
information  will  be  found  in  Professor  Unwin's  book  on  "  The 
Testing  of  Materials  of  Construction." 

The  machines  in  most  common  use  apply  tensile  load  to  the 
lower  end  of  a  test-piece  by  means  of  an  hydraulic  press.  The 
upper  end  is  pulled  by  means  of  a  lever  (whose  fulcrum  is  a  knife- 
edge),  over  which  a  weight  may  be  rolled  by  machinery  into  such 


Fig.  170A. 


APPLIED    MECHANICS. 


295 


H 


Fig.  171. 

positions  that  the  lever  is  kept  horizontal  ;*  the  position  of  the 
weight  measures  the  pull. 

Instruments  have  been  designed  which  register  on  a  sheet  of 
paper  (as  the  pencil  of  a  steam-engine  indicator  does)  the  load 
pulling  a  rod,  and  the  extension  which  it  produces.  A  little  brass 
cylinder  covered  with  paper  is  touched  by  a  pencil  on  the  end 
of  the  rod.  The  amount  of  rotation  of  the  barrel  is  regulated 
so  that  it  is  proportional  to  the  load.  By  this  means  curves  like 
that  of  Fig.  172  may  rapidly  be  drawn  as  the  load  on  the  rod  is 
gradually  made  to  increase  till  the  rod  breaks  (see  Art.  244). 

When  Young's  modulus  for  a  material  is  wanted,  my  advanced 
students  employ  Prof.  E  wing's  extensometer.  A  half -inch  bar  of 
iron  being  given,  even  a  load  of  40  IDS.  causes  sufficient  extension 


296  A1>PLIED 

of  the  distance  between  two  marks  about  a  foot  asunder  to  rte 
measured  with  sufficient  accuracy  for  the  determination  of  Young's 
modulus  by  this  beautiful  instrument,  shown  in  Fig.  170A. 

240.  When  we  speak  of  the  tensile  strain  in  the  wire,  and 
i^ant  to  use  the  term  strain  in  an  exact  sense,  we  mean  the 

fraction  of  itself  by  ivhich  a  b  lengthens.  Thus,  suppose  that 
a  b  was  50  feet  and  that  it  lengthens  1  foot,  we  say  that  the 
strain  is  -J^,  or  -02,  or  2  per  cent.  I  need  hardly  tell  you 
how  important  it  is  to  learn  the  exact  meaning  of  a  word 
like  this. 

241.  Stress. — If  you  take  another  wire  of  the  same  material, 
but  of  twice  the  sectional  area  of  this  one,  you  will  find  that 
it  needs  twice  as  much  load  to  produce  the  same  strain.     The 
reason  of  this  is  somehow  due  to  the  fact  that  you  have  at  any 
section  twice  as  many  particles  of  steel  resisting  the  pull.    The 
pull  produced  by  the  load  acts  at  every  cross-section  in  the 
same  way,  no  matter  how  long  the  wire  may  be  ;  *  but  if  the 
wire  is  thicker  at  one  place  than  another,  then  at  such  a  cross- 
section  the  pull  is  distributed  over  a  greater  number  of  pairs 
of  particles.     We  see,  then,  that  if  a  wire  or  rod  is  transmit- 
ting a  pull,  it  is  well  not  to  consider  the  total  load,  but  rather 
the  load  per  square  inch  of  section.   The  load  per  square  inch  is 
called  the  stress. 

*  One  important  result  of  St.  Venant's  investigation  (see  Art.  306)  is  that 
the  actual  distribution  of  the  load  on  a  small  area  is  not  important. 
At  a  point  not  very  near,  the  strain  will  be  the  same  whatever  the  distribu- 
tion of  load.  Near  the  gripping-places  and  places  of  rapid  changes  of  section 
in  our  specimens,  the  stress  cannot  be  expected  to  be  uniform  throughout  a 
cross-section  ;  hence  test-pieces  are  made  larger  near  the  grips,  as  we  do  not 
wish  to  break  or  study  the  specimen  at  a  place  where  the  distribution  of 
stress  is  unknown  to  us.  This  is  particularly  noticeable  at  a  narrow  neck  cut 
in  a  round  specimen.  There  is  great  stress  near  the  ends  of  the  neck, 
inducing  fracture  there  more  readily  than  elsewhere  if  the  material  is  hard, 
but  in  more  ductile  material  inducing  flowing  of  the  metal,  which  prevents 
the  section  of  the  neck  becoming  as  small  before  fracture  as  a  long  specimen 
would  become,  and  so  increasing  the  apparent  strength  of  the  material.  This 
causes  a  strip  of  boiler-plate  with  a  drilled  or  punched  hole  (the  plate  must 
be  annealed  after  punching  to  destroy  local  hardness)  to  seem  stronger  per 
square  inch  of  material  left  at  the  sides  of  the  hole.  If  a  specimen  is  too 
short,  it  does  not  seem  to  have  freedom  to  contract  as  much  in  section  before 
fracture,  and  therefore  the  breaking  stress  is  greater  and  the  fractional 
elongation  is  less ;  for  it  is  to  be  remembered  that  breaking  stress  is 
calculated  as  breaking  load  per  square  inch  of  the  original  section.  Very 
long1  rods  or  wires  give  too  small  a  breaking  stress  for  a  very  different  reason 
— namely,  because  of  the  greater  chance  of  the  existence  somewhere  of 
inferior  metal.  It  is  reasonable  to  suppose,  and  it  is  found  experimentally  to 
be  fairly  correct,  that  to  obtain  the  same  fractional  elongations  from  the  same 
material^  specimens,  if  of  diiierent  sizes,  ought  to  be  similar. 


APPLIED    MECHANICS.  297 

This  is  the  exact  meaning  which  we  give  to  the  word 
stress.  Much  of  the  difficulty  you  may  have  met  with  in  your 
reading  is  due  to  the  fact  that  you  have  not  made  a  proper 
distinction  between  the  meanings  of  these  two  words.  Stress 
is  the  load  per  square  inch  which  produces  a  fractional  altera- 
tion of  the  length  of  a  wire  or  rod,  and  this  fractional  alteration 
is  called  the  strain.  Suppose  your  load  to  be  6  Ibs.,  and  your 
wire  circular  in  section,  with  a  diameter  of  0*05  inch.  Then 
the  area  of  the  section  is  0-025  x  0-025  x  3-1416,  or  -00196 
square  inch.  The  stress  is  6  -4-  '00196,  or  3,061  Ibs.  per 
square  inch.  You  will  find  that  this  thin  wire  gets  the  same 
strain  with  a  total  load  of  6  Ibs.  as  a  rod  1  square  inch  in 
section  would  get  with  a  load  of  3,061  Ibs.  If  ever  you  get  a 
problem  to  work  out  relating  to  the  lengthening  of  a  wire  or 
rod  produced  by  a  load,  you  must  consider  not  the  total 
lengthening  of  the  wire  or  rod,  but  its  fractional  amount  of 
lengthening,  and  call  this  the  strain;  also  consider  not  the 
total  load,  but  the  load  per  square  inch  of  section,  and  call 
this  the  stress,  and  you  will  find  that  for  some  kinds  of 
wrought  iron  the  tensile  stress  =  the  tensile  strain  x  29,000,000. 

242.  It  is  usually  assumed  that  we  ought  to  expect  the  elongation 
and  strength  to  depend  upon  load  per  square  inch  and  not  upon 
the  shape  of  the  section.  To  me  it  seems  wonderful  that  mole- 
cules near  the  surface  should  not  behave  differently  from  molecules 
remote  from  the  surface.  It  is  possible  that  the  mere  shape  may 
have  some  small  effect  which  is  disguised  for  us  by  the  great 
differences  of  material  and  of  physical  state  in  the  specimens 
which  we  compare.  It  is  only  when  we  take  pains  to  obtain 
homogeneous  material  that  we  find  the  strain  very  nearly  propor- 
tional to  the  stress.  Any  departure  from  this  rule  may  be  traced 
to  initial  strains  in  the  specimens.  Unless  a  casting  or  cold- 
hammered  or  swaged  forging  is  annealed  by  heating  to  a  bright 
red  heat  and  a  slow  cooling,  it  is  heterogeneous.  Some  parts  take 
a  set  much  more  easily  than  others,  because  they  are  of  different 
material  or  because  they  are  already  strained..  Instead  of  Fig.  172, 
therefore,  we  ought  to  expect  a  figure  which  is  the  result  of  adding 
together  the  ordinates  of  a  great  number  of  curves  like  Fig.  17?, 
the  distances  o  N  being  very  different,  in  some  cases  0.  A  cast- 
iron  beam  takes  a  set  for  quite  small  loads,  and  it  is  often  loaded 
so  as  to  get  a  large  set  before  it  leaves  the  foundry.  Afterwards 
its  deflection  will  be  practically  proportional  to  load.  A  man  who 
puts  up  bells  in  a  house  or  a  telegraph  wire  "kills"  the  wire 
beforehand — that  is,  gives  it  a  permanent  set,  straining  it  to  its 
"yield  point"  indeed,  as  he  finds  that  after  this  operation  it  is 
harder,  will  stand  great  loads  without  taking  a  further  set,  and 
follows  the  laws  of  elasticity  better  for  pulling  forces.  This  is  like 
giving  a  good  set  to  a  piece  of  riveted  work,  which  means  that  tha 

K* 


298  APPLIED    MECHANICS. 

rivets  bed  better  into  their  holes.  When  a  wire  by  being  drawn 
through  a  die  is  reduced  to  a  smaller  size,  there  is  a  complete 
alteration  in  the  arrangement  of  its  particles  or  groups  of 
molecules,  and  yet  the  drawn  wire  has  usually  greater  strength 
than  it  had  originally.  Even  hardened  steel  wire  is  drawn  in  this 
way.  Metals  will,  in  fact,  flow  (the  filling  up  of  the  passages  in 
mines  shows  that  rocks  also  flow)  if  sufficient  stress  is  applied  to 
them,  and  at  the  end  of  the  operation  they  are  as  strong  or  stronger, 
but  less  plastic,  than  before.  If  they  are  harder  and  this  quality 
is  not  wanted,  it  may  usually  be  removed  by  annealing.  Plates 
of  iron  and  steel  rolled  cold  are  hardened ;  rolled  hot,  they  are 
gradually  annealed  as  they  leave  the  rolls.  In  constructing  a 
certain  magnetic  instrument,  I  find  it  necessary  to  anneal  a  certain 
piece  of  iron  from  so  high  a  temperature  that  the  nearly  pure  iron 
is  so  soft  that  it  almost  cannot  keep  in  shape.  If  this  is  scratched 
once  by  a  file,  sufficient  hardness  is  induced  to  make  the  instrument 


243.  It  is  somewhat  more  difficult  to  experiment  on  the 
shortening  of  a  strut  or  column  when  it  transmits  a  push, 
because  you  cannot  use  very  long  struts.  A  strut  tends  to  bend 
if  it  is  very  long ;  and  when  it  breaks,  unless  great  care  is 
taken  to  keep  it  straight,  it  breaks  more  easily  the  longer  it  is, 
The  bending  action  causes  the  load  to  act  more  on  one  part  of 
the  cross-section  than  another,  and  the  stress — or  the  pushing 
force  per  square  inch — is  greater  at  one  part  of  the  section 
than  at  another.  If  you  experiment,  therefore,  you  must  take 
care  to  use  struts  which  are  prevented  from  bending.  In 
Chap.  XVI.  we  shall  consider  the  bending  of  beams,  after  which 
you  will  better  understand  the  present  difficulty.  It  is  sufficient 
for  you  at  present  to  know  that  whereas  the  pull  in  a  tie-bar 
tends  to  make  it  straighter  if  possible,  the  push  in  a  strut  tends 
to  make  it  bend.  Hence,  in  an  iron  railway -bridge  or  rooj 
you  will  see  that  the  tie-bars  are  thin  solid  rods  usually,  and  they 
might  be  chains  or  ropes  ;  but  the  struts  must  not  merely  have 
a  proper  area  of  cross-section,  this  cross-section  must  also  be 
wide  in  every  direction.  Thus,  instead  of  a  solid  cast-iron 
column  you  always  see  a  hollow  one,  unless  the  column  is  very 
short.  Also,  a  thin  plate  of  iron  suffices  for  the  lower  boom 
or  flange  of  a  railway-girder  (because  it  resists  a  pull),  whereas 

the  top  boom  is  a  hollow  tube,  or  is  U-,  or  f|-,  or  i i-shaped, 

because  it  must  resist  a  push.  Long  struts,  therefore,  must  be 
considered  in  Chap.  XXI.,  after  we  have  investigated  the  bend- 
ing of  beams. 

If,  however,  we  prevent  bending,  the  laws  for  stiffness  and 
strength  of  long  struts  are  as  simple  as  those  of  tie-bars.  Thus 


APPLIED    MECHANICS.  299 

the  great  struts  of  the  Forth  Bridge  consist  merely  of  four  angle 
irons  arranged  as  the  four  parallel  edges  of  a  square  prism,  and 
they  are  fastened  to  one  another  laterally  by  very  light  bracing, 
whose  simple  function  is  to  prevent  the  bending  of  the  angle 
irons.  The  strength  and  stiffness  of  such  a  strut  are  to  be 
calculated  from  the  combined  sections  of  the  angle  irons. 

In  a  short  strut  or  a  long  strut  prevented  from  bending  (in 
Art.  373  it  will  be  shown  that  the  lateral  constraints  are  called 
upon  to  exert  only  very  small  forces  to  prevent  bending),  the 
load  per  square  inch  is  called  the  stress.  The  shortening  is  a 
fraction  of  the  whole  length  of  the  strut,  and  this  fraction  is 
called  the  strain.  You  will  find  from  your  experiments  that 
the  strain  is  proportional  to  the  stress.  Thus  for  wrought-iron 
struts  or  columns  the  compressive  stress  =  the  compressive 
strain  x  29,000,000.  The  multiplying  number  is  found  to  be 
the  same  for  the  same  material,  whether  it  resists  a  push  or  a 
pull.  This  number  is  called  "  Young's  Modulus  of  Elasticity  "; 
it  has  been  measured  for  various  materials,  and  is  given  in 
Table  XXII.,  p.  658.  In  using  it  you  must  remember  that 
the  stress  is  in  pounds  per  square  inch. 

Exercise  1. — By  how  much  would  a  round  bar  of  steel,  120 
feet  long,  whose  diameter  is  2  inches,  lengthen  with  a  pull  of 
30  tons  ?  Answer  :— 0-0855  foot. 

Exercise  2. — By  how  much  would  a  column  of  oak,  7  feet 
long  and  4  inches  square,  be  compressed  in  supporting  a  weight 
of  2  tons  1  Answer  :_0'0013  foot. 

244.  I  have  said  that  if  you  use  squared  paper  after  making 
your  experiments,  you  will  find  that  the  strain  is  proportional 
to  the  stress,  and  the  lengthening  of  a  tie  bar  is  proportional 
to  the  total  pulling  force.  But  you  will  find  that  this  law  is 
not  true  when  the  loads  become  too  great.  If  your  loads  are 
less  than  about  a  quarter  of  the  breaking  load,  you  will  find  on 
removing  them  that  the  wire  on  which  you*  are  experimenting 
goes  back  to  its  original  length.*  But  if  your  loads  much 

*  It  may  not  go  back  at  once  to  its  old  length,  but  in  a  few  minutes  it 
will  be  found  exactly  where  it  was  before  you  loaded  it.  Similarly,  when 
the  load  is  put  on  there  is  first  a  sudden  lengthening,  and  after  this  there  is  a 
slight  extension  going  on  so  long  as  the  load  remains,  but  it  practically  comes 
to  an  end  in  a  few  minutes.  This  after- action,  or  "  creeping,"  is  so  slight  that  I 
have  not  till  now  spoken  about  it,  although  we  have  reason  to  believe  that  its 
investigation  would  be  of  great  importance.  This  "creeping,"  which  seems 
connected  with  internal  friction  or  viscosity,  is  absent  in  the  quartz  fibres  of 
Professor  Boys  within  a  lai'ge  range  of  stress,  and  it  is  so  in  soft  metals  within 
a  small  range  ;  and  this  is  curious,  because  there  w  much  of  it  in  glass ;  and 


300 


APPLIED    MECHANICS, 


exceed  this  amount  it  will  be  found  that  the  wire  has  taken  a 
permanent  set ;  that  is,  if  you  remove  the  load  the  wire  will 
not  go  back  to  its  original  length.  It  remains  permanently 
longer  than  it  originally  was,  and  we  say  that  we  have  exceeded 
the  limits  of  elasticity.  The  load  which  produces  this  per- 
manent set  is  said  to  be  the  measure  of  the  elastic  strength  of 
the  wire,  for  although  it  does  not  break  the  wire  it  alters  it 
permanently.  Now  it  is  only  for  loads  less  than  this  that  the 
law  "  strain  is  proportional  to  stress  "  is  true.  Your  squared 
paper  for  experiments  on  a  steel  wire  would  give  a  straight  line 
becoming  a  curve,  like  Fig.  172. 

When  you  plot  your  results,  making  the  distance  m  n 
represent  the  extension  of  the  wire 
for  a  load  represented  by  the  dis- 
tance m  q,  to  any  scale  you  please, 
you  will  find  that  the  line  passing 
through  your  points  is  straight 
only  from  o  to  M,  say,  and  then 
it  curves.  The  distance  Q  M  repre- 
sents the  load  which  produces 
permanent  set.  For  greater  loads 
than  this,  the  extension  is  more 
than  proportional  to  the  load,  and 
increases  more  rapidly  until  we 
get  at  P  K  a  very  rapid  extension 
indeed,  and  the  elongations  there- 
after are  so  great  that  they  cannot 
.  be  represented  here  on  the  same 
*  scale  as  the  part  o  M.  P  is  called 
the  yield  point,  and  the  pheno- 
menon is  very  marked  in  wrought  iron  and  other  ductile 
materials.  It  seems  always  somewhat  higher  than  M,  the  true 
elastic  limit.  A  student  must  obtain  results  up  to  the  breaking 

certainly  within  the  limits  of  permanent  set  there  is  creeping  in  brittle  sub- 
stances, such  as  steel,  giving  trouble  in  measuring-instruments.  If  we  say  that 
there  is  perfect  elasticity  when  the  same  forces  are  permanently  required  to 
keep  a  body  in  any  particular  shape  at  the  same  temperature,  a  body  may  be 
perfectly  elastic,  and  yet  it  may  have  viscosity  (the  term  viscosity  indicates 
an  elasticity  which  is  affected  by  the  rate  of  change  of  strain),  and  disobey 
Hooke's  law.  But  there  is  imperfect  elasticity  if  the  forces  are  dependent 
not  only  on  the  shape  of  the  body,  but  on  its  previous  shapes,  its  past  history. 
The  only  perfectly  definite  limits  of  elasticity  in  nature  seem  to  be  those  at 
which,  at  constant  temperature,  a  vapour  becomes  a  liquid  and  a  liquid 
becomes  a  solid,  or  a  vapour  becomes  a  solid. 


Fig.  172. 


APPLIED    MECHANIC.  301 

of  a  wire  for  himself,  plotting  them  on  squared  paper ;  noting 
the  rapid  yield  after  P,  then  the  much  slower  but  still  rapid 
elongation,  becoming  again  rapid  as  the  specimen  gets  near 
breaking.  The  material  is  plastic  after  M.  He  will  note  that 
published  diagrams  are  not  of  great  value  unless  we  know  the 
rate  at  which  the  load  was  increased.  After  P  is  reached  the 
specimen  will  continue  to  increase  in  length  with  time,  even  when 
the  load  is  not  increased,  and  even  when  the  load  is  diminished. 

245.  These  great  plastic  elongations  are  permanent  (except 
for  the  very  small  elastic  part),  and  are  very  different  from  the 
great  non-permanent  compression  of  cork,  or  the  great  extension  or 
shortening  with  nearly  constant  volume  of  indiarubber.  When  there 
is  continuous  yielding  under  constant  forces,  the  solid  body  behaves 
like  a  fluid,  flowing  (as  lead  does  easily)  without  much  change 
of  density.      It  seems  from  M.  Tresca's  experiments  that  when  a 
round  punch  comes  down  upon  a  plate  of  lead  the  lead  flows 
rapidly  laterally  from  underneath  the  punch  into  the  rest  of  the 
plate,  until  the  shearing  suiiace  is  considerably  diminished,  because 
the   wad  is  much  less  thick  than  the  plate  from  which  it  is 
punched.     In  the  squeezing  of  metals  there  is  local  flow  wherever 
the  pressure  is  great. 

246.  A  fluid  is  defined  as  that  which  greatly  changes  its  shape 
under  the  action  of  indefinitely  small  forces.     This  is  why  we  call 
many  substances   (such  as  sealing-wax  and  pitch)  fluids.     Even 
their  own  weights,  acting  long  enough,  cause  them  to  flow.     Metal 
statues  and  brackets  several  thousands  of  years  old  prove  that  metals 
are  not  fluids.     Experimental  evidence  of  what  stresses  infinitely 
long  continued  will  just  cause  metals  to  flow  is  still  wanting.     Mr. 
Bottomley  found  that  seemingly  similar  wires  seemed  to  greatly 
increase  in  their  strength  if  the  increase  of  load  was  very  gradual 
indeed.     The  increase   continuing  in  one  case  for  a  month,  the 
strength  seemed  to  be  greater  by  27  per  cent,  than  on  the  specimen 
broken  in  the  ordinary  way. 

247.  When  after  the  yield-point,  a  load  is  left  upon  a  specimen,  the 
amount  of  hardness  and  increase  of  strength  produced  are  increased 
by  leaving  on  the  load  for  a  longer  time.     Even  when  the  load  is 
not  left  on,  if  the  specimen  is  left  unloaded  for  some  time,  it  seems 
to  become  harder  and  considerably  stronger  (luring  the  interval  of 
rest.     In  both  these  cases  by  hardness  I  mean  a  high  yield-point. 
As  a  matter  of  fact,  the  Young's  modulus  seems  to  diminish  4  to  9 
per  cent,  as  it  gets  what  I  now  call  harder,  and  annealing  increases 
it  again.     The  material  below  the  yield-point,  although  seeming 
very  elastic,  exhibits  time  plasticity.    The  nature  of  the  change  in 
the  character  of  these  tensile  and  compressive  phenomena  when 
other  stresses  (such  as  twisting  stresses)  are  acting  is  not  yet  well 
known ;  but  there  may  be  an  apparent  alteration  in  these  limits, 
as  deduced  from  the  bending  of  beams  or  the  twisting  of  shafts 
which  have  initial  strains.     Thus  James  Thomson  showed  that  a 
shaft   of    ductile  material  might  be  twisted  until  most  of  the 


302 


APPLIED   MECHANICS. 


material  had  taken  a  set.  The  shaft,  if  now  examined,  would  be 
found  to  take  a  set  very  much  more  readily  with  the  reversed  kind 
of  twisting  couple  than  with  one  of  the  kind  which  was  used  to 
give  it  its  set.  To  understand  this  properly,  a  student  ought  to 
make  a  diagram  showing  the  probable  stress  everywhere  in  the 
shaft  when  left  to  itself.  (See  Appendix.) 

248.  Moduli  of  elasticity  are  slightly  altered  by  manufacturing 
processes.  Lord  Kelvin  found  copper  and  iron  wire  to  alter  5  per 
cent,  in  rigidity  when  subjected  to  great  stretching.  Very  much 
greater  changes  were  produced  in  the  Young's  moduli  by  excessive 
twisting.  Lord  Kelvin  foiind  that  good  copper  wire,  annealed  by 
being  heated  to  redness  and  sudden  cooling  in  cold  water,  had  a 
modulus  of  rigidity  6'71  x  10°  Ibs.  per  square  inch.  The  same  kind 
of  wire  heated  to  redness  and  cooled  slowly,  so  that  it  was  brittle, 
had  a  modulus  5-58  x  106,  its  density  having  diminished  2f-  per  cent. 

It  is  interesting  to  note  that  a  watch  goes  faster  and  faster  for 
some  time  after  it  is  made,  but  at  the  end  of  some  months  the 
balance-spring  settles  down  into  a  state  which  does  not  much 
change  afterwards.  In  this  state,  then,  its  elasticity  is  greater 
than  it  was  in  the  beginning.  The  springs  of  chronometers  are, 
however,  often  laid  aside  as  useless  after  a  few  years'  service,  their 
elastic  condition  having  altered  so  much  since  the  beginning  that 
they  have  to  be  replaced. 

Young's  modulus  sometimes  diminishes  and  sometimes  increases 
with  temperature  ;  but  more  experiments  are  needed.  Wertheun's 
results  are  given  here  : — 

TABLE  III. 


Young's  Modulus  in  Millions  of  Ibs. 

per  Square  Inch. 

Metal. 

Specific  Gravity. 

At  15°  C. 

At  100°  0. 

At  200°  C. 

Lead 

11-232 

2-46 

2-32 

Gold          

18-035 

7-9 

7-5 

7-8 

Silver        ...         

10-304 

10-1 

10-3 

9-0 

Palladium            ...         ... 

11-225 

13-9 

Copper      ...         

8-936 

14-9 

13-3 

11  -2 

Platinum  ...         

21-083 

22-0 

20-1 

18-4 

Steel,  Drawn,  English  ... 

7-622 

24-6 

30-3 

27-3 

Cast  Steel            

7-919 

27-7 

27-0 

25-4 

Iron          

7'757 

29-6 

31-1 

25-2 

I  cannot  quite  believe  these  results,  because  I  have  never 
known  any  specimen  of  steel  to  have  a  less  E  than  28  x  106.  They 
are  reduced  from  the  numbers  as  selected  by  Lord  Kelvin  from 
Wertheim's  original  "Memoires";  and  I  must  say  that  Wertheiin's 
experiments  were  carefully  carried  out. 


APPLIED    MECHANICS.  303 

The  moduli  of  rigidity  of  iron  copper,  and  brass,  according  to 
Kohlrausch,  diminish  with  temperature.  The  cubic  modulus  of 
water  increases  15  per  cent,  for  53°  C.  rise  of  temperature ;  of 
alcohol  and  ether,  it  diminishes  with  temperature.  The  Young's 
modulus  of  indiarubber  increases  with  temperature.  The  con- 
clusions which  may  be  drawn  by  thermodynamic  reasoning  from 
facts  like  these  are  interesting. 

249.  Some  metals  are  believed  to  become  brittle  at  low  tempera- 
tures, much  as  "cold-short"  iron  or  steel  containing  phosphorus  is 
brittle.   In  many  cases  there  may  be  no  proper  foundation  for  this  be- 
lief.   There  is  a  slight  increase  of  strength  in  ordinary  iron  and  steel 
from  low  temperatures  to  near  200°  C.     Above  about  300°  0.  there 
is  a  great  lowering  of  strength  produced  by  rise  of  temperature. 
It  is  believed  that  iron  and  mild  steel  worked  at  a  "blue  heat" 
(lower  than  red  heat)  become  more  deteriorated  (brittle,  or  with  a 
tendency  to  become  brittle  afterwards)  than  if  worked  cold  or  at  a 
red  heat. 

250.  Loss  of  Energy  in  Change  of  Strain. — Even  in  fluids 
there  must  be  loss  of  thermodynamic  energy  in  change  of  volume, 
because  of  change  of  temperature.     But  the  loss  of  energy  in  the 
change  of  shape  of  solids  depending  on  the  rapidity  of  change,  is 
much  greater  than  can  be  accounted  for  thermodynamically  or  by 
motions  of  the  outside  air ;  and  it  must  be  put  down  to  internal 
friction  or  viscosity.     It  is  very  marked  in  zinc,  in  indiarubber, 
and    in    jellies.      The    reasons   for  the  viscosity   shown   in   the 
distortion  of  gases  and  liquids  are  known  to  us.     In  twisting  and 
untwisting  wires,  Lord  Kelvin  found  (1)  that  the  viscosity  does  not 
seem  to  be  proportional  to  the  velocity,  so  that  the  law  is  different 
from  that  of  the  viscosity  of  fluids.      (2)    Tension  in  the  wire 
produces  a  non-permanent  increase   in   the  viscosity.      (3)    The 
viscosity  at  any  time  depends  upon  the  history  of  the  specimen. 
The  viscosity  of  a  wire  on  Saturday,  after  a  week's  experimenting, 
was  greater  than  when  on  Monday  the  experiments  were  resumed 
after   a   Sunday's   rest.      Young's   modulus   in   iron   and   copper 
generally  diminishes   (in  bronze  it  increases)   with  repetition  of 
loading  if  there  is  little  pause.     The  effect  is  not  so  marked  if 
there  is  a  pause. 

251.  Strength.— Table  XXII.,  p.  658,  shows,  among  other 
things,  the  pulling  (or  tensile)  and  pushing  (or  compressive) 
stress  which  a  material  will  bear  before  breaking.  Probably  if 
these  stresses  were  allowed  to  act  on  the  material  for  some 
time  it  would  break  even  if  they  were  not  added  to.  They  are 
obtained  from  experiments  in  which  the  load  was  increased 
pretty  quickly,  and  yet  quietly — that  is,  without  any  jerking  or 
sudden  action.  The  numbers  in  the  table  are  taken  from  many 
sources,  and  must  in  general  only  be  regarded  as  giving  rough 
average  values.  Ultimate  crushing  stress  is  sometimes  badly 
denned,  because  the  materials  behave  as  if  gradually  becoming 
plastic.  Rolled  iroii  gets  a  fibrous  structure,  and  may  not  have 


304  APPLIED    MECHANICS. 

the  same  strengths  in  all  directions.  Rolled  ntccl  is  mom 
uniform.  The  smallest  stress  per  square  inch  which  will 
produce  a  permanent  set  in  the  material  is  sometimes  called  the 
elastic  strength.  The  working  stress  is  usually  a  fraction  oj 
this;  it  is  the  stress  which  experience  tells  us  to  calculate 
upon  for  loads  acting  for  a  long  time  on  materials,  and  which 
we  shall  be  sure  are  perfectly  safe  in  the  case,  of  such 
materials  as  are  supplied  from  foundries  and  forges. 

Exercise  1. — How  great  a  pull  will  a  round  rod  of  brass 
stand  before  it  breaks,  if  its  diameter  is  03  inch  t  What  pull 
would  produce  in  it  a  permanent  set,  and  what  is  the  safe 
working  pull  ?  Answers  :— 1,237 ;  484  ;  and  254  Ibs. 

Exercise  2. — A  short  hollow  cylindric  column  of  cast  iron 
is  8  inches  in  outer  diameter,  5  inches  inner  diameter.  What 
is  the  safe  load  and  what  load  will  produce  permanent  set  ? 
Answer: — The  area  of  cross-section  is  4  x  4  x  3*1416  minus 
2-5  x  2-5  x  3-1416,  or  30-63  square  inches;  30-63  x  21, 000  is 
643,230  Ibs.,  or  287  tons;  30-63  x  10,400  is  318,522  Ibs.,  or 
142  tons. 

252.  In  making  calculations  on  the  stiffness  and  strength  of 
structures  we  assume  that  strains  are  proportional  to  stresses. 
To  use  in  these  calculations  the  results  of  tests  on  materials 
beyond  the  elastic  limit  may  be  unscientific,  but  this  is  what 
we  do  :  employing  a  factor  of  safety,  which  is  the  ultimate 
strength  divided  by  the  working  strength.  That  is,  we  say 
that  we  may  use  a  certain  working  load  on  a  beam  because  we 
find  the  pushing  or  pulling  forces  which  it  produces  in  the 
various  parts  of  the  structure  ;  and  each  of  these  is,  say,  one 
quarter  of  the  pushing  or  pulling  force  which  would  break  that 
particular  bar ;  here  the  factor  of  safety  is  four.  We  do  not 
mean,  or,  rather,  we  ought  not  to  mean,  that  we  might  put  four 
times  our  working  load  upon  the  structure  before  breaking, 
because  we  have  no  theory  of  what  would  occur  in  the  structure 
if  the  elastic  limit  of  one  or  more  bars  were  passed.  The  factor 
of  safety  allows  also  for  contingencies  ;  there  is  the  risk  of 
defects  in  spite  of  precautions  and  the  possible  deterioration  with 
time;  sometimes  a  large  factor  is  used  because  we  suspect 
inaccuracy  in  our  theory  or  in  our  estimated  loads,  in  unfore- 
seen causes  of  shock  and  fatigue.  It  will  be  evident  from  this 
and  from  what  follows  that  great  judgment  is  needed  in  fixing 
a  factor  of  safety,  and  the  following  list  must  be  looked  upon 
merely  as  a  guide  in  setting  academic  problems  : — 


APPLIED    MECHANICS. 
TABLE  IV. 


305 


Steady  Load. 

Varying  Load. 

Structures  Subjected 
to  Shocks. 

Wrouglii  Iron 
Cast  Iron 

and  Steel 

H 

5 

5  to  8 
6  to  10 

10  to  12 
15 

Timber  

7 

10  to  15 

20 

Brickwork  and 

Masonry 

20 

30 

... 

Live  loads  are  usually  doubled  and  added  to  the  dead  loads. 
These  numbers  are  sometimes  called  "  factors  of  our  ignorance." 
The  Board  of  Trade  in  1858  adopted  the  rule  that  in  bridges 
the  stress  on  wrought  iron  must  not  exceed  5  tons  per  square 
inch,  and  later  for  steel  6J  tons  per  square  inch.  We  know 
now  from  Wohler's  experiments  that  the  range  of  stress  is  very 
important.  In  the  Conway  bridge,  in  which  the  range  is  not 
great  because  the  live  load  is  small  compared  with  the  dead  load, 
the  stresses  are  as  much  as  6  tons  to  the  square  inch. 

253.  The  load  per  square  inch  on  a  tie  or  strut  is  the  stress.  It 
is  obvious  that  as  the  section  of  a  tie-bar  becomes  smaller,  the  stress 
is  really  the  load  per  square  inch  of  the  diminished  section.  The 
change  of  area  of  section  in  our  ordinary  constructive  materials  is 
so  little  that  we  usually  consider  the  stress  to  be  load  -7-  original 
area  of  section,  as  this  is  very  convenient.  But  the  section  of  a 
bar  may  alter  very  greatly  when  being  fractured ;  and  although 
no  experimenter  yet  seems  to  have  stated  it,  I  am  sure  that  a  great 
deal  of  useful  information  is  lost  because  we  do  not  plot  tensile 
strain  and  actual  tensile  stress  up  to  the  breaking-point.  It  is- 
somewhat  difficult  to  measure  the  section  with  great  accuracy,  and 
indeed  within  the  elastic  limits  the  measurement  is  nearly  im- 
possible ;  but  it  ought  to  be  attempted  above  the  yield-point, 
because  the  section  alters  sometimes  50  per  cent,  before  fracture. 
The  stress  seems  to  increase  very  much  before  fracture. 

In  indiarubber  the  alterations  in  shape  with  good  elasticity  are 
so  very  great  in  comparison  with  what  we  find  in  ordinary 
substances  that  it  is  absolutely  necessary  to  measure  the  real 
stress  for  each  load  and  length.  Students  ought  to  do  this  for 
themselves,  as  observations  even  roughly  made  give  instructive 
results.  It  is  to  be  remembered  that  in  gases  the  cubical  increase 

of  stress  Sp  produces  the  cubical  strain ;  and  as  elasticity  =^ 

stress  -T-  strain,   the    elasticity  of    a  gas  is  —  v~ .      The  actual 
volume  at  the  time  is  to  be  taken.     If  we  define  Young's  modulus 


306  APPLIED    MECHANICS. 

of  the  material  as  l~  where  I  is  length  of  tie  or  strut,  and  p  is  the 

load  divided  by  the  actual  cross-section,  we  find  that  for  india- 
rubber  it  is  fairly  constant  for  loads  that  shorten  the  specimen  16 
per  cent,  or  lengthen  it  50  per  cent.  The  most  striking  thing  is 
the  rather  rapid  increase  of  the  resistance  to  elongation  in  the 
attenuated  specimen  before  it  breaks. 

254.  In  some  simple  structures  the  value  of  a  material  is  taken  to 
depend  upon  the  amount  of  change  of  shape  which  it  may  undergo 
before  it  breaks;   and  commercial  tests  often  consist  of  definite 
extreme  bending  or  twisting,  or  the  delivery  of  definite  blows  to 
definite  specimens. 

255.  At  first  the  contraction  of  section  occurs  pretty  uniformly 
along  the  specimen,  but  as  fracture  approaches,  the  contraction  (ex- 
cept in  hard  metals,  such  as  cast  iron  and  hard  steel)   becomes 
localised.     Hence  this  contraction,  and  not  the  lengthening  of  the 
whole  specimen,  ought  to  be  taken  as  a  measure  of  the  ductility. 
When  the  stretching  proceeds  slowly,  and  there  are  intervals  of 
rest,  there  is  greater  uniformity,  places  of  great  yielding  harden, 
and  the  yield  takes  place  afterwards  elsewhere.     But,  again,  there 
have  been  cases  where  the  loading  was  so  very  rapid  that  almost 
no   local   excess  of   yielding   occurred,  and  the   whole   specimen 
yielded  fairly  evenly,  the  specimen  extending  about  50  or  60  per 
cent.,  or  about  double  what  occurred  in  a  similar  specimen  tested 
in  the  ordinary  way. 

256.  It  is  interesting  to  examine  the  surface  of  the  fracture, 
noting  its  texture.     The  fracture  surface  may  be  a  flat  or  an  oblique 
cross-section  (see  Art.  290).     In  ductile  material  we  often  find  a 
combination  of  the  two  kinds,  sometimes  taking  the  form  of  conical 
and  flat  surfaces.     In  compression  in  short  specimens  the  usual 
way  of  applying  load  prevents  by  friction  the  lateral  enlargement 
of  the  ends.     The  material  takes  a  barrel  shape,  and  can  withstand 
enormous  loads,  because  the  internal  strain  is  greatly  of  the  nature 
of  cubical  strain,  and  it  is  only  shear  strain  that  seems  to  cause 
fracture  (see  Art.  291). 

257.  Art.  290  shows  that  the  shear  stress  in  a  tie-bar  or  strut  is 
greatest  on  a  plane  making  45  degrees  with  the  axis.     But  we  do 
not  find  that  oblique  sections  of  fracture  are  inclined  at  this  angle, 
possibly  for  the  reason  there  stated.      Tensile   stress  seems  to 
diminish  and  compressive  stress  to  increase  the  resistance  to  shear- 
ing.    When,  as  in  blocks  of  cast  iron  which  are  not  more  than 
half  as  high  again  as  the  width  of  base,  there  is  no  room  for  one 
shearing  surface  at  the  angle  preferred  by  the  material,  the  block 
fractures  at  a  number  of  surfaces,  cones  and  wedges  being  formed. 

258.  Attention  ought  to  be  paid  to  the  fact  that  local  strengthen- 
ing and  stiffening  of  a  structure  may  produce  general  weakness. 
When  a  hole  is  punched  in  an  iron  plate  there  is  a  local  hardening, 
and  probably  strength-increase  in  the  material  round  the  hole ;  but 
unless  the  plate  is  annealed  or  (as  a  partial  improvement)  the  hole 
is  rhymered  out  to  remove  the  hard  part,  when  load  is  applied  the 
hard  part  yields  less  readily  than  the  rest,  gets  an  undue  share  of 
load,  fractures,  and  then  the  softer  material  fractures. 


APPLIED    MECHANICS.  307 

259.  Live  Loads. — When  a  weight  is  suddenly  applied  to 
stretch  a  wire,  it  produces  greater  effects  than  when  slowly  and 
quietly  applied.  We  know  the  reason.  A  weight  which,  slowly 
applied,  would  produce  an  extension  of  1  inch,  will,  when  put 
on  and  let  go,  produce  an  extension  of  2  inches.  The  wire 
now  shortens  to  its  original  length,  then  extends  2  inches,  and 
continues  to  get  shorter  and  longer,  the  weight  vibrating.  As 
there  is  friction  of  some  kind  among  the  particles  of  the  wire, 
and  there  is  also  external  friction,  the  lengthenings  and 
shortenings  gradually  lessen  till,  in  a  short  time,  the  wire 
settles  down  into  the  same  state  as  it  would  have  been  in  if 
the  load  had  been  slowly  applied.  Now,  if  we  suppose  this 
wire,  when  stretched  2  inches,  to  be  strained  just  beyond  its 
elastic  strength,  it  is  evident  that  the  suddenly  applied  load 
does  harm  ;  whereas  the  same  load  slowly  applied  would  do 
no  harm.  The  harm  is  greater  if  the  weight,  besides  being 
applied  suddenly,  is  moving  before  it  begins  to  act  on  the  wire. 
Take  the  case  of  a  stone  which  is  being  removed  by  means  of  a 
crane.  If  the  stone,  happening  to  fall  a  little,  be  brought  up 
by  the  chain,  the  increase  in  the  stress  on  the  chain  depends 
on  the  height  from  which  the  stone  has  fallen,  and  is 
greater  the  less  the  chain  is  extended.  When  a  wire  is 
lengthened  •!  foot  by  a  weight  of  1,000  Ibs.,  which  has  been 
increased  gradually,  we  know  that  the  pull  on  the  wire  began 
with  0,  and,  as  the  wire  gradually  extended,  the  pull  became 
greater,  till  it  is  now  1,000  Ibs.  The  average  pull  was  500  Ibs., 
and  500  x  •!,  or  50  foot-pounds,  is  the  total  strain  energy 
stored  up  in  the  wire.  If  we  wish  to  give  more  energy  to  the 
wire,  we  must  strain  it  more ;  and  this  is  just  what  we  do 
when  we  let  the  weight  fall  suddenly.  The  extra  strains  due 
to  loads  being  live  depend  upon  the  mass  which  we  set  in 
motion  in  applying  the  load.  In  some  railway  bridges  it  has 
been  found  that  the  increased  deflection  is  14  per  cent,  greater, 
and  it  is  usual  to  add  50  or  even  75  per  cent,  to  such  live 
loads  and  treat  them  as  dead. 

260.  The  energy  stored  up  in  any  strained  body  may  be  cal- 
culated if  we  know  the  stress  and  the  strain.  The  mainspring  of  a 
watch  contains  a  store  of  energy  which  is  gradually  given  out  by 
the  spring  in  returning  to  an  unstrained  condition.  Each  strained 
portion  of  the  spring  contains  a  portion  of  the  store,  and  if  at  any 
place  in  the  body  there  is  too  great  a  store  the  body  will  break 
there. 

II  w  is  the  proof  load  (or  load  which  will  just  fall  short  of 


308  APPLIED    MECHANICS. 

producing  permanent  set)  on  a  tie-bar  of  length  I  and  cross-section 
A,  the  stress  being  W/A,  the  strain  is  W/AE,  the  elongation  is  W//AE, 
and  the  work  done  as  the  load  is  increased  from  0  to  w  is  ^w2^/AE. 
This  is  called  the  resilience  of  the  bar.  Now,  the  volume  of  the 
bar  is  I  A,  and  hence  the  resilience  per  unit  volume  is  -£w2/A2E. 
Replacing  W/A  by/,  the  proof  stress  which  the  material  will  stand, 
we  havo  the  resilience  per  cubic  inch  to  be  /2/2E.  If/  is 
the  proof  stress  in  compression,  we  have  the  same  expression 
for  struts. 

When  a  torque  M  produces  an  angular  change  SB,  the  work 
done  is  M  .  80  ;  and  when  a  twisting  moment  M  has  been  gradually 
increased  from  0  to  M,  the  twist  of  a  shaft  increasing  from  0  to  0, 
the  work  done  is  £M0. 

When  there  is  a  shear  stress/  and  a  corresponding  shear  strain 
//N,  the  shear  strain  energy  per  unit  volume  is  /2/2x  ;  and  if  /  is 
the  proof  shear  stress,  /2/2N  is  called  the  resilience  of  the  material 
per  unit  volume. 

The  resilience  is  the  strain  energy  which  material  may  store 
before  permanent  set  takes  place.  It  is  evidently  /2  -r  2E  per 
cubic  inch  in  tie-bars  and  struts,  if  /  is  the  tensile  or  compressive 
stress  which  would  produce  permanent  set  and  E  is  Young's 
modulus.  In  shear  the  resilience  is  /2  -f  2N,  if  /  is  the  limiting 
elastic  shear  stress  and  N  the  modulus  of  rigidity.  When  the  stress 
is  not  uniform,  as  in  beams  and  shafts,  the  average  resilience  per 
cubic  inch  is,  of  course,  less.  Shocks  due  to  blows,  as  of  falling 
weights,  will  often  cause  the  strain  energy  to  exceed  the  amount 
which  the  material  will  stand,  and  local  set  and  plastic  yielding 
may  take  place.  Much  depends  upon  the  rate  at  which  strain 
energy  is  earned  off  to  the  rest  of  the  material. 

261.  Let  us  consider  why  a  chisel  cuts  into  an  iron  plate.  When 
I  strike  the  head  of  a  chisel  with  a  hammer  I  give  to  the  chisel  in  a 
very  short  period  of  time  a  certain  amount  of  energy.  This  energy 
is  transmitted  very  quickly  to  the  plate  through  the  edge  of  the 
chisel.  The  shorter  and  more  rigid  the  chisel,  the  more  quickly 
ia  the  energy  sent  through  the  cutting  edge  into  a  portion  of  the 
plate.  If  it  is  not  conveyed  away  rapidly  from  the  edge,  the 
amount  contained  in  a  small  portion  of  material  just  under  the 
edge  is  very  great,  and  the  material  is  fractured  there.  As  the 
energy  of  strain  is  proportional  to  the  product  of  stress  and  strain 
or  to  the  square  of  either  stress  or  strain,  the  possibility  of  fracture 
for  a  material  is  represented  by  the  square  root  of  the  strain  energy 
it  contains  per  cubic  inch.  If  a  material  is  brittle,  there  is  a  sort  of 
instability  which  causes  fracture  at  one  place  to  extend  to  all  neigh- 
bouring places.  And  hence,  if  we  deliver  with  great  rapidity  to  a 
gmall  portion  of  such  a  material  a  moderate  supply  of  energy,  it  is 
sufficient  to  produce  a  large  fracture.  As  our  material  becomes 
less  and  less  brittle,  we  must  have,  over  a  larger  and  larger  part  of 
the  volume  in  which  we  want  fracture  to  occur,  a  sufficient  supply 
of  strain  energy  delivered.  Hence,  in  cutting  wood,  we  use  a 
wooden  mallet  and  a  more  or  less  lengthened  wooden-headed  chisel. 
The  mallet  and  chisel  act  as  a  reservoir  for  the  energy  of  the  blow 
which  is  delivered  to  the  wood  from  the  edge  of  the  chisel  with 


APPLIED    MECHANICS.  309 

comparative  slowness  and  just  in  sufficient  quantity  to  cause 
rupture  in  front  of  the  edge.  If  the  wood,  without  gaining  in 
strength,  became  more  rigid  so  as  to  be  able  to  carry  off  more 
rapidly  the  energy  given  to  it  by  the  chisel's  edge,  it  would  be- 
necessary  to  make  the  supply  more  rapid  by  using  a  more  rigid 
chisel  and  mallet,  and  as  we  do  this  we  must  take  care  that  the 
chisel  itself  near  the  edge  is  strong  enough  to  resist  fracture.  This 
is  one  way  of  considering  the  effect  of  a  blow.  The  exact  mathe- 
matical consideration  of  what  occurs  in  the  impact  of  elastic 
bodies  is  not  easy  even  for  spheres  and  cylinders  and  other  bodies 
of  simple  shape  (see  Art.  404). 

262.  As  we  have  just  seen,  the  extra  stresses  due  to  loads 
suddenly  applied  are  easy  enough  to  understand.  It  is  not 
so  easy  to  comprehend  why  quietly  varying  loads  which  pro- 
duce no  visible  vibrations  should  produce  what  we  call  fatigue. 
The  ordinary  kinds  of  test  as  to  strength  under  statical  load, 
ductility,  or  elongation  before  fracture,  applied  to  old  rails, 
tyres,  and  other  well-used  material,  show  no  great  difference 
from  what  we  obtain  with  new  non-ductile  material.  Sometimes 
flaws  are  found,  and  it  may  be  that  fatigue  somehow  acts  in  pro- 
ducing minute  flaws.  The  nature  of  the  fracture  in  Wohler's 
experiments  is  the  same  as  that  observed  in  old  tyres  and  axles ; 
it  shows  no  signs  of  ductility,  and  is  as  if  the  material  were 
brittle. 


263.  A  piston-rod  is  subjected  to  tensile  and  compressive 
stresses,  often  repeated.  It  is  found  that  its  breaking 
strength  is  not  45,000  Ibs.  per  square  inch,  which,  let  us  say, 
it  would  be  for  a  steady  pull  or  push,  but  15,000  Ibs.  per 
square  inch.  If,  instead  of  such  an  action,  we  have  a  tensile 
stress  which  varies  frequently  from  30,000  Ibs.  per  square  inch 
to  zero,  the  rod  will  break  after  a  time.  In  the  same  way, 
steel  which  will  bear  a  steady  stress  of  84,600  Ibs.  per  square 
inch  will  only  bear  46,500  Ibs.  per  square  inch  if  the  stress 
varies  between  this  and  zero,  but  is  always  of  the  same  kind ; 
whereas,  it  will  only  bear  25,000  Ibs.  per  square  inch  if  the 
stress  is  sometimes  a  pull  of  this  amount  and  is  sometimes  a 
push  of  the  same  amount. 


The  above  statement,  the  outcome  of  Wohler's  experiments 
begun  twenty-five  years  ago,  and  Fairbairn's  experiments  made  very 
much  earlier,  was  made  fifteen  years  ago  in  the  first  edition  of  this 
book.  The  experiments  made  since  give  results  of  much  the  same 
kind.  In  the  following  table  we  have  the  most  important  results 
arrived  at  up  to  the  present  time,  being  the  stresses  in  tons  which 
require  from  five  to  ten  millions  or  an  indefinitely  large  u.umber 
of  applications  of  the  load  to  cause  fracture. ;— 


310 


APPLIED    MECHANICS. 
TABLE   V. 


Material. 

lit 

Similar  Stresses. 

One  Stress 
Zero. 

Opposite  Stresses. 

Least. 

Greatest. 

Least. 

Greatest. 

Least. 

Greatest. 

Wrought  Iron. 

22-8 

12-0 

20-5 

0 

15-3 

-   8- 

+   8-6 

Krupp's     Axle 

I 

Steel  

52-0 

17-5 

37-8 

0 

26-5 

-14-1 

+  14-1 

a 

Untempered 

g 

Spring  Steel  . 

57-5 

12-5 

34-8 

0 

25-5 

-13-4 

+  13-4 

?~ 

Iron  Plate     ... 

22-8 

11-4 

19-2 

0 

13-1 

-  7-2 

+  7-2 

Bar  Iron 

26*6 

13-3 

22-0 

0 

14-4 

-   7'9 

+  7-9 

Bar  Iron 

26-4 

13-2 

21-9 

0 

15-8 

-  8-7 

+  8-7 

b 

Bessemer  Mild 

bO 

Steel  Plate  ... 

28-6 

14-3 

23-8 

0 

15-7 

-   8-6 

+  8-6 

.a 

Steel  Axle     ... 

40-0 

20-0 

32-1 

0 

19-7 

-  10-5 

-flO-5 

| 

Steel  Rail      ... 

39'0 

19-5 

30-9 

0 

18  4 

-  9-7 

+   9-7 

g) 

Mild       Steel 

i—  i 

Boiler  Plate  . 

26-6 

13-3 

22-6 

0 

158 

-  8-7 

+  87 

This  exceedingly  great  weakening  in  material  due  to  fatigue 
seems  almost  as  if  it  had  been  vaguely  known  to  English  engineers 
from  the  beginning,  and  justifies  the  larger  factors  of  safety  which 
were  wisely  used  in  this  country  fifty  years  ago  in  railway  bridges. 
Professor  James  Thomson  showed  that  the  two  elastic  limits  ought 
to  be  called  inferior  and  superior,  as  they  are  not  necessarily  equal 
positive  and  negative,  but  might  even  both  be  on  the  same  side  of 
the  zero  if  the  overstraining  were  great  enough,  and  that  variable 
displacements  outside  these  limits  would  produce  a  destructive 
succession  of  sets.  This  theoretical  deduction  from  the  considera- 
tion of  his  overtwisted  shaft  has  been  completely  verified  by  the 
experiments  of  Bauschinger,  and  may  be  said  to  completely  explain 
the  phenomena  of  fatigue.  Bauschinger  found  that  the  elastic 
range  does  not  alter  much,  although  either  end  of  it  may  be 
altered,  even,  indeed,  nearly  to  the  ordinary  breaking  stress.  This 
change  of  either  limit  also  is  not  very  permanent,  altering  with 
hammering  and  other  violent  treatment.  (See  Appendix.) 

264.  We  can  only  refer  to  a  few  of  the  ways  in  which  our 
principles  are  applied.  When  bolts  tightly  screwed  up  fasten 
two  things — the  flanges  of  a  cylinder  cover,  for  example — the 
bolts  get  lengthened  and  other  things  (let  us  call  them  the 
cushion)  get  shortened.  If  extra  forces  are  now  applied  it  may 


APPLIED    MECHANICS.  311 

be  that  for  a  comparatively  small  extra  lengthening  of  the  bolts 
the  cushion  is  nearly  relieved  and  does  not  add  its  force  as  it 
did  previously  ]  whereas  in  other  cases,  when  the  cushion  is 
more  springy,  there  may  be  almost  no  relief,  and  the  old  forces 
due  to  the  cushion  may  act  along  with  the  new  forces.  Very 
often  this  initial  tightening  up  of  bolts  is  too  great,  and  we 
have  rule  of  thumb  methods  of  designing  sizes  based  upon  an 
experience  of  the  carelessness  of  workmen  which  it  is  difficult 
to  express  algebraically.  One  rule  based  on  experience  is  that 
the  length  of  a  spanner  shall  never  much  exceed  fifteen  times 
the  diameter  of  the  bolt.  Another,  that,  in  certain  kinds  of 
machinery,  bolts  of  less  than  a  certain  size  shall  not  be  used. 
Students  ought  to  make  careful  sketches  of  the  various  kinds 
of  bolts  and  nuts,  including  the  usual  forms  of  lock  nuts  and 
ways  of  locking,  and  also  other  kinds  of  fastening.  In  study- 
ing some  of  the  proportions  of  these  important  details  of 
machinery  our  theories  are  useful  in  suggestion ;  in  some 
hands  the  theory  is  only  a  snare.  Every  true  engineer  must 
respect  the  proportions  which  have  been  arrived  at  by  the  fit 
and  try,  the  failure  and  success  methods  of  generations  of 
engineers.  When  a  very  novel  thing  has  to  be  made  the  good 
engineer  sighs  for  practical  guidance,  and  he  is  very  cautious 
in  using  his  theory. 

265.  It  is  found  then  that  when  a  rod  is  pulled,  with  however 
small  a  force,  not  only  does  it  get  longer,  but  its  diameter  gets 
less.  When,  for  example,  a  rod  of  glass  is  pulled  so  that  its  lenglh 
increases  by  the  one-thousandth  of  itself,  it  is  found  that  its 
diameter  gets  less  by  the  one  three-thousandth  of  itself.  When  a 
strut  shortens,  it  also  swells  laterally.  The  ratio  of  the  lateral  to 
the  axial  strains  in  compression  or  in  tension  is  called  Poisson's 
ratio.  It  is  of  great  importance  in  the  theory  of  structures. 

The  nature  of  the  strain  in  a  wire  which  is  being  extended,  or 
in  a  column  which  is  being  compressed,  cannot  be  said  to  be 
simple.  If  all  lines  in  one  direction,  and  in  one  direction  only, 
became  shorter  or  longer,  the  strain  would  be  called  simple,  but  it 
needs  rather  a  complicated  system  of  external  compression  or 
extension  to  produce  this  effect.  No  matter  how  a  body  is 
strained,  if  we  consider  a  small  portion  of  it  we  shall  find  that, 
besides  angular  changes,  any  strain  simply  consists  of  extensions 
and  compressions  in  different  directions.  In  fact,  imagine  a  very 
small  spherical  portion  of  the  body  before  it  is  strained.  The 
effect  of  strain  is  to  convert  the  little  sphere  into  a  figure  called  an 
ellipsoid  (that  is,  a  figure  every  section  of  which  is  an  ellipse)  or  a 
circle.  Eemember  that  every  section  of  a  sphere  is  a  circle.  It 
may  be  proved  that  there  were  three  diameters  of  the  sphere  at 
right  angles  to  one  another,  which  remain  at  right  angles  to  one 


312 


APPLIED    MECHANICS. 


another  in  the  ellipsoid,  and  are  known  as  the  principal  axes  of  (he 
ellipsoid.  These  directions  are  now  called  the  principal  axes  of  the 
strain  existing  at  that  part  of  the  strained  body.  Along  one  of 
these  directions  the  contraction  or  extension  is  less,  and  in  another 
greater,  than  in  any  other  direction  whatever. 

Example.— Thus  if  M'N'    (Fig.  173)    is  part  of  a  long  wire 
subjected  to  a  pull,  the  portion  of  matter  which  was  enclosed  in 

the  very  small  imaginary 
spherical  surface  A  B  c  D  before 
the  pull  was  applied  is  now 
enclosed  in  the  ellipsoidal 
spherical  surface  A'  c'  B'  D'. 
The  sphere  has  become  an 
ellipsoid  of  revolution;  A  B 
becomes  A'  B',  c  D  becomes  c'  D'. 
The  strain  in  the  direction 

.     A'B'  —  AB        ,    .,  .     . 
A  B  is  — —        -    and  this  is 
AB       , 

equal  to  the  pull  in  the  wire 
F'g- 173.  per  square  inch  divided  by 

Young's  modulus  of  elasticity, 

IE.  As,  however,  it  is  often  more  convenient  to  use  a  multiplier 
than  a  divisor,  we  are  in  the  habit  of  using  the  reciprocal  of  E,  and 
•denoting  it  by  the  letter  a.  Thus,  if  the  pull  per  square  inch  is p 
pounds,  it  produces  a  strain  of  the  amount  pa.  in  the  direction 

A  B  ;  the   lateral   contraction  of  the   material  is  ,   and 

CD 

in  this  case  is  usually  denoted  by  pft,  the  ratio  ft/a  being  Poisson'a 
ratio. 

266.  In  the  following  exercises  on  struts,  the  load  is  sup- 
f  posed  to  be  carefully  applied  so  that  there  is  the  same  stress 
at  every  point  of  the  section  : — 

1.  Find  the  area  of  the  base  of  a  sandstone  column  carrying  a  dead 
i  load  of  6  tons.     Take  the  ultimate  crushing  stress  as  3,600  Ibs.  per  square 
:  inch,  arid  use  a  factor  of  safety  of  20.        Ans.,  75  square  inches  nearly. 

2.  Find  the  least  safe  sectional  area  of  a  short  cast-iron  strut  to  bear  a 
1  load  of  12  tons.     Take  the  safe  stress  as  15,000  Ibs.  per  square  inch. 

Ans.,  l-S  square  inch. 

3.  The  external  and  internal  diameters   of  a  short  hollow  cast-iron 
column  are  10  inches  and  8  inches  respectively.  If  the  safe  working  stress 
be  15,500  Ibs.  per  square  inch,  find  what  load  it  will  safely  bear. 

If  the  external  diameter  had  been  7  inches,  what  ought  the  thickness 
to  have  been  to  bear  a  load  of  100  tons  ?  Ans.,  196  tons  ;  -73  inch. 

4.  A  tie-rod   made   from  f-inch  wrought-iron  plate   has   to   sustain 
a  load  of  15  tons.  TVhat  should  be  its  width,  allowing  a  working  stress  of 
7,000  Ibs.  per  square  inch  ?  Ans.,  6'4  inches. 

5.  The  steam  pressure  in  a  locomotive  boiler  is   175  Ibs.  per  square 
inch ;    the  stay-bolts  which  connect  the  flat  sides  of  the  firebox  with  the 
end  plate  of  the  boiler  are  placed  4  inches  from  centre  to  centre,  vertically 
tand  horizontally.     \Yh$t  i?  t^e  tensile  force  in  each  stavrbolt,  and  what 


ALLIED    ME<5HAt>TlS&.  31 3 

niubt  be  the  diameter  of  each,  the  metal  not  being  subjected  to  a  greater 
stress  th;m  5  tons  per  square  inch  of  the  section  of  the  bolts  ? 

Ana.,  2,800  Ibs. ;  0'54  inch. 

6.  What  are  the  working  and  breaking  loads  cf  pianoforte  wire  TV  inch 
diameter  ?     And  if  the  wire  hangs  vertically  5  miles,  what  weight  at  the 
end  will,  together  with  the  weight  of  the  wire,  produce  the  working  stress 
at  the  top  ?    What  is  now  the  average  load  and  the  average  stress  in  the 
wire  ?     By  how  much  will  the  wire  lengthen  when  subjected  to  it  P 

Am.,  297  Ibs. ;  1,040  Ibs. ;  27  Ibs. ;  162  Ibs. ;  16,460  Ibs.  per  sq.  in. ;  20  yds. 

7.  A  strut  100  feet  long  is  made  up  of  four  angle  irons  of  wrought 
iron  (3|  +  3|)  \,  which  are  prevented  from  bending.     Find  the  ultimate 
proof   and   working   loads.      How   much   shortening    occurs  under   the 
working  load?  Ans.t  working  load  91,000  Ibs. ;  0'28  inch. 

8.  A  piston-rod  of  mild  steel  4  inches  diameter,  6  feet  long,  the  piston 
30  inches  diameter ;  what  maximum  pressure  of  steam  may  be  used  (1)  if 
the  engine  is  double  acting;    (2)  if  the  engine  is  single  acting?      (See 
Art.    263.)       What    lengthening    and    shortening    occur    under    these 
pressures?  Am.,  53  Ibs. ;  160  Ibs. ;  (1)  0-08  inch  ;  (2)  0-12  inch. 

9.  Columns  about  8  feet  high  of  brickwork,  sandstone,  and  granite,  14 
inches  square ;  what  working  loads  will  they  carry  ?    If  they  are  6  feet 
apart,  and  carry  a  wall  of  brickwork  28  inches  thick,  to  what  heights 
may  it  be  carried?     The  brick  columns  being  replaced  by  cast-iron 
cylinders  6  inches  diameter,  what  ought  to  be  their  thickness?    How 
much  do  they  shorten  under  the  load  ? 

Am.,  15|  tons  each;  30ft.;  0' 12  inch. 

10.  The  tight  side  of  a  gearing- chain  taking  a  pull  p,  let  it  act  at  5 
inches  from  the  centre  of  the  wheel,  and  transmit  20  horse-power  at  100 
revolutions  per  minute ;  find  p.     Each  of  a  pair  of  links  takes  this  pull. 
If  the  thickness  of  a  link  is  one-third  of  the  diameter  of  the  pin,  and  if 
its  breadth  is  two  and  a  half  times  the  diameter  of  the  pin,  find  the  section 
of  each  link.     It  is  a  good  exercise  to  design  the  chain. 

Ans.,  5,044  Ibs. 

11.  A  single  (a  little  over  0'2  inch  thick)  leather  belt  will  stand  an 
average  pull  of  1,000  Ibs.  per  inch  of  its  width.     The  weakness  of  the 
average  fastener  reduces  this  to  about  300  Ibs.  per  inch  x>f  width,  and  it  is 
usual  to  take  66  Ibs.  as  a  working  load.     The  pull  on  the  tight  side  of  a 
belt  is  two  and  a  half  times  that  on  the  slack  side.     The  pulley  is  30 
inches  diameter,  150  revolutions  per  minute ;  find  the  breadth  of  a  single 
belt  to  transmit  5  horse-power.     (See  also  the  exercises  Art.  185). 

.     Ans.,  3 '5  inches. 

12.  Columns  of  different  material,  constrained  to  keep  of  the  same 
length,  of  cross-sections  AX,  A2,  A3,  etc,  of  coefficients  of  expansion  ol5  o^, 
a.,,  etc.,  unstressed,  at  a  particular  temperature,  are  raised  t  degrees  in. 
temperature.  What  is  the  fractional  elongation?  And  what  are  the? 
stresses  in  them  ? 

Ans.,  If  unconstrained,  their  fractional  elongations  would  be  ta^. 
tea,  etc.  If  x  is  the  real  elongation,  each  has  a  compressive  strain, 
tei  —  x,  ta^  —  x,  etc.  Their  compressive  stresses  are  EI  (t^  —  a?),, 
E2  (ta^  —  »),  etc.,  if  EJ,  E3,  etc.,  are  their  Young's  moduli,  and  the< 
total  push  in  each  is  AjEx  (tai  -  x},  A2E2  (ta^  —  x),  etc.  Thei 
Sum  of  these  pushes  is  o  (some  of  them  being  negative ;  that  is^, 
tensile  forces),  or  A^  (ta^  -  a;)  +  AgEo  (ta^  —  ar)  +  etc.  =  o,  or 


314  APPLIED    MECHANICS. 


Aj  E,   +  A2  E2  +  etc. 

etc.,  can  be  found  when  x  is  known. 

Thus,  for  example,  suppose  there  are  columns  equal  in 
section,  two  of  them  of  brass,  one  of  cast  iron.  Then  AX  =  2  A2, 
BI=  19  X  10 ~6,  a2=  11  x  10-6  for  centigrade  degrees;  EJ  = 
9-2  xlO6,  E2=17xl06, 

2  x  9-2  x  106  x   19  x  10-6  +  17  x  10s  x  11  x  10~6 

~~2~x  9-2  x  106  +  17  X  106 

=  Zx  15-15  x  10~6.  The  compressive  stress  in  the  brass  is  9'2  x  10r> 
(M9xlO-6~*  15-5  x!0-6)  =  £x  30-2  Ibs.  per  square  inch.  The 
stress  in  the  cast  iron  is  17xl06  (t  llxlO-6-*  15'5  x  10~6)  = 

Thus,  if  t  is  30  centigrade  degrees,  there  is  a  compressive  stress 
of  900  Ibs.  per  square  inch  in  the  brass,  and  a  tensile  stress  of 
2,300  Ibs.  per  square  inch  in  the  iron. 

13.  A  bar  of  wrought  iron  25  feet  long  and   1'75  inch  diameter  is 
heated  to  180°  C.     While  in  this  condition  it  is  made  to  connect  (by 
means  of  nuts  screwed  on  the  ends)  the  two  side  walls  of  a  building 
which  have  fallen  outwards  from  the  perpendicular.     If  the  walls  do  not 
yield  against  the  tendency  of  the  bar  to  contract,  find  the  pull  between 
them  when  the  bar  has  cooled  down  to  80°  C.     Take  the  mean  coefficient 
of  linear  expansion  of  wrought  iron  as  -0000124  for  1°  C. 

Am.,  92,350  Ibs. 

14.  Two  bars  of  copper  with  a  bar  of  wrought  iron  between  them,  all 
of  .the  same  section  and  length,  have  their  ends  rigidly  connected  together. 
If  the  bars  are  heated  from  15°  C.  to  100°  C.,   find  the  stresses  in  the  bars, 
and  their  fractional  elongation,  the  coefficients  of  expansion  for  copper 
and  wrought  iron  being  taken  at  -0000172  and  -0000124  respectively. 

Ans.,  copper,  2,940;  iron,  6,000  Ibs.;  0-00125. 

15.  A  tie-rod  100  feet  long  and  2  square  inches  in  sectional  area 
carries  a  load  of  32,000  Ibs.,  by  which  it  is  stretched  f  inch.     Find  the 
stress,  strain,  and  E. 

Am.,  16,000  Ibs.  per  square  inch;  -000625  ;  25,600,000  Ibs.  per  square 
inch. 

16.  A  wooden  tie  40  feet  long,  12  inches  broad,  7  inches  thick,  was 
tested  with  a  pull  of  130  tons,  which  stretched  it  1-28  inches.     Find  the 
value  of  E  for  the  timber.  Ans.,  1,300,000  Ibs.  per  square  inch. 

17.  A  vertical  wrought-iron  tie-rod  200  feet  long  has  to  lift  a  weight 
of  2  tons.     Find  the  area  of  the  section  and  the  diameter  if  the  greatest 
strain  is  -0005  and  E  =  30,000,000.     Neglect  the  weight  of  the  rod. 

Ans.,  0-298  square  inch;  0'616  inch. 

18.  Steam  at  a  pressure  of  200  Ibs.   per  square  inch  is  suddenly 
admitted  upon  a  piston  18  inches  in  diameter.     If  the  piston-rod  be  3 
inches  in  diameter  and  7  feet  long,  what  is  the  compression  and  strain 
energy  in  the  rod  at  maximum  compression?    E  =  30,000,000.     Find, 
also,  the  maximum  stress  in  the  rod. 

Ans.,  "04  inch;  171  foot-pounds  ;  14,400  Ibs.  per  square  inch. 

19.  A  ship  is  moored  by  two  cables  of  90  feet  and  100  feet  in  length, 


APPLIED    MECHANICS. 


315 


respectively.     The  first  cable  stretches  2f  inches,  and  the  second  stretches 
3  inches  under  the  pull  of  the  ship..     Find  the  strain  of  each  cable. 

Ans.,  -0024 ;   -0025. 

The  weight  of  a  rope  in  pounds  per  foot  is  taken  as  ag^  where  g 
is  its  girth  in  inches,  and  its  breaking  weight  in  pounds  is  bg* 
where  a  and  b  have  about  the  following  values  : — 


a 

b 

Tarred  Hemp         
White  Hemp         
Iron  Wire  ... 

•04 
•04 
•13 

900 
1,300 
4,000 

Steel  Wire 

•13 

6  700 

20.  What  lengths  of  thamselves  will  each  of  those  kinds  of  rope  carry  ? 

Ans.,  22,500,  32,500,  30,770,  51,540  ft. 

21.  Compare  the  weights  and  strengths  of  iron-  and  steel- wire  ropes 
with  iron  and  steel  wires  of  the  same  circumference.      Ans.t  '49,  -4,  '49,  -65. 

The  Admiralty  rule  for  the  proof -load  P  in  tons  of  the  ordinary 
close-link  chain  of  welded  iron  is  12rf2,  where  d  is  the  diameter 
of  the  iron  in  inches ;  this  means  nearly  8  tons  per  square  inch  in 
the  iron.  For  studded  chain-cables  it  is  P  =  ISd2,  which  means 
11|  tons  per  square  inch  in  the  iron.  The  working  load  is  from 
half  to  one-quarter  of  this,  depending  on  circumstances.  The 
weight  of  either  chain  is  about  IQd2  Ibs.  per  foot.  Hemp  rope  of 
girth  y  is  taken  as  being  of  about  the  same  strength  as  a  chain  if 
q  =  lOdto  lid. 

22.  Two  close-link  chains,  each  making  an  angle  of  50  degrees  with 
the  vertical,  are  to  support  a  working  load  of  10  tons ;  what  is  the  proper 
size  of  the  iron  ?  Ans.,  0'8  inch  diameter. 

23.  A  ship  of  2,000  tons  (take  it  that  one-quarter  as  much  mass  of 
water  moves  as  the  ship  moves)  is  moving  at  O'l  knots,  and  is  brought 
to  rest  in  three  seconds,  the  law  of  the  motion  during  stoppage  being 
v  =  0-1  cos.  Jet,  where  t  is  time  and  k  a  constant,  and  v  is  in  knots.     The 
pull  comes  directly  on  a  studded  chain.     If  the  chain  gets  its  proof  load, 
what  is  the  diameter  of  the  iron  ? 


When  three  seconds  elapse,  cos.  kt  is  0,  or  cos.  3k 


or   k 


I   knot  = 


In  feet-second  units,   v  =  0-1689   cos.   -  t,   because 

G 

' 


6Q,  or  1-6889  feet  per   second,  and  acceleration 

ie>  —  -1689  x  -  sin.  -^t,  being  numerically  -1689  x  -|,  or  0-5305  feet 

per  second  per  second  where  greatest.  The  mass  is  2,500  x  2,240  •&• 
32-2,  or  173,900  in  engineer's  units.  Hence  the  greatest  force  IB 
92,300  Ibs. 

24.  Find  the  work  which  may  be  stored  up  in  a  pound  of  hard  spring 


316 


APPLIED    MECHANICS. 


steel  when  stretched  to  its  elastic  limit,  taking  the  modulus  of  elasticity 
as  35  x  10°  Ibs.  per  square  inch;  the  elastic  limit,  100,000  Ibs.  per  square 
inch;  and  the  weight  of  one  cubic  inch,  "29  Ib.  Ans.,  410 '5  ft.-lbs. 

25.  A  cylindrical  rod  of  copper  \  inch  diameter  and  4  feet  long,  and 
one  of  wrought  iron  |-  inch  diameter  and  3  feet  long,  are  to  be  stretched 
the  same  amount.     Compare  the  forces  necessary  to  do  this,  the  values  of 
R  for  copper  and  wrought  iron  being  17,000,000  Ibs.  per  square  inch  and 
29,000,000  Ibs.  per  square  inch  respectively.     Compare  also  the  amounts 
of  work  expended  in  each  case.  Ans.,  0'28  :  1,  0-28  :  1. 

26.  What  would  be  the  resilience  of  a  steel  tie-bar  1  inch  in  diameter 
and  4  feet  long  if  the  bar  becomes  permanently  stretched  under  a  load  of 
10  tons,  the  modulus  of  elasticity  being  32,000,000  Ibs.  per  square  inch? 

Ans.,  479  inch-lbs. 

267.  Exercise. — Find  every  number  in  the  following  table. 
Values  of  tensile  (/t2/2E)  or  compressive  (/C3/2E)  resiliences. 
These  numbers  express  the  relative  values  of  the  following 
materials  for  the  making  of  springs  in  which  elongation, 
compression,  or  bending  occurs.  In  bending,  the  smaller  of 
the  two  values,  or  possibly  an  intermediate  value,  must  be 
taken.  The  numbers/// 2 N,  or  the  shear  resiliences  per  cubic 
inch,  express  the  relative  values  of  the  following  materials  for 
the  making  of  springs,  such  as  spiral  springs,  in  which  shearing 
is  most  important.  The  numbers  in  each  case  show  the 
amount  of  energy  (in  inch-pounds)  which  it  is  possible  to  store 
in  each  cubic  inch  of  the  material  in  the  most  carefully  con- 
structed springs.  In  Art.  518  we  give  a  statement  showing  how 
much  less  these  stores  usually  are  in  ordinary  springs.  The 
work  actually  done  upon  ductile  materials  before  fracture  is 
often  1,000  times  as  great  as  the  resilience,  and  in  hard  steel 
it  is  150  times,  in  cast  iron  being  twenty  times  the  resilience. 


/*2/2E 

/C2/2E 

/52/2  v 

Cast  Iron 

3 

12 

5 

Wrought  Iron 

10 

10 

19 

Mild  Steel  (Hardened) 
Best  Hard  Steel        

83 
600 

83 

128 
810 

Copper  (Rolled 

or  Drawn)  .  . 

0-5 

6-5 

0-73 

Fir 

4 

Oak    

6 

... 

... 

268.  The  diminution  in  bulk  of  a  substance  when  it  is  sub- 
jected to  pressure  uniform  all  round,  as,  for  instance,  when  it  is 
surrounded  by  water  in  an  hydraulic  press,  or  sunk  in  the  sea, 


APPLIED    MECHANICS. 


317 


ing 


has  been  experimented  upon.  The  lessening  in  the  bulk  per  cubic 
inch  is  called  the  cubical  strain  of  the  substance.  The  pressure 
in  pounds  per  square  inch  all  over  its  surface  represents  the  stress, 
and  it  is  found  that  the  strain  is  proportional  to  the  stress.  In 
fact,  in  any  substance  the  stress  is  equal  to  the  strain  multiplied 
by  a  certain  number,  for  which  the  letter  K  is  usually  employed, 
called  the  Modulus  of  Elasticity  of  bulk. 

If  an  increase  of  pressure  Sp  causes  the  volume  v  to  become 
v  +  5v  (5v  is  usually  negative),  the  stress  being  Sp  and  the  strain 
—  Sv/v,  the  elasticity  K.  is  denned  by  Sp  =  —  K  Sv/v,  or  K  =• 
-  v  .  dp/dv.  In  solids  it  is  found  that,  whether  the  change  takes 
place  quickly  (at  constant  entropy,  as  it  is  called  in  thermodynamics) 
or  slowly  (at  constant  temperature),  there  is  no  very  great  differ- 
ence ;  but  in  gases  and  liquids  it  is  very  important  to  specify 
under  what  circumstances  the  rate  of  relative  change  of  p  and  v  is 
measured.  The  ratio  is  in  air  1-41;  water,  1-004;  alcohol,  1-22; 
ether,  1-58;  mercury,  1-38;  flint-glass,  1-004;  drawn  brass,  1-028; 
iron,  1-019;  copper,  1-043.* 

The  table,  page  657,  of  slow  moduli  of  elasticity  of  bulk  is  in 
pounds  per  square  inch. 

269.  A  cube  1  inch  in  each  edge  (Fig.  174),  subjected  to  a  uniform 
compressive  force  of  1  Ib.  per  square  inch  on  the  opposite  faces 
A  DBF  and  BCLO.  Evidently  the  edges 
AB,  CD,  LE,  and  GF,  become  1  —  a  inch 
in  length,  a  being  the  reciprocal  of  Young's 
modulus  used  above.  Also  the  edges  AD, 
B  c,  o  L,  and  F  E  get  the  length  1  +  )8  inch. 
If  now  we  give  to  the  faces  ABCD  and 
E  F  G  L  of  this  cube  compressive  forces  1  Ib. 
per(square  inch,  it  is  the  edges  AI*,  etc., 
which  shorten,  and  the  edges  AB,  etc., 
which  lengthen.  Again,  give  the  com- 
pressive forces  to  the  third  pair  of  opposite 
faces,  A  B  G  F  and  c  D  E  L,  and  we  have  the 
edges  AD,  etc.,  shortening  and  BG,  etc., 
lengthening.  If,  now,  all  three  sets  of 
compressive  forces  act  at  the  same  time  (that  is,  the  cube  gets  on 
every  face  a  pressure  of  1  Ib.  per  square  inch),  as  the  compressions 
and  extensions  are  exceedingly  small,  each  edge  shortens  by  the 
amount  a  and  lengthens  by  the  amount  2  #.  Hence  the  edge 
which  used  to  be  1  inch  is  now  1  —  a  +«2  )8  inch.  The  cubic 
contents  used  to  be  1  cubic  inch  ;  it  is  now  1  —  3  (a  —  2  £)  with 
great  exactitude.  Hence  3  (a  -  2  #)  is  the  amount  of  cubical 
strain  produced  by  1  Ib.  per  square  inch.  That  is,  the  Modulus  oj 
Elasticity  of  bulk, 

K_       _J  __  . 
3  («  -2  0)' 
and  if  we  know  a  and  ft  it  may  be  calculated. 

Lord  Kelvin  (article  on  elasticity)  gives  the  above,  and  also  the  follow- 
numbers.    The  ratios  of  the  quick  to  the  slow  Young's   moduli  are  : 


1-008;  tin,  T0036  ;   silver,   1-00315;   copper,   1*00325;   lead,  1  '00310: 
glass,  1-0006;  iron,  1*0026;  platinum,  1-00X3. 


318  APPLIED    MECHANICS. 

Even  very  porous  bodies,  such  as  cork,  have  some  elasticity  oi 
bulk.  Fluids  and  homogeneous  solids,  such  as  crystals,  are 

C'  ably  perfectly  elastic  as  to  bulk  even  at  enormous  pressures, 
ufactured  metals  are  generally  porous,  and  alter  (not 
necessarily  increasing)  in  density  after  they  have  been  hammered 
or  drawn. 

Experiments  on  metals  with  great  negative  stress  in  all 
directions  are  wanting.  Liquids  do  not  seem  capable  of  resisting 
great  negative  pressures,  and  the  contrast  between  them  and  solids 
in  this  respect  is  remarkable. 

270.  Students  now  know  enough  to  make  calculations  on  the  stiff- 
ness and  strength  of  ties  and  struts  [only  when  the  struts  are  kept 
from  bending].  Before  going  on  to  other  structures,  even  such 
simple  structures  as  boilers  and  pipes,  I  wish  to  make  a  few 
remarks  on  the  application  of  theory. 

The  engineer  must  have  some  sort  of  theory  to  work  upon.  I 
shall  give  the  theory  recognised  by  men  like  Rankine,  and  also  by 
the  most  successful  practical  engineers.  It  is  simple  as  I  shall 
give  it,  and  fits  fairly  well  a  great  number  of  practical  conditions. 
It  is  founded  on  the  assumption  that  materials  are  homogeneous 
and  not  loaded  beyond  elastic  limits,  and  yet  it  gives  us  knowledge 
which  the  judicious  man  finds  useful  beyond  elastic  limits.  It  is 
also  founded  on  the  assumption  that  certain  things,  too  difficult  to 
calculate,  are  negligible,  and  hence  its  mathematical  results  ought 
to  be  tested  by  experiment  when  this  is  possible. 

To  give  an  example.  Our  theory  of  bending  is  founded  on  the 
assumption  that  the  plane  cross-section  of  a  beam  remains  plane 
after  bending.  Every  mathematical  result  seems  to  agree  with 
every  experiment  made  on  beams,  seeming  discrepancies  being 
always  explainable  by  the  tests  not  having  been  confined  between 
the  elastic  limits  of  the  materials.  Again,  the  more  elaborate 
theory  of  St.  Venant  (Art.  311),  involving  fewer  hypotheses,  gives 
results  that  are  practically  in  agreement  with  us.  Hence  we 
regard  this  easy  theory  which  I  shall  give  as  one  which  may  be 
depended  upon  in  long  beams,  in  spite  of  the  fact  that  a  plane 
section  of  a  beam  does  not  remain  exactly  plane.  When  it  is 
applied  in  new  cases  not  yet  tested  by  St.  Yenant's  theory  or  by 
experiment,  we  use  it  only  as  a  fairly  trustworthy  guide  in  our 
practical  work. 

Another  example.  In  the  great  twisting,  which  might  occur 
without  fracture  in  indiarubber  shafts,  it  was  known  that  the 
plane  section  of  a  square  shaft  did  not  remain  plane.  Nevertheless, 
the  warping  in  ordinary  shafts  being  small,  a  simple  theory  was 
adopted,  in  which  there  was  the  assumption  of  no  warping. 
Results  of  experiments  on  round  shafts  agreed  with  the  simple 
theory.  Results  from  other  shafts  did  not  agree,  and  St.  Venant 
has  shown  us  why  there  is  a  discrepancy.  Although  his  investiga- 
tion is  difficult  to  follow  mathematically,  his  results  are  easy  enough 
to  comprehend.  I  find  it  necessary,  therefore,  to  give  not  merely 
the  simple  theory  of  the  engineer,  but  an  account  of  St.  Tenant's 
results,  and  also,  for  advanced  students,  a  short  account  of  St. 
Venant's  theory  of  beams  and  shafts.  It  is  to  be  remembered  that 


APPLIED    MECHANICS. 


319 


our  theory  assumes  perfect  elasticity  of  the  material.  We  shall 
see  that  at  the  bottom  edge  of  a  key-way  in  a  shaft  we  have  a 
place  where  the  stress  becomes  very  great  indeed.  I  have  made 
experiments  on  such  a  shaft  with  a  key- way,  and  I  find  that  it  is 
by  no  means  such  a  great  source  of  weakness  as  the  theory  supposes, 
and  this  leads  us  to  consider  how  the  results  of  the  theory  are 
modified  by  the  flow  of  the  material,  instead  of  its  fracture,  under 
great  stress. 

271.  In  thin-shelled  vessels,  such  as  boilers  and  pipes,  sub- 
jected to  fluid  pressure  p  inside,  we  assume  that  the  tensile  stress 
f  is  the  same  throughout  the  thickness  ;  so  that  if  a  is  the  area 
of  metal  cut  through  at  any  plane  section  of  the  boiler,  af  is  the 
resistance  of  the  metal  to  the  bursting  of  the  boiler  at  that 
section.  Now  the  equal  and  opposite  force  due  to  the  fluid  is 
A  p  if  A  is  the  whole  area  of  this  plane  section  of  the  boiler. 
Hence  Ap  =  af  and  p  =  af/& ...  (1)  gives  us  the  bursting  or 
working  pressure  if  f  is  the  ultimate  or  working  stress.  To 
prove  this : — 

In  Fig.  175  let  p  B  c  D  E  be  part  of  a  boiler  whose  separation 
from  the  rest  by  a  plane  section  at  F  E  we  are  now  studying. 
Arrow-heads  are  drawn  showing  the  forces 
with  which  the  fluid  everywhere  acts  nor- 
mally on  the  shell.  We  want  to  know  the 
resultant  of  these  forces.  Imagine  a  boiler 
made  with  the  part  F  »  c  D  E  and  a  rigid  flat 
plate  P  E  closing  it.  If  we  neglect  the 
weight  of  the  fluid,  all  the  pressure  forces 
on  the  shell  balance  one  another.  This  is 
Newton's  law  of  motion.  The  mutual 
forces  of  parts  of  a  system  cannot  affect  the 
motion  of  the  centre  of  inertia  of  the  whole 
system.  (See  Art.  482.)  If  the  above  boiler 
were  placed  upon  a  truck  with  frictionless 
wheels,  there  will  be  no  more  tendency  to 
move  on  a  level  road  when  there  is  great 
pressure  inside  than  when  there  is  little.  The  force  due 
to  pressure  on  any  one  little  portion  of  the  surface  balances 
the  forces  on  all  the  rest  of  the  surface.  Hence  it  is  that  if 
we  make  a  hole  there  is  a  want  of  balance  and  our  truck 
will  move.  When  we  make  the  hole  the  pressure  everywhere 
changes  because  of  the  motion  of  the  fluid,  and  hence  we  can 
only  calculate  the  unbalanced  force  by  knowing  the  momentum 
of  the  fluid  which  leaves  the  vessel  per  second.  In  the  closed 


Fig.  175. 


320  APPLIED    MECHANICS. 

vessel  of  Fig.  175  we  know  that  the 'resultant  pressure  on  the 
flat  surface  F  E  is  A  /?,  therefore  the  equal  and  opposite  force  on 
F  B  c  D  E  is  also  A  p.  As  an  example  of  this  we  saw  in  Art. 
122  that  the  resultant  force  axially  on  the  ram  of  an  hydraulic 
press  is  exactly  the  sajne  whatever  be  the  shape  of  the  end  of  it. 

Example. — Spherical  boiler  of  diameter  D.  Any  plane  sec- 
tion is  a  circle.  If  we  use  the  above  rule  for  any  such  section 
we  find  that  less  pressure  will  burst  the  boiler  if  the  section  is 

diametral.     The  area  of  such  a  section  is  A  =  '  d2,  and  the  area 
of  metal  laid  bare  is  a  =  IT  d  t.     Hence  (1)  becomes  p  =  v  d  tj 
V  -d*~or  4  if  Id.  .  .  (2). 

272.  In  a  long  cylindric  boiler  or  pipe  it  is  easy  to  show 
that  (neglecting  the  strength  of  the  ends)  the  tendency  to  burst 
laterally  is  twice  as  great  as  the  tendency  to  burst  endwise. 

Endwise,  the  area  of  a  circular  cross-section  is  A  =  -7  d2,  and 

the  metal  laid  bare  is  ir  d  t,  so  that  (1)  becomes  p  =  4///d, 
just  as  in  a  spherical  boiler.  But  laterally,  at  a  plane  passing 
through  the  axis  of  the  boiler,  A  =  I  d  if  Us  the  length,  and 
a  =  %lt  if  we  neglect  the  metal  at  the  ends;  and  hence  (1) 
becomes  p  =  %ftjd  .  .  .  (3).  The  bursting  pressure  endwise 
being  twice  as  great  as  this,  we  always  take  (3)  as  the  formula 
for  the  strength  of  a  cylindric  pipe  or  boiler. 

Readers  will  now  understand  why  in  cylindric  boilers  the 
longitudinal  seams  are  always  much  stronger  than  the  girth 
seams.  When  the  boiler  has  riveted  joints  we  must,  of  course, 
regard  the  material  as  weaker  than  if  it  could  resist  tensile 
stress  everywhere  like  a  continuous  boiler-plate.  The  working 
/"for  copper  ought  not  to  be  taken  greater  than  2,400  Ibs.  per 
square  inch  for  steam  pipes.  In  cast-iron  pipes  and  in  steam- 
engine  cylinders  it  has  to  be  remembered  that  the  difficulty  in 
getting  castings  which  are  of  the  same  thickness  everywhere, 
and  the  allowance  that  must  be  made  for  tendency  to  cross- 
breaking  when  the  pipes  are  handled,  as  well  as  the  great  allow- 
ance that  must  be  made  in  steam-engine  cylinders  for  stiffness, 
the  difficulty  of  casting,  and  boring  out,  cause  such  calculations 
as  the  above  to  be  somewhat  useless.  Thus  it  will  usually  be 
found  that,  whereas  a  large  cast-iron  water-pipe  is  not  much 
thicker  than  the  above  calculation  would  lead  us  to  expect, 


APPLIED    MECHANICS.  321 

because  it  is  usually  carefully  moulded  in  loam,  yet  a  thin  cast- 
iron  pipe  is  often  of  more  than  twice  such  a  thickness  on  the 
average,  and  it  is  our  rule  never  to  attempt  casting  a  9-foot 
length  of  pipe  of  less  than  f  inch  thick.  In  these  cases  the 
maximum /for  cast  iron  is  taken  as  1,500  Ibs.  per  square  inch, 
whereas  for  large  pipes  we  usually  take  3,000. 

EXERCISES. 

1.  What  must  be  the  thickness  of  the  plates  used  in  the  construction 
of  a  boiler  10  feet  in  diameter  working  under  a  pressure  of  120  Ibs.  per 
square  inch,  taking  the  efficiency  of  the  joints  to  be  70  per  cent,  and  the 
safe  stress  at  10,000  Ibs.  per  square  inch?  Ans.,  1-03  inch. 

2.  A  copper  pipe  is  4  inches  diameter  and  f  inch  thick.     What  is  the 
working  pressure?    Take/  =  2,000  Ibs.  per  square  inch. 

Ans.,  375  Ibs.  per  sq.  in. 

3.  A  vertical  cast-iron  pipe  is  4  inches  in  internal  diameter.     The 
pressure  at  a  certain  place  is  50  Ibs.  per  square  inch.     At  this  place,  and 
at  places  100,  200,  300,  etc.,  feet  lower  in  level,  find  the  proper  thickness 
of  the  metal  if  the  working  stress/  is  taken  as  3,000  Ibs.  per  square  inch. 

Am.,  -033  inch;  -062  inch;  -091  inch;  -149  inch,  etc. 

4.  The  rule  used  for  loam-moulded  cast-iron  water  mains  is  t  =  ^  + 
hd  ~-  13,000  where  h  is  head  of  water  in  feet,  d  diameter  of  pipe  in  inches, 
t  the  thickness  in  inches.     A  pipe  3  feet  diameter,  1  inch  thick ;  find  h. 
Find  the  corresponding  pressure  in  pounds  per  square  inch.     Find  what 
value  of  /  the  working  stress  will  cause  the   ordinary  rule  for  thin 
cylinders  to  give  the  same  answer. 

Ans.,  316  ft.,  136*7  Ibs.  per  sq.  in.,  2,460  Ibs.  per  sq.  in. 

5.  A  wrought-iron  pipe  2  feet  diameter,  £  inch  thick ;    its  working 
stress  is  5  tons  to  the  square  inch,  but  strength  of  plate  is  diminished  30 
per  cent,  because  of  riveted  joint.    What  is  the  working  pressure  ? 

Ans.,  326  Ibs.  per  sq.  in. 

6.  A  cylindrical  boiler  12  feet  diameter  is  constructed  of  -ff  inch  steel 
plate.     The  test  pressure  applied  is  245  Ibs.  per  square  inch.     Find  the 
stress  produced  in  the  plate,  and  hence  deduce  the  stress  in  the  plate 
between  the  rivet  holes,  the  sectional  area  being  there  reduced  to  -77  of 
the  solid.  Ans.,  19,500  and  25,300  Ibs.  per  square  inch. 

273.  Storage  of  Energy  in  Fluids. — The  volume  of  a  cylin- 
dric  vessel  being  v  and  the  safe  pressure  being  p,  we  may  take 
vp  as  proportional  to  the  energy  which  may  be  stored.  If  the 
diameter  is  d  and  thickness  t  and  length  I,  the  volume  is 

v  =  -T  dzl.  Assume  what  is  known  to  be  true  (see  Art.  272), 
that  the  safe  pressure  for  a  long  cylindric  vessel  is  (neglecting 
the  strength  of  the  ends)  p  =  —v,  where  /  is  the  safe  stress 

which  the  material  will  stand.  The  weight  of  the  metal, 
neglecting  the  ends,  is  w  =  irdtlw,  if  w  is  the  weight  of  unit 

L 


322  APPLIED    MECHANICS. 

volume  of  the  material.     The  surface  of  the  vessel  is  s  =  irdl. 
In  all  cases  we  neglect  the  ends. 

Then  the  storage   capacity  of  energy  per  unit  weight  of 

vessel   is    -j  dPl  -j-  -4-  irdtlw,    or    ~— .     So  we  see  that   it   is 

independent  of  the  diameter. 

In  water-tube  boilers,  therefore,  which  must  store  energy  in 
this  way,  and  where  it  is  of  importance  that  there  should  be 
great  surface,  we  must  consider  surface  -=-  vp.  This  is  2/£/*or 
4/pd.  Hence  the  thinner  the  tubes  are,  and,  if  the  pressure 
is  fixed,  the  smaller  they  are,  the  more  surface  they  have  as 
compared  with  their  storage  capacity  for  energy. 

The  same  considerations  cause  us  to  use  thin  tubes  for 
surface  condensation  and  other  purposes,  and  there  is  the 
further  consideration  that  accidents  are  less  likely  to  be  serious. 

In  cases  where  energy  is  stored  in  hot  water  and  steam, 
the  rate  of  waste  of  energy  is  proportional  to  the  surface,  and 
this  requires  just  the  opposite  conditions. 

EXERCISES. 

1.  Sixty  tubes  of  wrought  iron  4  inches  inside  diameter,  10  feet  long, 
J  inch  thick.     Find  volume,  weight,  internal  area  in  square  inches,  and 
working  pressure  if  working /=  10,000  Ibs.  per  square  inch.     Neglect 
the  ends.     How  many  tubes  22  inches  diameter,  10  feet  long,  will  have 
the  same  volume?    And  find  the   thickness  suitable  for  the  working 
pressure.     Find  also  the  area  and  weight. 

Ans.,  52-36  cub.  ft.,  6,729  Ibs.  wt.,  12-54  sq.  in.,  1,250  Ibs.  persq.  in., 
2,  1-375  inches,  380  sq.  in.,  6,525  Ibs. 

2.  Cylindric  boiler  of  mild  steel  5  feet  6  inches  in  diameter,  at  ekstic 
limit ;  pressure  300  Ibs.  per  square  inch.     What  is  the  thickness  ?     Joints 
supposed  to  be  of  60  per  cent,  of  strength  of  unhurt  plate/.     Replace  this 
boiler  with  tubes  5  inches  diameter  of  same  length.     How  many  tubes 
are  needed  to  make  up  the  same  volume  ?     What  will  be  their  thickness 
(no  seams)?    Their  weight?    Now  replace  with  3-inch  tubes,  finding 
thickness  and  weight. 

Ans.,  -47  inch;  174  ;  -021  inch  ;  193  Ibs.  per  foot  length  ;  -013  inch; 
199  Ibs.  per  foot. 

274.  When  a  belt  or  rope  of  weight  w  Ibs.  per  inch  of  its  length 
is  moving  with  a  velocity  of  v  feet  per  second  in  a  curved  path  of 
radius  r,  the  centrifugal  force  on  a  small  length  of  it,  r .  86,  is 

t?2 

w  .  r .  89  —•    Now,  if  a  sketch  be  made  showing  how  this  force  is 

balanced  by  tensile  forces  T  at  the  ends  of  the  small  length  r .  89, 
it  will  be  seen  that  the  centrifugal  force  is  equal  to  T  .  89  if  89  is 
very  small ;  so  that  T  =  wv*lg,  being  independent  of  the  radius. 
This,  then,  is  the  tensile  force  which  acts  in  a  belt  or  rope  when  in 
motion — an  addition  to  the  tensile  force  which  acts  when  the  rope 


APPLIED    MECHANICS.  323 

is  at  rest — and  it  must  be  taken  into  account  in  considering  the 
strength  of  a  belt.  Notice  also  that  if  a  is  the  section  of  the  rim 
of  a  pulley  of  wrought  iron,  the  weight  of  it  is  -28a  Ib.  per  inch  of 
its  length.  Hence  the  tensile  force  is  -28av2/^>  or  '28v2/#  Ib.  per 
square  inch  is  the  tensile  stress  induced  in  the  rim  by  centrifugal 
force  when  it  moves  at  v  feet  per  second.  Taking  the  working 
tensile  stress  in  wrought  iron  of  a  pulley  as  6,000  Ibs.  per  square 
inch,  the  rim  speed  of  a  wrought-iron  pulley  ought  not  to  exceed  850 
feet  per  second.  The  usual  limiting  speed  of  cast-iron  pulley  rims 
is  80  feet  per  second.  Arms,  if  numerous,  serve  to  diminish  this 
action.  If  an  arm  of  uniform  cross-section  moves  at  n  turns  per 

second,  the  limiting  length  L  of  it  is  —  A  /  — /,  if  w  is  the 

2irw  V      w 

weight  per  cubic  inch.  Thus,  if  we  take  /=  6,000,  w  =.  '28, 
L  =  — .  Thus,  if  n  =  50  revolutions  per  second,  L  is  about 

18  inches.  If  such  arm  has  a  section  a  at  the  distance  r  from  the 
centre,  it  is  easy  to  show  that  if  each  section  has  simply  to  with- 
stand a  pull  due  to  the  centrifugal  force  of  the  part  outside  it 
a  =  #0e-fo'a,  where,  if  r  is  in  inches,  b  =  7r%'n2r2/193/. 

The  condition  as  to  strain  of  a  rotating  disc  has  been  investi- 
gated by  Dr.  Chree  (see  Art.  307). 

As  the  tensile  force  in  a  perfectly  flexible  rope  due  to  its  motion 
is  independent  of  the  shape  of  the  path  in  which  a  particle  is 
moving,  a  very  rapidly-moving  rope,  if  no  external  force  such 
as  those  due  to  gravity  act  upon  it,  has  no  tendency  to  any 
alteration  of  shape ;  each  particle  follows  the  path  of  every 
other,  like  particles  of  water  in  a  stream-line  in  a  state  of  steady 
motion.  We  can  impulsively  alter  the  shape  of  such  a  rope  at  any 
place,  and  then  the  new  shape  will  be  retained.  Vibrations  will 
be  transmitted  by  such  a  rope  as  if  along  a  naturally  straight  rope 
not  moving  in  the  direction  of  its  length  in  which  there  is  the  same 
tension  as  is  here  produced  by  motion.  A  thinking  student  will 
from  these  facts  readily  see  that  there  is  a  quasi-rigidity  produced 
in  the  rope  by  the  rapid  motion. 

EXEEOISES. 

1,  Two  pulleys,  3  feet  6  inches  in  diameter,  running  at  150  revolu- 
tions per  minute,    are  connected  by  a  ^leather  belt  weighing  0'6  Ib. 
per  foot  in  length.      Taking  p.  =  -3,  find  the  greatest  tension  in  belt 
when  transmitting  7J  horse-power.  Ans.,  260  Ibs. 

2.  In  a  travelling  crane  the  driving  rope  runs  at  5,000  feet  per 
minute.     Find  the  tension  due  to  centrifugal  action,  having  given  that  a 
rope  1  inch  diameter  weighs  0'28  Ib.  per  foot  of  length.     Am.,  60-4  Ibs. 

275.  Strength  of  Thick  Cylinders.— Let  the  inside  and  outside 
radii  be  rl  and  r0,  and  the  inside  and  outside  fluid  pressures  be 
pl  and  p0.  ^  Consider  an  elementary  ring  of  metal,  1  inch  parallel 
to  the  axis,  inside  radius  r,  outside  radius  r  +  Sr.  Imagine  a 
compressive  stress  p  inside  it  and  p  +  tip  outside  (5p  is  usually 
negative  in  our  examples),  and  a  compressive  stress  q  in  the 


324  APPLIED    MECHANICS. 

material  (q  is  usually  negative  or  the  stress  is  tensile)  at  right 
angles  to  the  radius,  p  x  2r  is  a  force  tending  to  fracture  this 
ring  at  a  diametrical  plane ;  2  (p  +  Sp)  (r  -\-  br)  —  2q .  8r  is  the 
force  tending  to  prevent  fracture.  Note  that  there  is  a  possible 
shear  stress  on  the  sides  of  this  strip  that  we  are  neglecting,  and  a 
careful  student  will  give  thought  to  the  matter.  Our  j  ustifi cation  for 
neglecting  it  lies  in  this,  that  the  strength  of  our  cylinder  cannot  be 
imagined  to  depend  upon  its  length ;  and  if  we  consider  a  very  long 
cylindric  strip,  end  effects  are  negligible.  Balancing  the  forces,  we 
have  the  well-known  rule  for  the  strength  of  a  thin  cylinder.  Divide 
by  the  thickness  5r,  and  imagine  Sr  smaller  and  smaller,  and  we 

find  p  +  r  -j-  =  q  .  .  .  .  (1).     As    the    material  is  subjected    to 

crushing  stresses  p  and  q  in  two  directions  at  right  angles  to  one 
another  in  a  cross-sectional  plane,  the  dimensions  parallel  to  the 
axis  of  the  cylinder  elongate  by  an  amount  which  is  proportional 
top  +  q.  We  must  imagine  this  to  remain  constant  if  a  plane 
cross- section  is  to  remain  a  plane,  and  we  make  this  reasonable 
assumption.  Hence  (1)  has  to  be  combined  with  p  +  q  =  2a. . .  (2), 
where  20  is  a  constant.  Substituting  q  from  (2),  in  (1)  we  get 

-J?  =  —  —  —  ....  (3),  and  we  find  by  trial  that  the  solution 
dr  r  r 

is  p  =  a  -f  -y  and  q  =  a  -  -^  .  .  .  .  (4),  where  the  constants 

a  and  b  are  to  be  found  by  the  conditions  of  any  problem.  Thus,  in 
the  case  of  a  gun  or  hydraulic  press,  let  the  pressure  inside  be  ^?, 
where  r  =  rv  and  let  pn  =  0  where  r  =  r0.  If  we  insert  these 
conditions  in  (4),  we  find  (q  is  everywhere  negative,  and  I  shall 
use/,  the  tensile  stress,  to  replace  —  q} 


/  is  greatest  at  r  = 

The  student  is  to  note  that  the  circular  compressive  strain  at 
any  place  is  qa  -p&.  This  is  the  fractional  diminution  of  the  radius  r.* 

A  student  ought  to  take  an  example  such  as :  An  hydraulic 
press  has  an  inside  radius  ^  =  4  inches ;  the  stress  is  not  to 
exceed  5,000  Ibs.  per  square  inch;  find  the  greatest  possible 
pressure  pi,  first,  if  the  thickness  is  1  inch,  then  if  the  thickness  is 
2  inches,  and  so  on.  Note  how  little  gain  there  is  by  increasing 
the  thickness  more  than  a  certain  amount ;  and  it  may  be  well  to 
write  out  a  list  of  numbers  for  various  thicknesses,  showing  among 
other  things  the  gain  in  weight. 

EXERCISES. 

1.  Tn  a  certain  kind  of  work  either  one  cylindric  hydraulic 
press  of  24  inches  diameter  or  four  are  needed,  of  same  aggregate  area 
and  sifcme  material,  to  stand  the  same  pressure.  Compare  the  square  area 
on  which  the  two  arrangements  will  stand.  Observe  that  the  ratios  of 
internal  to  external  radii  will  be  the  same  in  the  small  and  large  presses. 

*  See  Appendix. 


APPLIED    MECHANICS.  325 

It  the  student  figures  it  out,  he  will  find  that  the  four  will  just  occupy  the 
same  square  space  as  the  single  press,  and  they  will  weigh  just  the  same. 

2.  Thin  Cylinder.  —  Take  r-^  =  \d  and  r0  =  \ci  +  t,  where  d  is  the 
inside  diameter  of  a  cylindric  boiler  and  t  is  its  thickness,  and  assume  that 
t  is  small.  Then  (6)  becomes 


A  first  approximation  which  is  generally  used,  and  has  been  given  above, 
*•/—  '"ff  ----  (7).    A  second  is/=f^  +  £p  ----  (8).  (See  Appendix.) 

3.  A  gun  of  12  inches  internal  and  24  inches  external  diameter  is 
subjected  to  a  maximum  internal  pressure  of  40,000  Ibs.  per  square  inch. 
Find  the  stress  produced  at  r  =  6,  7|,  9,  10^,  and  12  inches.     Find  what 
was  the  initial  stress  everywhere  if  it  was  just  sufficient  to  cause  the  final 
stress  everywhere  to  be  the  mean  of  the  stress  produced  at  r  =  6  and 
r  =  12.     Now  make  diagrams  showing  the  state  of  stress  when  p,  = 
60,000,  50,000,  40,000,  30,000,  20,000,  and  10,000  Ibs.  per  square  inch, 

Ans.,  66,666  ;  47,466  ;  37,037  ;  30,748  ;  4,444  Ibs.  per  square  inch. 

4.  Pipes   of  a  water-pressure  supply  company  are  to  withstand  a 
possible  pressure  of  1,000  Ibs.  per  square  inch;   they  are  of  6  inches 
internal  diameter.     What  is  the  outside  diameter,  the  safe  tensile  stress  of 
the  metal  being  3,000  Ibs.  per  square  inch?  Ans.,  8-485  inches. 

5.  Pipes  are  to  withstand  a  working  pressure  of  1,000  Ibs.  per  square 
inch.     If  their  internal  diameters  d  are  2,  3,  4,  5,  and  6  inches  in  each 
case,  find  the  thickness.     Find  in  each  case  the  weight  w  per  foot  length. 
Draw  a  curve  showing  the  relative  values  of  w  and  d. 

Ant.,  thicknesses  0-41,  0-62,  0-83,  1-04  inch. 

6.  A  cast-iron  water-main  is  30  inches  internal  diameter  and  1  inch 
thick.    What  is  the  greatest  head  of  water  that  it  ought  to  be  subjected  to  ? 
Safe  tensile  stress,  3,000  Ibs.  per  square  inch.     Numbers  to  recollect  are  : 
34  feet  of  head  represent  1  atmosphere,  or  head  in  feet  -f-  2-3  =  pressure 
in  Ibs.  per  square  inch.     If  the  pipes  are  wrought  iron,  what  ought  their 
thickness  to  be  if  safe  /=  10,000  Ibs.  per  square  inch,  and  if  the  longi- 
tudinal seams  are  of  60  per  cent,  of  the  strength  of  the  unhurt  plate  ? 

Ans.,  460  ft.;  -5  inch. 

276.  In  the  above  theory  we  have  considered  the  material 
initially  unstrained  ;  or,  rather,  the  stresses  and  strains  calculated 
by  us  are  additional  to  any  initial  stresses  and  strains  in  the 
material. 

The  student  will  see  why  the  outer  material  of  a  thick  cylinder 
is  comparatively  useless  if  he  shows  in  a  curve  /for  various  values 
of  r,  calculating  from  (5),  for  /decreases  as  the  inverse  square  of  r. 
If,  when  pn  =  pl  =  0,  there  are  already  strains  and  stresses  in  the 
material,  the  stresses  given  in  (5)  are  algebraically  added  to  those 
already  existing  at  any  place.  Hence,  in  casting  an  hydraulic 
press,  we  chill  it  internally,  cold  water  circulating  in  a  metal  core 
painted  with  loam  ;  and  in  making  a  gun  we  build  it  of  tubes,  each 
of  which  squeezes  those  inside  it.  So  that  there  is  considerable 
compressive  stress  where  r  =  r^  and  considerable  tensile  stress 
where  r  =  r0  before  any  pressure  comes  on  inside.  We  try  to 


326  APPLIED    MECHANICS. 

create  such  initial  stresses  that  when  there  is  the  maximum 
pressure  pl  the  material  has  about  the  same  tensile  stress  in  it 
everywhere.  Much  knowledge  is  needed  to  produce  this  result  in 
guns. 

Exercise,  —  Thick  spherical  shell  subjected  to  internal  fluid 
pressure.  If  p  is  the  radial  compressive  stress  at  a  point  at  the 
distance  r  from  the  centre,  and  q  is  the  tangential  tensile  stress 
there,  show  after  the  manner  of  Art.  275  that  p  =  A  -f-  2B/r3, 
q  =  A  —  B^r3,  where  A  and  B  are  constants,  which  may  be 
found  if  the  internal  pressure  and  the  inner  and  outer  radii  are 
stated. 

277.  The  above  theory  of  the  strength  of  thick  cylinders  seems 
to  agree  with  our  practical  experience  for  suck  ratios  of  r0  and  7-j  as 
we  find  in  the  pipes  and  presses  used  by  engineers.  But  all  the 
rules  given  above  show  that  a  flat  plate  has  no  strength.  The 
neglected  terms  in  our  theory  become  important  in  this  case.  In 
truth,  the  mathematical  theory  of  a  shell  is  so  troublesome  that  we 
cannot  say  there  is  yet  a  satisfactory  treatment  of  it.  The  strength 
of  a  flat  plate  has,  however,  been  investigated  in  a  number  of 
cases,  and  we  are  led  to  the  following  results  :  —  For  a  circular 
plate  of  thickness  t  and  diameter  d  supported  all  round  its  edge, 
with  a  normal  load  of  p  Ibs.  per  square  inch,  if  /  is  the  greatest 
stress  in  the  material,  /  =  5?>2p/6£2.  If  the  plate  is  fixed  all  round 
its  edge,  /=  2r2p/3fi.  A  square  plate  of  side  s,  fixed  at  the  edges, 
/=  s§?/4£a.  A  rectangular  plate  of  length  I  and  breadth  b,  fixed 
round  the  edges,  f=!Wp/2P(l*  +  b4}.  A  round  plate  with  a 
load  w  in  the  middle,  supported  at  the  edges,  /  =  w/?r£2.  For 
stays  in  square  formation,  distance  asunder  being  s  ;  each  stay 
has  a  load  psz,  and  the  greatest  stress  in  the  plate  of  thickness  t 
is 


278.  When  the  pressure  is  greatest  outside  a  thin  shell,  its 
strength  to  resist  collapse  ought  evidently  to  follow  the  law  (1), 
which  becomes  (3)  if  the  vessel  is  cylindric  ;  but  it  is  in  the  very  way 
in  which  a  strut  may  be  relied  upon  if  the  slight  lateral  restraints 
are  provided  which  prevent  bending.  So  also  the  thin  shell  of  a 
boiler  flue  must  be  provided  with  certain  restraints  against 
buckling  ;  and  just  as  we  have  (Art.  372)  a  theory  of  laterally 
unsupported  struts,  so  we  have  a  theory  of  long  boiler  flues. 
Beyond  a  certain  length  for  a  given  pressure  there  is  instability, 
and  hence  flues  are  either  corrugated  or  furnished  with  a  number 
of  rings.  The  most  important  practical  outcome  of  the  theory  is 
that  the  distance  between  two  rings  divided  by  Vdt  must  not 
exceed  a  certain  limit.  The  rules  usually  followed  by  boiler- 
makers  are  given  in  the  new  edition  of  my  book  on  Steam. 

Exercise.—  In  a  corrugated  flue  of  3  feet  mean  diameter  the 
plate  is  -i  inch  thick,  but  the  corrugations,  being  longer  than  the 
axial  length,  make  it  virtually  f  inch  thick.  What  is  the  working 
pressure  if  the  working  compressive  stress  (we  allow  for  corrosion, 
etc.)  in  the  material  is  3,000  Ibs.  per  square  inch  ? 

Ans.,  125  Ibs.  per  square  inch. 


APPLIED    MECHANICS. 


327 


MORE  DIFFICULT  EXERCISES  ON  THICK  CYLINDERS. 

1.  A  tube  of  wrought  iron,  inside  radius  3  inches,  outside  4  inches, 
outside  pressure  0.  What  is  the  inside  pressure  P  to  produce  a  maximum 
tensile  stress  of  15,000  Ibs.  per  square  inch  ?  Find  the  fractional  increase 
in  size  of  the  inside  radius.  Here  p  =  0  where  r  =  4  ;  p  =  P,  and  q  = 
-  15,000  where  r  =  3.  Inserting  these  values  in  (4)  Art.  275,  we  find 

15,000  = 


-  15,000  =  a  - 


b  =  86,400 


a  =  -  5,400 


P  =  -  5,400  +  9,600  =  4,200  Ibs.  per  square  inch. 
Let  the  student  calculate  —  q  for  several  values  of  r  from  3  to  4,  and 
plot  his  results  on  squared  paper. 


r 

3 

3'25 

3-5 

3-75 

4 

-9 

15,000 

13,580 

12,453 

11,544 

10,800 

The  fractional  diminution  in  size  of  any  radius  is  q  a  -  p  0,  or 
(o  -  0)  a  -  ^2  (a  +  j8).  Taking  a  =  |  x  10  ~  7,  0  =  ^  10  ~  7,  the  frac- 
tional  diminution  of  the  3  inch  radius  is 

^10~7«-gi210~7=s~  5,350  x  10  ~  7< 

That  is,  the  3  inch  radius  increases,  becoming  3  x  5,350  x  10-?,  or  O00161 
inch  larger. 

2.  A  tube  of  wrought  iron,  inside  radius  2  inches,  outside  3  inches,  no 
pressure  inside ;  pressure^  =  4,200  Ibs.  per  square  inch  outside.  Find  the 
circular  compressive  stress  everywhere,  and  also  tHe  diminution  of  the 
outer  radius. 

b  b 

Here  in  P  =  a  +  -j2>Q=a--j$''''  (*)• 

Fractional  diminution )  b  .  In. 

of  the  3  inch  radius    )  =  («  ~  0)  a  ~  9  (a  +  #' <2) 

We    have   p  =  4,200  where  r  =  3 ;  p  =  o  where  r  =  2. 

4,200  =  a  +  ^ 

— >•*! 


_  30,240,  a  =  7,660 


328  APPLIED    MECHANICS. 

Hence  0  =  7,560  +  30,240/r2.    Thus  for  the  following  values  of  r  we  have 


r 

q. 

2 

2-25 

2-5 

2-75 

3-0 

15,120 

13,533 

12,400 

11,560 

10,920 

And  the  3  -inch  radius  decreases  by  the 
|-  x  7,560  x  10~7  + 

fraction  of  itself 

9       12  ] 

or  3-29    x    10  ~4,    so  that  the  3  inch  radius   becomes    '00099   inch 
smaller. 

3.  A  tube  of  radius  4  inches  outside,  and  radius   2-9984  inside,  is 
squeezed  in  some  way  upon  a  tube  2  inches  inside  radius  and  3-00098 
inches  outside  radius.     Find  the  compressive  circular  stress  at  all  points 
in  both  tubes. 

Am.,  It  will  be  seen  that  I  have  taken  just  the  sizes  necessary  to 
produce  the  states  of  Exercises  1  and  2.  It  is  evident  that  we  may  take 
the  outside  radius  of  the  inner  tube  as  3  inches,  and  the  inside  radius 
of  the  outer  tube  2 '99 74  inches.  Observe  that  as  the  coefficient  of  ex- 
pansion of  iron  is  1*2  x  10  ,  to  produce  a  fractional  increase  in  size 
•0026  -f-  3,  the  outer  tube  must  be  raised  in  temperature  more  than 

•J)026_  ^  x  10  ~  5),  or  72  Centigrade  degrees,  before  it  will  slip  over 

a 
the  other. 

4.  A  tube  of  wrought  iron,  2  inches  radius  inside,  4  inches  outside,  is 
subjected  to  a  fluid  pressure  of  50,000  Ib.  per  square  inch  inside  and  no 
pressure  outside.     Find  the  tensile  stress  everywhere  in  the  material  or 
the  values  of  -  q. 

Here  in  p  =  a  +  -£,q  =  a  — %>   inser*  P  —  50,000    where    r  =  2, 


p  =  0  where  r  =  4,  and  so  find 
-  q  is  the  tensile  stress  — 

a       266'667    1    16667 

Of  course 

q  „           -|      1O,OO/. 

r 

2 

2i 

3 

H 

4 

-9 

83,333 

59,333 

46,300 

38,440 

33,333 

5.  To  sum  up  our  results.  The  built-up  tube  of  Exercise  3,  with  initial 
tensile  stress/  or  -  q  of  E.T.  3  taken  from  the  tables  of  Exercises  1  and  2,  is 
subjected  to  the  internal  fluid  pressure  of  50,000  Ibs.  per  square  inch  of 
Exercise  4,  there  being  no  pressure  outside.  We  have  seen  that  /'or  -  q 
of  Ex.  4  shows  the  tensile  stress  produced  by  the  fluid  pressure,  and 
hence/",  which  is  /' +/,  is  the  real  tensile  stress  every  where  in  the 
compound  tube.  Students  ought  to  draw  curves  showing/,/',  and/". 


APPLIED    MECHANICS. 


329 


r 

2 

u 

2j 

2J 

3 

3 

3J 

3* 

3J 

4 

/ 

-15,120 

-  3,533 

-12,400 

-  11,560 

-  10,920 

15,003 

13,580 

12,453 

11,544 

10,800 

f 

83,333 

69,337 

59,333 

51,937 

46,300 

46,300 

41,317 

38,440 

35,727 

33,333 

r 

68,213 

55,804 

46,933 

40,377 

35,380 

61,300 

54,897 

50,893 

47,271 

44,133 

279.  These  exercises  will  enable  the  student  to  understand  the 
usual  calculations  of  shrinkage  which  must  be  made  in  building  up 
a  gun.  He  will  have  no  great  difficulty  in  working  out  all  the 
necessary  formulae  himself,  but  a  man  interested  in  gun-making 
ought  to  refer  to  three  articles  published  in  Nature  about  August, 
1890,  by  Prof.  Greenhill.  It  will  be  noticed  that  when  a  gun  is 
built  of  tubes  there  is  a  sudden  change  in  the  tensile  stress  at  the 
common  surface  of  two  tubes.  To  get  a  more  uniform  stress 
throughout,  instead  of  using  many  thin  tubes  we  now  use  an  inner 
tube  of  steel  strong  enough  to  withstand  the  longitudinal  forces, 
then  a  thick  layer  of  steel  wire  wound  on  with  varying  tension, 
and  then  a  covering  tube,  which  is  almost  unstrained  initially. 
Probably  the  hydraulic  presses  of  the  future  will  also  be  built  up 
in  this  way. 

Prof.  Greenhill  says :  "  Mr.  Longridge's  principle  of  strengthen- 
ing a  tube  with  wire  wound  with  appropriately  varying  tension 
will  be  found  useful  in  peace  and  in  war.  He  can  claim  credit 
that  a  gun  strengthened  on  this  principle  (the  9-2  inch  wire  gun) 
was  chosen,  from  its  great  strength,  to  test  the  extreme  range  of 
modern  artillery  in  1888,  with  what  were  called  the  "Jubilee 
rounds,"  when,  with  an  elevation  of  about  40  degrees,  a  range  of 
21,000  yards,  or  12  miles,  was  attained,  the  projectile  weighing 
380  Ibs.,  and  the  muzzle  velocity  being  about  2,360  feet  per 
second.". 

280.  Exercise  upon  a  Tube  Gun.— Write  out  how  for  any  tube 
when  given  p9  the  outer  pressure  at  r0  and  /j  the  inner  tensile 
stress  at  r±  we  find  p^.  Now  beginning  with-  the  outermost  tube 
pQ  rr  0,  /j  =  18  tons  per  square  inch,  let  us  say,  find  pv  This 
pl  is.the^0  of  the  next  tube ;  studying  this  next  tube  we  have^>0 
and/j  =  18  and  find^.  We  thus  study  the  tubes  in  succession 
till  we  find  the  powder  pressure.  Now  find  p'  and  f  every- 
where for  the  powder  pressure  alone.  Having  tabulated  all 
values  we  find  p"  =p—p,  f"  =f—f  for  the  gun  before  firing. 
Thus  let  the  radii  of  tubes  in  a  gun  in  inches  be  5,  7,  9,  11,  13, 
16,  and  we  find  the  following  figures,  which  students  ought  to 
work  out  and  also  show  in  curves.  The  column  headed  "  Shrink- 
ages "  shows  the  sizes  of  the  tubes  before  they  are  shrunk. 


330 


APPLIED    MECHANICS. 


FIRING. 

POWDER     ALONE. 

BEFORE     FIRING. 

SHRINK- 

AGES. 

r 

P 

/ 

y 

/ 

P" 

/" 

5 

7 

31-97 
19-73 

18 
5-76 

31-97 
14-62 

38-89 
21-54 

0 
5-11 

-20-89 
-15-77 

5-0083 
7-0078 

7 

19-73 

18 

14-62 

21-54 

5-11 

-3-54 

7-0012 

9 

12-28 

10-55 

7-48 

14-40 

4-80 

-3-85 

9-0019 

9 

12-28 

18 

7-48 

14-40 

4-80 

3-61 

8-9966 

11 

7-27 

12-99 

3-86 

10-78 

3-41 

2-22 

10-9973 

11 

7-27 

18 

3-86 

10-78 

3-41 

7-22 

10-9929 

13 

3-68 

14-4 

1-78 

8-70 

1-90 

5-71 

12-9935 

13 

3-68 

18 

1-78 

8-70 

1-90 

9-30 

12-9899 

16 

0 

14-32 

0 

6-92 

0 

7-40 

15-9905 

Exercise  on  a  Wire  Gun. — Assume  that  the  covering  tube  gives 
no  lateral  strength.  Let  the  tensile  stress  /  be  constant  and 
equal  to  T  when  firing  takes  place.  Then  if  rQ  and  rl  are  the 
outer  and  inner  radii  of  wire  winding,  T  (r0  —  r^  =:  p  r, 
p  =  T  (^0/r  ~~  1)'  Pi  — -  T  (ro/ri~  !)•  Thus  we  find  the  pressure 
on  the  outside  of  the  inner  solid  tube.  Assuming  a  tensile  stress 
at  the  inside  of  this  tube  we  calculate  the  powder  pressure  P. 
(Gunmakers  prefer  to  have  the  innermost  part  of  the  inner  tube 
just  in  compression  or  with  no  stress.  They  say  that  it  makes 
the  metal  wear  better.  Longridge  made  the  inner  tube  thin  and 
of  cast  iron  ;  Schulz  made  it  thick  and  of  steel.)  Now  find  the 
effect  of  p  on  the  whole  gun  producing  p'  and  f  everywhere  in 
the  wire.  Thus  T-/'  =  /"  the  tensile  stress  in  the  wire  before 
firing  and  p  -  p'  =  p"  is  the  radial  pressure  before  firing.  Now 
if  we  have  a  cylinder  of  outside  radius  r  and  inner  R,  a  com- 
pressive  stress  p"  on  its  outside,  p,  being  o,  will  produce  com- 
pressive  hoop  stress  r  of  the  amount 

y'C^  +  RiW -*!*>• 

This  is  the  amount  by  which  the  present  tensile  stress  f"  at  r 
is  less  than  the  stress  in  the  wire  when  it  was  wound  on. 

Thus  take  radii  of  inner  tube  7  and  10  inches ;  of  wire 
10  and  14  inches.  Take  T  =  30  tons  per  square  inch. 

p  = 30,  P!  =  12.     It  is   easy  to  get  the  figures  headed 

Firing  in  the  following  table  from  r  =  14  to  r  =  10.  Now  for 
the  figures  above  these,  the  inner  tube,  we  have  pQ  =  12  at 
r0  •=  10,  and  /=  o  at  r  =  7.  Assuming  p  =  a  -f  bjr2  and 

/=  —  a  -\ — ^  and  finding  a  and  b  we   calculate  the  figures 


APPLIED    MECHANICS. 


331 


tabulated.  We  find  the  powder  pressure  to  be  16-1  tons  per 
square  inch.  To  find  the  effect  of  this  powder  pressure  alone 
p'  =  o  where  r  =  14  and  p'  =  16*1  where  r  =  7,  and  this 
enables  us  to  find  p'  and  /'.  A  radial  compressive  stress^/'  at  r 
with  no  pressure  at  r  =  7  produces  a  hoop  compressive  stress 
p"  (f-  +  49)/(r»  -  49)  at  r ;  this  added  to  /"  is  the  winding 
stress  in  the  wire. 


FIRING. 

POWDER     ALONE. 

FINISHED    GUN. 

7* 

P 

f 

P' 

f 

o> 

i 

WINDING 

STRESS 

7 

16-1 

0 

16-1 

26-84 

0 

-26-8 

8 

14-22 

-1-89 

11-1 

21-8 

3-1 

-23-7 

9 

12-92 

-3-18 

7-6 

18-4 

5-3 

-21-5 

10 

12 

-4-10 

5-2 

15-9 

6-8 

-20 

10 

12 

30 

5-2 

15-9 

6-8 

14-1 

34-1 

11 

8-2 

30 

3-3 

14-1 

4-9 

15-9 

27-4 

12 

5 

30 

1-9 

12-7 

3-07 

17-3 

236 

13 

2-3 

30 

0-9 

11-6 

1-45 

18-4 

21-0 

14 

0 

30 

0 

10-7 

0 

19-3 

19-3 

Such  calculations  as  these  must  be  incorrect  as,  the  wire 
(usually  strip  of  rectangular  section),  however  closely  wound, 
cannot  behave  like  solid  metal.  Again,  it  is  all  very  well  to 
say  that  the  outside  of  one  tube  is  to  be  7'0078  inches,  and  the 
inside  of  another  is  to  be  7-0012  inches,  but  it  is  not  possible  to 
bore  and  turn  long  lai'ge  tubes  with  the  necessary  accuracy. 
It  is  probable  that  in  all  cases  where  the  stresses  become  too 
great  the  material  in  guns  yields  permanently  and  adjusts 
itself  in  ways  about  which  we  may  speculate  but  we  cannot 
calculate. 


332 


CHAPTER  XIV. 


SHEAR     AND      TWIST. 


281.  LET  c  D  (Fig.  176)  be  the  top  of  a  firm  table,  F  H  a  long 
prism  of  indiarubber  glued  to  the  table,  A  B  a  flat  piece  of  wood 
glued  along  the  upper  side  of  the  indiarubber.  We  try  in  this 
way  to  apply  a  horizontal  force  to  the  whole  upper  surface  of 
the  indiarubber,  so  that  if,  for  instance,  the  pull  in  the  cord  is 


Pig.  178. 


20  Ibs.,  and  the  upper  surface  of  the  indiarubber  is  10  square 
inches  in  area,  there  will  be  a  force  of  2  Ibs.  per  square  inch 
acting  at  every  part  of  the  surface,  and  this  force  will  be 
transmitted  through  the  indiarubber  to  the  table.  When  the 
length  of  the  prism  is  great  compared  with  z  F,  we  may  suppose 
that  the  bending  in  it  is  very  small,  and  in  this  case  we  say  that 
the  indiarubber  is  being  subjected  to  a  simple  shear  strain  ;  the 
force  per  square  inch  acting  on  its  surface  is  also  acting  from 
each  horizontal  layer  to  the  next,  and  is  called  the  shear  stress. 
If  you  had  drawn  vertical  lines  like  Y'  x  before  the  cord  was 
pulled,  you  would  now  find  them  sloping  like  Y  x.  Thus, 
making  a  magnified  drawing  of  Y  x  in  Fig.  176,  the  point  Y' 
has  gone  to  Y,  and  any  point  like  M  has  gone  to  N.  Points 
touching  the  table  cannot  move,  but  the  farther  a  point  is  away 
from  this  fixed  part  the  farther  it  can  move.  Now  suppose 
that  Y'  Y  is  O'Ol  inch,  and  we  know  that  x  Y'  is  2  inches,  what 
is  the  amount  of  motion  of  M  if  M  x  is  1  -7  inch  ?  Evidently 


APPLIED   MECHANICS.  333 

Y'  Y  is  greater  than  M  N  just  in  the  proportion  of  Y'  x  to  M  x,  or 
2  to  1-7  ;  hence  MN  is  0-0085  inch.  Thus  the  motion  of  any 
point  is  simply  proportional  to  its  distance  above  the  fixed 
plane,  and  if  we  know  the  amount  of  motion  at,  say,  a  distance 
of  1  inch,  we  can  calculate  what  it  must  be  anywhere  else. 
The  amount  of  motion  at  one  inch  above  the  fixed  plane  is  called 
the  shear  strain.  The  motion  is  small,  and  it  is  evident  that 
the  shear  strain  is  the  angle  Y'  x  Y  in  radians.  In  this  case  we 
have  supposed  the  force  on  F  G  to  be  2  Ibs.  per  square  inch. 
This  is  said  to  be  the  amount  of  the  shear  stress,  and  it  pro- 
duces or  is  produced  by  a  shear  strain  whose  amount  is  '005 
inch  per  inch.  If  the  shear  stress  were  4  Ibs.  per  square  inch, 
you  would  find  the  strain  to  be  '01 ;  if  the  stress  were  8  Ibs.  per 
square  inch  the  strain  would  be  -02.  In  -fact,  we  find  experi- 
mentally that  the  stress  and  strain  are  proportional  to  one 
another.  Thus  if,  instead  of  indiarubber,  we  had  a  block  of 
tempered  steel,  we  should  find  that  the  force  in  pounds  per 
square  inch  is  equal  to  13,000,000  times  the  strain.  This 
number  is  called  the  modulus  of  rigidity  for  steel ;  it  is  given 
in  Table  XX.  It  seems  a  pity  that  the  name  shearing  elasticity 
is  not  given  to  this  number. 

282.  However  long  we  may  make  our  block  of  indiarubber  in 
Fig.  176,  we  shall  still  have  some  tending  in  it ;  that  is,  the  stress 
will  not  be  uniformly  distributed  over  each  horizontal  layer  (see 
Chapter  XXI.).  To  prevent  this  bending  effect,  and  to  produce  a 
really  simple  shear  strain,  we  ought  to  have  force  distributed  over 
the  ends  F  z  and  o  H 
of  the  same  amount 
per  square  inch  as 
we  have  now  acting 
over  P  G  and  z  H. 
These  are  shown  in 
Fig.  177.  Where  P  ,  I 
is  the  pull  in  the  cord  ^  p  -  / 
of  Fig.  176,  P'  is  the 
equal  and  opposite 

force  exerted  by  the  «.    ,«. 

table  on  the  glued 
underside  of  the  india- 
rubber,  and  F  and  F'  are  equal  and  opposite  forces  distributed 
over  the  ends,  such  that  the  couple  F  F  is  able  to  balance 
the  couple  PP'.  There  can  now  be  no  bending  moment  at 
any  place.  As  F  multiplied  by  _  the  length  of  the  prism  is  the 
moment  of  the  couple  FF',  and  is  equal  to  P  multiplied  by  the 
vertical  dimension,  we  see  that  p  distributed  over  the  horizontal 
surface  is  the  same  stress  per  square  inch  as  F  distributed  over  the 


334  APPLIED    MECHANICS. 

ends.  From  such  a  material  then,  if  we  cut  a  cubical  block  A 
(Fig.  178),  its  horizontal  faces  xy  and  xa?  are  acted  upon  by 
equal  and  opposite  tangential  forces,  and  its  faces  YX  y  a?  are 
acted  upon  by  forces  of  exactly  the  same  amount.  The  faces 
parallel  to  the  paper  have  no  forces  acting  on 
them.  This  will  give  you  the  best  idea  of  simple 
shear  stress.  The  material  in  Fig.  176  near  the 
ends  of  the  block  does  not  get  a  simple  shear ; 
but  if  the  block  is  very  long,  then  at  the  middle 
there  is  a  nearly  simple  shear  acting. 

In  Fig.  178  the  cube  x  Y'  y'  so  has  become 
xYt/a?.     Suppose  the  side  of  this  cube  to  "be  1 
Fig.  178.  inch,  then  Y'  Y  is  the  shear  strain,  which  I  shall 

call  s.  The  tangential  force  distributed  over 
Y  y  is  p  Ibs.,  let  us  say.  Then,  if  we  denote  by  the  letter  N  the 
modulus  of  the  rigidity  of  the  material,  p  =  N  «. 

283.  Example. — A  beam  of  steel  has  one  end  fixed,  and  at 
the  other  is  a  weight  of  20  tons.     The  cross  section  of  the  beam 
is  2  square  inches  in  area,  and  the  length  of  the  beam  is  5  inches. 
Besides  the  deflection  of  this  beam  due  to  bending,  there  is  a 
certain  deflection  due  to  shearing ;  how  much  is  it  \   Answer  : 
the  shear  stress  is   10  tons,  or  22,400  Ibs.  per  square  inch. 
This  produces  a  shear  strain  of  22,400  -f  13,000,000,  or  '00172. 
This  is  the  amount  of  yielding  at  1  inch  from  the  fixed  end,  and 
at  5  inches  the  yielding  must  be  5  x  '00172,  or  -0086  inch. 

In  a  short  beam  like  this,  or  in  one  20  inches  long,  if  we  con- 
sider, for  example,  that  it  is  1  inch  broad  and  2  inches  deep,  we  may 
calculate  the  deflections  due-  to  bending  and  to  shear,  and  reflect 
upon  the  fact  that  in  very  short  beams  the  yield  due  to  shear 
is  much  more  important  than  the  yield  to  bending ;  whereas  in 
long  beams  we  may,  and  indeed  always  do,  neglect  altogether 
the  deflection  due  to  shear.  These  reflections  are  in  the  main 
correct,  but  the  actual  distribution  of  shear  stress  over  the 
sections  of  a  beam  not  short  and  not  long  is  unknown  to  us. 
Its  distribution  in  the  section  of  a  long  beam  will  be  given  in 
Art.  369,  and  it  is  obvious  that  the  calculation  of  the  deflection 
due  to  shear  is  not  so  simple  as  in  very  short  beams. 

284.  The  shear  stress  which  will  produce  rupture  is  not 
well  known  for  any  substance  except  cast-  and  wrought-iron,  but 
the  shear  stress  which  will  produce  permanent  set  is  fairly 
well  known,  and  we  are  also  agreed  as  to  the  ordinary  working 
shear  stress  of  materials.     For  wrought  iron  and  mild  steel  it  is 
usually  regarded  as  from  75  to  85  per  cent,  of  the  tensile  stress  ; 
but  in  a  single-riveted  lap  joint  in  boiler-plates,  as  the  holes  are 
usually  punched  (and  this  weakens  the  metal),  and  as  rivet  iron 


APPLIED    MECHANICS.  335 

is  usually  of  a  better  quality  than  plate,  the  cross  section  of 
the  iron  which  is  left,  which  is  resisting  pull,  is  made  to  have  the 
same  area  as  the  cross  sections  of  all  the  rivets,  which,  of  course, 
resist  shearing.  Besides  breaking  by  either  a  tensile  or  a  shear 
stress,  a  riveted  joint  may  give  way  by  the  rivet  crushing  or 
being  crushed  by  the  side  of  its  hole.  Again,  in  many  riveted 
joints,  when  the  rivets  are  long,  as  they  tend  to  contract  in 
cooling  and  are  prevented  by  the  plates,  so  much  tension  may 
remain  permanently  in  them  that  they  are  greatly  weakened. 
In  bolts  there  is  usually  some  bending,  and  consequently  a  want 
of  perfectly  uniform  distribution  of  the  shear  stress,  and  they 
are  made  larger  than  rivets  in  the  same  positions. 

285.  In  the  punching  of  rivet-holes  it  may  be  taken  that  a 
shearing  force  v  acts  on  the  material ;  the  area  of  the  curved 
.side  of  the  hole,  multiplied  by  the  breaking  shear  stress  of  the 
material  per  square  inch,  represents  tJie  force  with  which  the 
punch  must  be  pressed  down  on  the  plate.  The  punch  must  be 
able  to  resist  this  force  as  a  compressive  stress  on  its  own 
material.  Experiments  made  on  punching-machines  show  that 
about  24  tons  per  square  inch  is  the  average  shearing  force 
required.  This  pressure  has  to  be  exerted  through  a  very 
short  distance.  In  shearing-machines,  if  the  entire  edges  of 
the  shears  coincided  with  the  plate  as  soon  as  they  touched 
anywhere  there  would  be  the  same  sort  of  effect  produced  ; 
but  by  inclining  the  edges  the  shearing  action  does  not  occur 
instantaneously  at  every  place,  and  the  rupture  being  more 
gradual  than  in  punching,  the  shearing  resistance  is  usually 
from  10  to  30  per  cent.  less.  It  is  very  probable  that  the 
power  lost  in  punching-  and  shearing-machines  is  wasted  rather 
in  the  friction  of  the  heavy  parts  of  the  mechanism  than  in  the 
almost  instantaneous  effort  of  cutting  the  material.  The  effort 
required  seems  rather  that  of  an  impact  .than  of  the  more 
gradual  action  to  be  found  in  most  existing  machines.  At  the 
same  time  we  must  remember  that  M.  Tresca's  experiments 
indicated  a  flowing  of  the  metal.  Machines  such  as  hydraulic 
bears  and  shears  may  be  uneconomical  as  to  mere  energy,  but 
they  produce  the  higher  economy  due  to  convenience  and 
certainty.  A  man  can  stop  the  motion  of  the  hydraulic  punch 
very  rapidly  if  he  sees  that  the  plate  is  not  quite  right  in  posi- 
tion. In  the  fly  presses  used  for  hand  punching,  and  used  largely 
in  coining,  the  idea  of  an  impact  is  already  in  use  ;  it  will  come 
much  more  into  use  in  large  machines  when  engineers  become 


336 


APPLIED    MECHANICS. 


better   acquainted    with   the   distinction    between   force   and 
energy. 

286.  Riveted  Joints. — Students  are  supposed  to  have  made 
sketches  of  a  number  of  well-proportioned  joints.  In  applying 
the  results  of  experiment  arid  our  imperfect  theory  to  actual 
structures  we  must  remember  that  our  practical  conditions  some- 
times closely  agree  with  our  hypothetical  conditions,  and  then 
our  calculations  may  be  fairly  exact ;  whereas  in  other  cases  our 
theory  is  only  an  imperfect  sort  of  guide  in  our  workshop 
systems  of  seeking  for  success  and  avoiding  failure.  It  is  only 


r 


_ 


~~ 


Pig.  179. 


a  very  capable  engineer  who  knows  exactly  what  weight  ought 
to  be  given  to  his  theory  in  every  case. 

If  we  had  a  perfect  theory  we  need  only  consider  the  various 
ways  in  which  a  riveted  joint  may  fracture.  Then  we  should 
state  the  algebraic  conditions  that  the  joint  shall  be  equally 
ready  to  fracture  in  these  various  ways,  and  we  have  at  once 
the  right  proportions. 

Thus,  take  the  strip  which  illustrates  the  single  riveted  lap 
joint  of  a  boiler  plate,  Fig.  179  (1),  or  the  single  riveted  butt 
joint  of  Fig.  179  (5).  A  B  =  />,  the  pitch  or  distance  from  centre 
to  centre  of  rivets  ;  t  is  thickness  of  plate ;  \  is  D  c,  the  overlap 


APPLIED   MECHANICS.  337 

minus  radius  of  rivet ;  /8  the  shearing  stress  which  the  rivet 
will  stand  ;  /t  the  tensile  stress  which  the  plate  will  stand. 
If  fracture  occurs  as  in  (1),  the  tensile  force  p  which  the 

joint  will  stand  is  p  =  -  d*Ja  ....  (1).     If  in  double  shear,  as 

in  (5),  P  =  \  d^  .  .  .  .  (5).    If,  as  in  (2),  p=(p  -d)tft....  (2). 

If  as  in  (3),  when  the  iron  of  the  plate  is  crushed  by  the  rivet 
or  the  rivet  is  crushed  by  the  iron,  even  the  most  absurd 
person  will  hesitate  when  he  puts  p=fctd  ....  (3),/c  being 
some  kind  of  surface  crushing  resistance  per  square  inch,  and 
td  being  assumed  somehow  to  represent  the  surface  at  which 
crushing  may  take  place.  If,  as  in  (4),  we  assume  that  the  part 
EOF  breaks  like  a  beam  fixed  at  the  ends  and  loaded  in  the 
middle,  we  have  a  much  wilder  assumption,  giving  P  =  X2£/*t  -4- 
\d  .  .  .  .  (4).  Now  we  cannot  assume  that  p  has  any  of  the 
above  values.  There  is  always  great  friction  at  the  joint ;  the 
rivet  is  in  a  state  of  unknown  tension,  and  we  have  no  inform- 
ation as  to  fa  under  such  circumstances,  and  in  a  lap  joint 
there  is  evidently  a  bending  action  due  to  the  plates  not  being 
in  line.  Nevertheless,  all  the  above  formulae  and  similar 
formulae  easily  made  out  for  all  joints  may  be  made  use  of  to 
guide  us  in  obtaining  information  when  we  vary  proportions 
and  make  tests. 

For  example,  tests  may  be  made  upon  X  and  d,  and  it  has 
been  found  that  A.  from  the  edge  of  the  hole  to  the  edge  of  the 
plate  must  be  at  least  equal  to  d,  and  considerations  of  economy 
and  workmanship  guide  us  in  adding  about  a  quarter  of  an  inch 
when  only  half-inch  rivets  are  used.  Now  if  we  put  (4)  equal 

to  (3),  or  \*tft  +  \d  —  fctd (6),  we  have  X  =  d A/| ^c. 

•  ft 

It  is  dangerous  to  follow  this  any  further  ;  we  have  reached  a 
rule  (by  taking  convenient  values  for  the  stresses)  that  agrees 
so  well  with  the  practice  of  the  best  makers  that  we  shall  be 
apt  to  think  our  theory  more  valuable  than  it  is.  It  will  be 
noticed  that  if  we  write  (3)  equal  to  (1)  we  get  a  rule  which 
also  agrees  fairly  well  with  practice ;  but  if  we  write  (4)  equal  tc 
(5),  which  we  have  just  as  good  a  right  to  do,  we  find  that  X 
ought  to  be  v/2  times  as  great  when  the  rivets  are  in  double 
shear,  and  this  is  certainly  not  in  agreement  with  good  practice. 
If  we  imagine  that  instead  of  (4)  we  use  the  idea  of  a  beam 


338  APPLIED    MECHANICS. 

breaking  by  shearing,  we  again  find  that  \  ought  to  be  propor- 
tional to  d.  But  the  result  of  a  careful  consideration  of  this 
and  many  other  points  in  machine  design  is  that  possibly  such 
formulae  as  (4)  are  more  likely  to  be  misleading  than  useful. 
A  complete  theory  to  replace  (4)  is  perfectly  possible,  but  it  has 
not  yet  had  the  services  of  the  necessary  good  mathematician. 
With  (1)  and  (2)  we  feel  much  safer  than  with  the  others. 

Putting  them  equal,  ^  dtyB  =  (p  -  d)tft  ....  (7).    We  need  some 

other  equation  evidently.  Now,  in  all  design  we  try  to  obtain 
maximum  advantage  of  some  kind.  To  get  maximum  economy 
of  material  we  were  obviously  right  in  trying  to  have  a  joint 
equally  ready  to  break  in  various  ways. 

But  there  is  a  more  general  kindof  economy,  not  at  all  easy  to 
express  algebraically  (see  my  Calculus  for  Engineers,  Art.  37), 
which  tells  us  to  punch  holes  in  thin  plates  up  to  J"  thick  and 
drill  them  in  thick  plates.  For  thin  plates,  then,  we  have  such 
a  rule  as  d  oc  t,  or  taking  into  account  cost  of  riveting,  say 
d  =  f  £  +  J  if  we  are  not  to  have  too  much  expense  in  the 
fracture  of  punches  [compressive  strength  of  punch  more  than 
equal  to  shearing  resistance  of  plate],  whereas  we  have  only  ons 
of  these  things  to  consider  in  drilling  holes.  We  may  take  the 
rule  d  =  1.2  \/ 1  .  .  .  .  (8)  as  probably  agreeing  with  the  best 
practice  from  t  =  \  to  t  =  1  inch.  If  this  rule  were  followed, 

then,  using  (7)  and  (8),  we  find  ^  l'44/8=  (p  —  d)ft,  or 
p  —  d  =  '36irfs/ft  =  A,  say.  The  strength  of  the  joint  is  the 
strength  of  the  unhurt  plate  multiplied  by  ^- ,  or  by 

A  P 

-— 3  ....  (9)  where  A  stands  for  p  —  d.     For  a  rivet  in  double 

shear  we  use  (2),  (5),  and  (8),  and  find  p  -  d='72irfjft,  and  we 
may  call  this  A  in  the  formula.  It  is  just  twice  the  last. 

Now  for  complicated  kinds  of  joint  we  must  make  assump- 
tions which  are  less  likely  than  the  above.  It  is  usual  to 
calculate,  instead  of  p  —  d  in  (2),  a  width  of  strip  w  equivalent 

7T 

in  tensile  strength  to  one  rivet  in  single  shear,  or  w  =  -j  d^fB/tft. 

Draw  round  each  rivet  a  circle  of  diameter  d  +  w,  and  let  lines 
come  to  the  circles  dividing  the  plate  up  into  strips  of  the  breadth 
w  ;  thus  we  allot  a  strip  of  plate  to  each  rivet.  Students  ought 


APPLIED    MECHANICS. 


339 


fco  scheme  for  themselves  examples  such  as  are  figured  in 
books  on  "  Machine  Design."  In  such  books  they  will  also  find 
sketches  of  many  riveted  joints  which  are  well  worth  study  in 
case  there  are  no  actual  specimens  to  look  at.  The  result  of 
the  calculation  is  a  formula  like  (9).  * 

The  results  of  actual  tests  show  that  instead  of  A  in  the 
numerator  of  (9),  we  have  a  different  number  k  A,  and  students 
who  have  read  Chap.  XIII.  carefully,  and  also  what  I  have 
here  given,  will  know  fairly  well  why  k  is  not  equal  to 
unity.  We  have,  then,  d  =  1-2  */T,  p  =  A  +  d  .  .  .  .  (10). 

k  A 
Strength  of  joint  =  — — — r  x  strength  of   unhurt   plate. .  .  . 

(11),  where  k  is  given  in  the  following  table  : — 


Iron  Plates. 

Steel  Plates. 

Single  Riveted 

Drilled     .. 

•88 

1-0 

»                        J5 

Punched  .. 

•77 

0-9 

Double      „ 

Drilled     .. 

•95 

1-06 

Punched  .. 

•85 

1-0 

Treble       „ 

Drilled     .. 

... 

1-08 

And  A  is  given  in  the  following  table : — 


Iron  Plates  and  Iron 
Kivets. 

Steel  Plates  and  Steel 
Rivets. 

Drilled 
Holes. 

Punched 
Holes. 

Drilled 
Holes. 

Punched 
Holes. 

Lap  Joint  or  Butt 
Joint  with  One 
Covering  Plate 

)  Single  Riveted 
[  Double      „ 
Treble       „ 

1-20 
2-22 
3-23 

1-47 
2-66 

0-9 
1-7 
2-5 

1-08 
1-93 

All  these  values  of  A  are  to  be  doubled  for  butt  joints  with 
two  covering  plates. 

It  is  wise  to  slightly  round  the  outside  sharp  edges  of 
drilled  holes.  It  is  to  be  remembered  that  a  punched  hole  is 
to  be  called  a  drilled  hole  if  the  plate  has  been  annealed,  or  if 
the  hole  has  been  rhymered  out  after  punching.* 

*   See  Appendix. 


340  APPLIED   MECHANICS. 

The  suitable  pressure  p  Ibs.  per  square  inch  of  a  boiler  is 

then  P  =  t =  /  -f-  R  if  t  and  R  are  thickness  of  plato 

A  +  1-2  v/r 

and  radius  of  boiler  in  inches,  and  f  is  the  suitable  tensile 
stress  in  pounds  per  square  inch  of  the  unhurt  plate,  and 
where  A  and  k  are  the  numbers  given  in  the  above  tables  for 
the  longitudinal  seams. 

In  the  following  exercises  the  working  f  is  to  be  taken  as 
10,000  Ibs.  per  square  inch  for  wrought  iron,  and  12,000  Ibs. 
per  square  inch  for  mild  steel. 

EXERCISES. 

1.  Cylindric  boiler  10  feet  diameter,  1^-inch  steel  plates,  with  steel 
rivets,  drilled  holes ;  butt-joint,  with  two  covering  plates ;  treble  riveted. 
What  is  the  working  pressure  ?    What  are  the  pitch  and  the  diameter  of 
the  rivets?      Ans.,  p  =  213  Ibs.  per  sq.  m.  p  =  6'34  inches.  d=l'34  inch. 

2.  Cylindric  boiler  5  feet  diameter,  iron  plates,  iron  rivets,  double 
riveted  butt-joint,  one  covering  plate,  drilled  holes;  working  pressure, 
150  Ibs.  per  square  inch.     Find  t.  Am.,  t  =  0-67  inch, 

3.  Find  the  dimensions  and  strength  of  a  treble  riveted  butt-joint 
with  double  covering  plates  for  plates  f  inch  thick.     Plates  and  rivets 
are  of  steel.      Ans.,  d  =  1-04  inch,    p  =  6-04  inches.     Efficiency  =  -895. 

4.  Determine  the  pitch  of  the  rivets  for  a  single  riveted  lap-joint  of 
|  inch  plate,  so  that  the  joint  may  be  equally  strong  to  resist  tearing  and 
shearing.    Diameter  of  rivets  is  |  inch.    Safe  shearing  strength  7,800  Ibs. 
per  square  inch  ;  safe  tensile  strength  is  10,000  Ibs.  per  square  inch. 

Ans.,  1-8 13  inch. 

5.  In  a  single  riveted  lap-joint  the  plates  are  £  inch  thick,  and  the 
rivets  |  inch  diameter ;  calculate  the  pitch  for  the  greatest  strength  of 
joint,  the  shearing  resistance  of  the  rivet  being  three-quarters  of  the 
tiensle  resistance  of  the  plate  per  square  inch  of  section. 

Ans.,  p  =  l-78  inch. 

6.  The  steel  plates  of  a  boiler  are  •&  inch  thick,  and  connected  by 
longitudinal  double  riveted  butt-joints,  with  covering  plates  on  each  side  ; 
find  suitable  proportions  for  a  joint  and  calculate  its  efficiency. 

Ans.,  d  =  0-9  inch,    p  =  4-3  inches.    Efficiency  =  '79. 

7.  In  a  single  riveted  lap-joint,  assume  that  the  shearing  strength  per 
unit  of  area  of  the  rivets  is  equal  to  the  tearing  strength  of  plates  per 
unit  area.     If  the  plates  are  |  inch  thick,  diameter  of  rivets  I  inch,  and 
the  pitch  2£  inches,  will  the  joint  yield  by  tearing  or  by  shearing? 
Calculate  the  efficiency  of  the  joints. 

Ans.,  Tearing.    Efficiency,  -69; 

8.  Two  lengths  of  mild  steel  tie-rods,  7  inches  x  1  inch,  are  to  bo 
connected  with  double  butt-straps.     Determine  dimensions  and  efficiency. 

Ans.,  d  =  1-3  inch.     Three  rivets  each  side.     Straps,  t  =  0-647  inch. 
Efficiency,  0-81. 

287.  Nature  of  Shear  Strain.— When  i/  moves  to  y,  Fig.  178,  the 
diagonal  x  y'  becomes  extended  to  x  y.      Imagine  the  angle  yf  a  y, 


APPLIED    MECHANICS.  34] 

as  shown,  to  be  an  exaggeration  of  about  one  hundred  times  what 
it  ever  is  in  iron  or  steel.  Its  original  length  was  A/2~  (the 
diagonal  of  a  square  whose  side  is  1),  and  its  new  length  is 

V2  -\-  "Tr,  afc  we  see  very  easily.  Hence  the  diagonal  x  y',  and 
(til  lines  parallel  to  this  diagonal,  have  a  tensile  strain,  whose 

amount  is  — — — ^-1 TV   °r    ~/z  -5-  \/2J  and  this  is  — .     Again. 

original  length        v2  2 

in  the  same  way  we  find  that  the  diagonal  y'a?,  and  all  lines 
parallel  to  it,  have  a  compressive  strain  whose  amount  is  ~.  Thus 

it  has  become  quite  clear  that  a  shear  strain  simply  consists  of  a 
compressive  strain  in  one  direction,  accompanied  by  a  tensile 
strain  in  the  perpendicular  direction,  these  strains  being  each  half 
the  shear  strain. 

We  are  led  to  speculate  on  the  behaviour. of  material  sub- 
jected to  equal  compressive  and  tensile  stresses  crossing  inter- 
faces at  right  angles  to  one  another.  Consider  a  small  right- 
angled  prism  of  material,  shown  in  Fig.  178,  of  which  M  F  K 
(Fig.  180)  is  a  magnified  cross-section. 
Neglect  the  weight  of  the  prism,  and 
for  ease  of  calculation  make  M  F,  say, 
1  inch,  M  K  the  same,  and  let  the  length 
of  the  prism  at  right  angles  to  the  paper 
be  also  1  inch.  This  prism  is  kept  at 
rest  by  the  matter  outside  it  acting  on 
its  three  faces.  Face  M  F  is  pushed  by 
a  normal  force  of  p'  Ib.  per  square  inch, 
and  as  its  area  is  just  1  square  inch, 
the  resultant  push  is  p'  Ib.  acting  at  the 
centre  of  the  square  face  M  F.  Similarly 
the  face  M  K  is  pulled  by  a  normal  force  of  p'  Jb.  And  also 
the  face  F  K  is  acted  on  by  tangential  forces  of  p  Ib.  per  square 
inch,  and  as  its  area  is  V2  square  inch,  the  total  amount  of 
shearing  force  acting  on  F  K  is  p  V2  Ib.  Now,  when  three  forces 
keep  a  body  in  equilibrium,  and  two  of  them  are  at  right  angles, 
the  sum  of  their  squares  is  equal  to  the  square  of  the  third  force 
(this  is  easily  seen  if  we  draw  the  triangle  of  forces) ;  hence  the 
square  of  p  v%  which  is  2p%,  is  equal  to  p'%  -\-  p'2  or  2jo'2.  Hence 
p  —  p' ;  and  we  have  proved  that  the  compressive  and  tensile 
stresses  which  occur  in  simple  shear  strain  are  numerically  equal 
to  what  we  called  the  shear  stress.  We  have  also  to  recollect  that 
we  cannot  have  shear  stress  parallel  to  the  paper  in  planes 
at  right  angles  to  the  paper  without  having  equal  shear 
stress  parallel  to  the  paper  in  a  set  of  planes  at  right  angles 
to  the  first  and  also  to  the  paper  (see  Art.  282). 

Suppose  we  cut  a  cube  ABCD  (Fig.  181)  of  1  inch  side, 
from  a  material  subjected  to  simple  shear  strain,  and  let  the 
faces  of  the  cube  parallel  to  the  paper  have,  as  before,  no 


342  APPLIED    MECHANICS. 

stress  upon  them,  the  other  faces  being  at  right  angles  to  the  direc- 
tions of  compression  and  extension.  Shear  occurs  parallel  to  the  face 

A  c.  Let  us  consider 
the  motion  of  the  point 
D  relative  to  AC;  in 
fact,  regard  A  c  as 
fixed.  Under  the  sole 
S  action  of  the  pushes 

jf-P  on    A  D    and    B  c  we 

know  that  the  side  D  c 
shortens  by  the  small 
amount  pa  (see  Art. 
265).  Let  us  set  this 
off  from  D  to  M.  It  is 
greatly  magnified,  as 
shown  in  Fig.  181. 
But  when  this  occurs 
the  side  A  D  lengthens 
by  the  amount  pfi  ;  set 
this  off  from  M  to  D'. 
Hence  the  pushing 
forces  on  A  D  and  B  c 
cause  D  to  move  to 
D'.  Again,  the  pulling 
forces  on  D  c  and  A  B 

further  lengthen  A  D  by  the  distance  pa,  which  we  set  off  from  D' 
to  L,  and  shorten  D  c  by  the  distance  ^/3,  which  we  set  off  from  L 
to  D".  Hence  the  motion  of  D  due  to  the  pulls  and  pushes  acting 
together  is  D  D",  and  we  see  that  this  is 

(DM  -}-  LD")  \/2  or  (a  -f  /3) pV2. 

But  s,  the  amount  of  shear,  is  D  D"  -r  D  o,  and  as  D  o  =  -^=  inch, 


'4. 


Fig.  .81. 


as  A  D  is  1  inch,  we  have 

*  =  (a 


=  or  2p  (a 


That  is,  shear  strain  =  shear  stress  multiplied  by  2  (o  +  0).  So 
that  the  reciprocal  of  2  (o  +  0)  is  what  we  called  N,  the  modulus  of 
rigidity  of  the  material. 

289.  General    Results.—  Referring  back  to  Arts.  269  and  288, 
you  will  see  that  we  have 


Modulus  of  rigidity     N  = 

Modulus  of  elasticity  of  bulk     ...     x  = 


Young's  modulus  of  elasticity   ...     B  =  - 


and  you  will  also  see  that  if  we  know  two  of  these  for  any 
material,  we  can  find  the  third. 

These  results  are  so  important  that  we  put  them  aJ so  thus:— 


APPLIED    MECHANICS.  343 

a-1.1       E  -     9NK 
*   -«?     9k'       -3I+S* 

Some  French  mathematicians  have  thought  that  the  ratio  of  0 
'  to  a  (called  Poisson's  ratio,  and  always  denoted  by  the  letter  <r), 
and  therefore  the  ratios  of  N,  K,  and  E  to  one  another,  are  constant 
for  isotropic  substances,  a  being  always  four  times  /3.  Experiment 
Las  shown  that  this  is  not  the  case,  the  ratio  of  a  to  $  being  3  to 
2-5  in  glass  or  brass,  3  '3  in  iron,  4'4  to  2-2  in  copper,  and  in  other 
substances  varying  from  these  values  very  much  indeed. 

Just  as  Young's  modulus  is  seldom  found  from  experiments  on 
the  extension  of  wires,  but  rather  from  the  bending  of  beams,  so 
the  modulus  of  rigidity  is  seldom  found  from  experiments  like  that 
of  Fig.  176,  but  rather  from  experiments  on  the  torsion  of  rods  or 
wires. 

In  all  the  cases  of  simple  shear  which  we  have  described  we 
had  more  than  simple  compression  and  extension.  There  was  a 
rotation  of  the  lines  in  which  the  compression  and  extension  took 
place.  When  there  is  no  rotation,  and  this  is  very  easy  to  imagine 
(let  A  c  not  be  imagined  fixed  in  Fig.  181,  but  imagine  the  lines  at 
45°  to  the  horizontal  to  remain  fixed  in  direction),  so  that  the 
principal  directions  of  strain,  as  they  are  called,  remain  the  same  in 
direction,  the  strain  is  said  to  \>%pure  (or  irrotational),  otherwise  it 
is  said  to  be  "  rotational." 

EXERCISES. 

1.  For  mild  steel  E  =  30  x  106,  N  =  12  x  106  ;  find  o,  ft,  and  K. 

An*">    a==BQ  x  10«'    0  —  12  x  10?'    K  =  2°  x  10<J- 

2.  The  halves  of  a  flange  coupling  are  bolted  together  with  six  bolts 
at  6  inches  from  the  centre,  60  horse-power  transmitted,  100  revolutions 
per  minute  ;  safe  shear  stress,  8,000  Ibs.  per  square  inch.   Find  the  proper 
diameter  of  each  bolt.  Ans.,  -41  inches. 

3.  A  3-inch  square  steel  bar,  fixed  at  one  end,  loaded  with  80  Ibs.  at 
its  other  end;  find  the  deflections,  due  to  shearing  in  the  two  cases; 
length  3  inches,  length  50  inches;  according  to  Art.  281,  and  compare 
with  the  correct  method  of  Art.  369. 

4.  Ten  cubic  inches  of  wrought  iron  and  10  cubic  inches  of  water  are 
subjected  to  fluid  pressure  of  3  tons  per  square  inch;    find  the  new 
volumes.     If  the  iron  is  spherical,  what  are  the  old  .and  new  diameters  ? 

Am.,  Iron,  9  '99664  cubic  inches  ;  water,  9'79  cubic  inches. 
Iron,  diameters  2'673  and  2-6727  inches;  water,  2'673 
and  2  '671  inches. 

5.  For  wrought  iron  E  =  29,000,000,  K  =  20,000,000  ;   find  a,  ft,  N, 
and  calculate  the  value  of  Poisson's  ratio. 

23  x  10':  -263- 


6.  A  cube  of  copper  3  inches  edge  is  subjected  to  a  hydrostatic  pres- 
sure of  4  tons  to  the  square  inch  ;  find  its  new  volume.    K  =  24,000,000. 

An*.,  26-99  cubic  inches. 


344 


APPLIED    MECHANICS. 


7.  For  copper  N  =  5,600,000,  K  =  24, 000,000 ;  calculate  the  value  of  B 
and  Poisson's  ratio.  Am.>  15'6  X  106  ;  -385. 

8.  A  steel  punch  \  inch  in  diameter  is  employed  to  punch  a  hole  in  a 
plate  |  inch  in  thickness ;   what  will  be  the  least  pressure  necessary  to 
drive  a  punch  through  the  plate  when  the  shearing  strength  of  the 
material  is  35  tons  per  square  inch  ?  Ans,,  51-56  tons. 

9.  In  a  fly  press  for  punching  holes  in  iron  plates  the  two  balls 
weigh  30  Ibs.  each,  and  are  placed  at  a  radius  of  30  inches  from  the  axis 
of  the  screw,  the  screw  itself  having  a  pitch  of  1  inch.     What  diameter 
of  hole  could  be  punched  by  such  a  press  in  a  wrought-iron  plate  of 
\  inch  in  thickness,  the  shearing  strength  of  which  is  22-5  tons  per  square 
inch  ?    Assume  that  the  balls  are  revolving  at  the  rate  of  60  revolutions 
per  minute  when  the  punch  comes  into  contact  with  the  plate,  and  that 
the  resistance  is  overcome  in  the  first  sixteenth  of  an  inch  of  the  thick- 
ness of  the  plate.*  Ans.,  1-136  inch. 

290.  We  have  been  in  the  habit  of  testing  material  under  tensile 
stress  only,  or  compressive  stress  only,  or  shearing  stress  only. 
Tests  are  now  greatly  required  of  the  strength  of  material  under 
combined  tensile  and  compressive  stress,  these  not  being  equal  in 
amount  as  they  are  in  shear. 

In  an  indirect  manner  we  have  evidence  that  if  on  an  interface 
there  is  shear  stress  q  and  compressive 
stress  p,  the  strength  of  the  material  in 
resisting  fracture  at  this  section  by 
shearing  is  as  if  a  shear  stress  acted  of 
the  amount  q  —  pp,  where  /*  is  a  constant 
for  the  material.  In  fact,  the  compres- 
sive stress  strengthens  the  material  in 
its  resistance  to  shearing.  It  is  as  if  ju 
were  a  co-efficient  as  of  friction,  pre- 
venting sliding.  The  evidence  is  the 
fact  that  when  struts  of  cast  iron, 
stone,  brick,  and  cement  are  crushed, 
fracture  usually  occurs  at  a  section  which 
makes  an  angle  greater  than  45°  with 
the  cross-section. 

In  Fig.  182,  A  B  is    the    cross-section 
of   area   A   of    a    tie-bar    or   strut,    and 

CD  is    a    sloping    section    making    an  angle   6  with  the  cross- 
section  ;  so  that  the  area  of  c  D   is   A  sec.  0.     An   axial  load  w 

produces  tensile  or  compressive  stress^  =  ^  on  the  cross- section, 

\v> 
and or  pl  cos.  B  in  an  axial  direction   on   the   sloping 

A  SCC.  u 

*  I  suppose  that  we  shall  always  have  academic  exercises  like  this  and 
like  Ex.  6  of  Art.  172.  The  advanced  student  will  notice  (see  Art.  486)  that 
It  is  just  as  misleading  as  those  absurd  pile-driver  problems  in  which  force  ia 
calculated  as  the  kinetic  energy  of  the  pile-driver  divided  by  the  distance 
through  which  the  pile  is  driven,  so  that,  if  the  pile  is  not  driven  any 
distance  into  the  ground,  the  force  is  infinite.  In  impact  questions  like  these 
safety  is  only  to  be  found  in  considering  force  as  the  time  rate  of  transfer  of 
momentum. 


Fig.  182. 


APPLIED    MECHANICS. 


345 


section.  The  stress  on  the  sloping  section  consists,  therefore,  of 
a  normal  tensile  or  compressive  stress  pl  cos.2  0  and  a  shear  stress 
Pi  cos.  0  .  sin.  0.  The  material  does  not  tend  to  break  at  c 
r>  by  the  normal  stress  if  the  material  is  isotropic,  as  pl  cos.2  0 
is  less  than  pr  The  shear  stress  is  a  maximum,  and  equal  to 
^pl  if  0  is  45°.  Now,  in  materials  like  cast  iron,  stone,  brick, 
and  cement,  the  ultimate  shear  stress  is  less  than  half  the 
ultimate  compressive  stress,  and  it  might  be  expected  that  the 
fracture  of  a  strut  would  be  at  a  section  making  an  angle  of  45° 
with  the  cross-section.  But  in  every  case  we  find  the  angle 
greater  than  45°,  being  nearly  constant  for  the  same 
material.  The  assumption  that  there  is  a  resistance  to 
shearing  of  the  nature  of  friction — that  is,  proportional  to  the 
normal  compressive  stress — is  not  only  a  reasonable  -  looking 
hypothesis,  but  it  agrees  quite  well  with  the  observed 
phenomena.  For  a  shear  stress  pl  cos.  0  sin.  0  and  a  compres- 
sive stress  PI  cos.2  0  are  by  our  assumption  to  be  replaced  by  a 
shear  stress  j?x  (cos.  0  sin.  0  —  /*  cos.2  0)  ....  (1)  ;  and  this  is 

a  maximum  on  the  plane  for  which  0  =  45°  +  ^  if  /t  =  tan.  </>... (2 

In  cast  iron,  fracture  usually  occurs  on  a  plane  making  54f °  with 
the  cross-section.  If  our  hypothesis  were  really  correct,  our 
co-efficient  of  internal  friction  would  therefore  be  *35  for  cast  iron. 

291.  There  is  much  published  information  on  the  fracture  by 
compression  of  blocks  of  stone,  cement,  and  bricks.     In  almost 
every  case  care  is  taken  in  loading  the  usually  short  specimens  that 
friction  at  the  ends  shall  prevent  the  material  swelling  laterally. 
When  sheet-lead  is  inserted  at  the  ends  it  gives  a  small  amount  of 
lateral  freedom,  and  in  every  case  the  breaking  load  is  lessened 
by  its  use,  and  therefore  it  is  said  to  be  wrong  to  use  lead.     I 
consider  all  this  published  information  to  be  nearly  valueless, 
except  that  there  is   some  proba- 
bility that  half   the  usually  pub- 
lished ultimate  compressive  strength 

for  a  cube  is  the  true  resistance  to 
compression  in  the  material.  Hence, 
when  the  published  strength  of 
stone  is  given  as  between  250  and 
1,600  tons  per  square  foot,  we 
ought  possibly  to  take  it  as  truly 
from  125  to  800.  In  masonry 
structures  the  working  stress  is 
probably  never  as  great  as  50  tons, 
and  generally  it  does  not  reach  10 
tons  per  square  foot. 

292.  To    see     to    what    extent 
we  can,  by  means  of  lateral  fluid 
pressure  on  a  strut,  strengthen  it, 

we  may  provisionally  use  the  theory  Fig.  188. 

above  given. 

If  there  is  compressive  stress  in  a  strut,  plt  and   a  lateral 
normal  pressure  p*  and  the  material  breaks  by  shearing,  let  the 


346  APPLIED    MECHANICS. 

plane  B  c  be  part  of  the  cross-section,  with  axial  compressive  stress 
acting  through  it ;  let  the  plane  A  c  have  normal  stress  ^;2  through 
it ;  let  the  interface  A  B  have  normal  compressive  stress  p  and 
tangential  stress  q  upon  it ;  consider  the  equilibrium  of  a  prism 
1  inch  at  right  angles  to  the  paper,  and  let  A  c  =  1  inch.  The 
normal  pressure  forces  on  the  faces  parallel  to  the  paper  balance 
one  another  independently.  Equating  the  resolved  parts  of  the 
forces  vertically  and  horizontally,  we  find 

p  =  PI  .  cos.20  +  p2  sin.20,   q  =  (pi  —  p2)  cos.0  .  sin.0. 
We  shall  use  the  hypothesis  that  q  will  not  produce  fracture  until 
it  exceeds  /  by  the  amount  of  the  friction  pp ;  that  is,  we  may  take 
it  that  the  value  of  the  shear  stress  as  a  producer  of  rupture  is 
only  q  —  up,  and  it  is  easy  to  show  that  this  is  greatest  when 

6  =  45  +  ^  .  .  .  .  (1).     This  angle,  therefore,  is  independent  of  p\ 

and  jt?2-  it  q  —  ftp  is  put  equal  to  /s,  we  see  that  the  least  value  of 
Pi  to  resist  fracture  is 

_  pl  (cos.0  sin.0  -  fj.  cos.20)  -/« 

cos.0  sin.0  +  p  sin.20 

or,  indeed,  we  may  say  that  the  principal  compressive  stresses  pl 
and  p2  will  produce  fracture  if 

pi  (cos.  0  sin.  6  —  ft  cos.20)  —  pz  (cos.  0  sin.  0  +  /*  sin.2  0)  >/a...(3), 
where  0  has  the  value  given  in  (1).  p.2  is  supposed  to  be  less  than 
pi.  If  jt?2  is  a  tensile  stress,  we  have  only  to  give  it  a  negative 
sign  in  (3).  (3)  may  be  written  mp\  —  npz>fs  .  .  .  .  (4),  where 
m  and  n  are  constants  for  the  material. 

Now  let  pv  py,  and  p  be  tensile  stresses,  and  reverse  the 
arrowhead  on  q.  We  have  the  same  numerical  relations  between 
Pi  9i  PD  Py>  an^  fracture  occurs  by  shearing  when  fs  has  the  value 
q  +  pp,  so  that  fracture  occurs  when 

pl  (cos.0  sin.0  +  p.  cos.20)  —  p2  (cos.0  sin.0  —  p  sin.20)  >  fs, 
if  6  is  taken  of  such  a  value  as  to  make  q  +  pp  a  maximum;  that 
is,  if  0  =  45°  +  0/2  ....  (2).  In  this  p%  is  supposed  to  be  less 
than  pi.  .  If  jp2  is  a  compressive  stress,  we  have  only  to  give  it  a 
negative  sign  in  (1).  (1)  may  be  written  m*p\  —  nlp2>fs,  where 
ml  and  nl  are  constants  for  the  material.  Let  p.2  =  0.  If  pl  is 

compressive  stress,  0  =  45  +  |,  and  it  will  be  found  that 
.f±  =  sec.0  -  tan.<p (4), 


If  j*>j  is  a  tensile  stress,  6  =  45  -  |,  and  it  will  be  found  that 


*  =  Sec.0  +  tan.*  ----  (6), 

-  ...a, 


APPLIED    MECHANICS.  347 

using  /c  and  /t  as  the  usual  crushing  and  tearing  stresses  of  the 
material  and  eliminating  <f>  "between. 

Given  compressive  stress  pl  and  the  value  of  /  to  find  the  least 

value  of  j92  for  resistance  to  fracture  by  shearing  ;  take  6  =  45°  +  ^ 

as  giving  the  plane  on  which  the  tendency  to  shear  is  greatest, 
whatever  p%  may  be,  and  we  find 

_  -2fs  . 
* 


Thus,  for  example,  taking  cast  iron,  let  us  suppose  that 
<f>  =  28°  and  p.  =  0*35,  as  already  found,  if  the  ultimate  fc  = 
95,000,  and/s  =  28,500,  then  sin.</>  =  '47,  p%  =  —  34,000  +  .  36  pl 
(that  is,  pl  increases  27,800  Ibs.  per  square  inch  for  every  increase 
of  10,000  Ibs.  per  square  inch  in  pz). 

If  for  any  material  /c,  ft,  fs  are  the  numerical  values  (treated 
all  as  positive  quantities)  of  the  three  stresses  which  the  material 
will  stand,  2/s  =/c  (sec.<J>  —  tan.<J>)  =ft  (sec.<f>  +  tan.<f>). 


Now  in  cast  iron  the  proof  -stresses  given  in  Table  XXII.  are 
/c  =  21,000,  ft=  10,500,  /s=  8,000,  |  =  2  =  144^,  and 

hence  </>  =  sin."1  £  =  19°|,  tan.<f>  =  -35,  0C  =  54°|,  6t  =  35°£. 
With  this  value  of  <J>,  fs  ought  to  be  '356/c,  whereas  it  really  is 
•38/c.  This  is  a  discrepancy  very  allowable,  and  we  may  take  it  as 
some  sort  of  verification  of  the  theory.  These  were  the  first  numbers 
I  tried,  but  I  have  since  found  that  other  published  numbers  are 
less  satisfactory  in  their  support.  For  example,  fc  ought  to  be 
greater  than  ft  for  all  materials  if  the  theory  is  correct.  It  is 
evident  that  special  experiments  are  required  as  a  test. 

Tresca  and  Darwin  have  propounded  the  hypothesis  that 
Pi  —  Pz  is  constant  ;  and  it  ought  to  be  easy,  by  experiment,  to 
decide  whether  this  is  constant,  or  whether,  as  by  my  theory,  it 
increases  as  pl  and  p%  increase.  For  example,  I  make  pl  —  p2 
equal  to  94,000  when  p%  =  0  in  cast  iron,  and  equal  to  272,000 
when  p%  =  100,000.  These  are  from  the  ultimate  values.  As  to 
the  permanent  set  strength  of  cast  iron,  I  make  j^  —  j?2  =  21,000 
when  p2  =  0,  and  121,000  when  p2  =  100,000.  In  fact,  instead  of 
making  p\  —  j?2  =  94,000,  I  make  p\  —  2*78  p2  =  94,000. 

The  hypothesis  of  Poncelet  and  St.  Tenant  is  that  the  material 
is  fractured  when  the  strain  exceeds  a  certain  amount,  and  this 
hypothesis  is  often  used  to  give  ultimate  strength  conditions  when 
the  strains  have  been  investigated  mathematically.  But  in  a 
wrought-iron  bar  we  have  sometimes  before  fracture  strains  which 
are  several  hundreds  of  times  as  great  as  the  greatest  strains  to 
which  the  calculations  apply,  and  it  seems  to  me,  therefore,  that 
Poncelet's  theory  has  no  probability  of  correctness  in  its  favour. 

If  we  assume  that  in  any  kind  of  earth  there  is  a  coefficient  of 


348 


APPLIED    MECHANICS. 


friction  n  between  layers,  but  no  permanent  resistance  to  true  shear- 
ing  (that  is,/,  =  0),  we  get  Rankine's  theory  of  earth  pressure 
from  the  above  theory.  The  lateral  pressure  p%  necessary  to  prevent 
a  direct  pressure  p-±  from  causing  motion  or  fracture  is  given  by 
p^_  1  -  sm.<^  ^  a  the  stati(j  coefficieilt  of  friction  in 
pl  1  +  sin.<f> 

certain  kind  of  earth  is  '9,  so  that  <J>  is  42°,  then  pjjh  is  ~~  .669r 
or,  say,  £. 

Example. — The  weight  of  a  building  is  107  Ibs.  The  area  of 
the  concrete  bottom  of  the  foundations  is  2,000  square  feet.  At 
what  depth  ought  it  to  be  below  the  level  of  the  soil,  if  the  soil  is 
such  that  $  =  42°  ? 

Am.,  the  pressure  p\  Ibs.  per  square  foot  is  5,000  Ibs.  To 
resist  this,  a  horizontal  pressure  n*  of  1,000  Ibs.  is  needed. 
Regarding  the  horizontal  pressure  01  1,000  Ibs.  as  a  new  plt  it 
needs  a  vertical  pressure  of  200  Ibs.  per  square  foot,  due  to  the 
weight  of  outside  earth,  to  balance  it.  If  the  earth  weighs  100  Ibs. 
per  cubic  foot,  the  depth  of  the  foundation  must  be  at  least  2  feet. 

Assuming  that  the  theory  is  right,  the  difficulty  in  carrying 
out  a  rule  of  this  kind  is  that  we  do  not  know  #,  the  angle  of 
repose  of  a  particular  kind  of  earth.  Rankine  assumed  that  a  long 
mound  of  earth  would,  after  much  weathering  and  rest,  get  to 
have  a  natural  slope  <|>.  If  the  natural  surface  of  earth  is  hori- 
zontal, it  is  easy  to  find  on  the  above  theory  the  stress  on  any 
interface  when  motion  is  about  to  take  place,  and  particularly  on 
a  vertical  interface,  and  so  the  reason  for  part  of  the  following 
rule  of  Rankine's  is  known.  The  study  of  the  stresses  when 
the  ground  is  sloping  must  be  left  as  an  exercise  for  students. 

293.  Rankine's  Rule  for  Earth.— Draw  an  angle  x  o  R,  Fig. 
184,  to  represent  the  static  permanent  angle  of  repose  of  the  kind 
of  earth.  Describe  Y  R  x,  a  semicircle  touching  OR.  If  A  B, 
Fig.  1 85,  is  the  vertical  face  of  a  wall  sustaining  a  bank  of  this 
earth,  whose  slope  is  A  c,  make  the  angle  x  o  P  equal  to  the 
inclination  of  A  c  to  the  horizon.  Find  B  D  =  o  Q.A  B/O  P.  Then 


Fig.  184. 


Fig.  185. 


A  B  D  is  a  wedge  of  earth  whose  weight  represents  the  total 
pressure  acting  on  A  B.     The  pressures  act  in  directions  parallel 


APPLIED    MECHANICS. 


849 


to  A  C,  and  the  resultant  force,  representing  the  total  pressure, 
acts  a  third  of  the  way  up  from  B  to  A. 

This  rule  may  be  compared  with  the  rule  of  Art.  173  for 
water  pressure  against  a  vertical  wall.  Rankine  neglects  fric- 
tion against  the  wall  face.  Students  of  this  subject  are  directed 
to  a  paper  by  Sir  B.  Baker,  Proc.  Inst.  O.E.,  Vol.  65,  and  its 
discussion. 

When  grain  is  stored  in  vertical  prismatic  cylinders,  the 
average  pressure  in  pounds  per  square  foot  on  the  flat  bottom  is 
c  d  w  where  d  is  the  diameter  in  feet,  w  the  weight  of  a  cubic 
foot,  and  c  is  0*84  for  wheat,  0'96  for  peas.  At  greater  heights 
than  three  times  the  breadth  of  section  of  a  bin  the  pressure 
on  the  sides  is  constant,  being  about  50  Ibs.  per  square 
foot  for  all  kinds  of  grain  until  we  get  near  the  surface. 

294.  Twisting.— In  Fig. 
186,  AB  represents  a  wire  held 
firmly  at  A.  At  B  there  is  a 
pulley  fixed  firmly  to  the 
wire,  and  this  pulley  is  acted 
upon  by  two  cords,  which 
tend  to  turn  it  without 
moving  its  centre  sideways. 
In  fact,  they  act  on  the 
pulley  with  a  turning  mo- 
ment merely.  But  the  pulley 
can  only  turn  by  giving  a 
twist  to  the  wire,  and  the 
amount  of  motion  it  gets 
tells  us  how  much  the  twist 
is.  A  little  pointer  fastened 
at  c  moves  over  a  cardboard 
dial,  and  tells  us  accurately 
how  much  twist  is  given 
to  the  wire.  The  angle 
turned  through  by  the 
pointer  is  called  the  total 
angle  of  twist  at  c.  If 
we  had  a  pointer  at  each 
of  the  places  G,  H,  and  c, 
and  if  A,  G,  H,  and  C  were 
one  foot  apart  from  one  another,  we  should  find  that  the 


Fig.  186. 


350  APPLIED    MECHANICS. 

angles  of  twist  at  G,  H,  and  c  are  as  1  :  2  :  3  ;  in  fact,  the  angl* 
of  twist  is  proportional  to  the  length  of  ivire  twisted. 

You  will  find  that  if  a  twisting  moment  of  10  pound-feet 
produces  a  twist  of  4°,  then  a  twisting  moment  of  20  pound- 
feet  produces  a  twist  of  8°,  and,  in  fact,  the  twist  is  proportional 
to  the  twisting  moment  which  is  applied.  You  will  also  find 
that  if  you  try  different  sizes  of  wire  of  the  same  material,  say 
wire  whose  diameters  are  in  the  proportion  of  1,2,  3,  &c.,  and 
to  each  of  them  you  apply  the  same  twisting  moment,  the 
amount  of  twist  produced  in  them  will  be  in  the  proportion  1, 

— ,  — ,  &c. ;  that  is,  inversely  as  the  fourth  power  of  the  diameter 
16  81 

of  the  wire.  Lastly,  taking  wires  all  of  the  same  diameters  and 
lengths,  but  of  different  materials,  and  applying  to  them  the 
same  twisting  moment,  the  amount  of  twist  will  be  inversely 
proportional  to  the  number  which  we  call  the  modulus  of  rigidity 
of  the  material.  The  exact  rules  are  given  in  (1)  and  (2),  and 
the  values  of  N  given  in  Table  XX.  may  be  relied  upon  in  such 
calculations,  because  they  have  all  been  determined  from 
experiments  on  the  twisting  of  wires  and  shafts. 

Exercise. — A  brass  wire  20  inches  long,  0*1  inch  diameter, 
twists  through  a  total  angle  of  130  degrees  when  a  twisting 
moment  of  4  pound-inches  is  applied.  Find  N  for  the  material. 
Answer  : — 3*6  X  106  Ibs.  per  square  inch. 

Exercise. — What  would  be  the  twist  of  a  shaft  of  the  same 
material  with  a  twisting  moment  of  600  pound-inches,  20  feet 
long,  1'2  inch  diameter  1  Answer  : — 7 '8  degrees. 

It  is,  however,  well  to  notice  that  the  drawn  brass  will 
probably  have  a  diflerent  value  for  its  N  than  the  brass  of  a 
much  larger  shaft. 

It  will  be  seen  that  the  strain  is  a  shear  strain.     Con- 
sider M  H  G  (Fig.  187)  to  be  a  cross-section  of  the  wire  ;  then' 
a  point   which  is  at  H  before  the  twist 
occurs  is  found  to  be  at  G  when  there  is 
a  twist  in  the  wire,  and  a  point  such  as 
p'  moves  to  p,  but  a  point  o  in  the  centre 
\c  of  the  wire  does  not  move.     Now  there  is 
no  such  motion  at  the  fixed  place  A,  Fig. 
186,  and  in  each  section  there  is  more  of 
this  motion  the  farther  it  is  away  from  A ; 
in  fact,  the  motion  is  just  as  it  was  in  the 
Fig.  187.  indiarubber  of  Fig.  176,  only  that  it  varies 


APPLIED    MECHANICS.  351 

in  the  section,  the  motion  being  greatest  at  the  outside 
of  the  wire,  and  nothing  at  the  centre.  The  material 
breaks  when  the  shear  stress  at  the  surface  becomes  too 
great,  and  the  rule  found  by  experiment  is  that  for  any 
material,  whatever  the  length  of  the  wire,  the  twisting  moment 
which  will  cause  rupture  is  proportional  to  the  cube  of  the 
diameter.  It  is  well  known  that  when  a  shaft  is  transmitting 
power,  the  horse-power  transmitted  is  proportional  to  the 
twisting  moment  or  torque  in  the  shaft  multiplied  by  the 
number  of  revolutions  made  by  it  per  minute.  The  rule  used 
by  engineers  is  this  : — d  =  3'3  \/H/«  ....  (1),  as  giving  the 
safe  diameter  of  a  wroughtiron  shaft  at  n  revolutions  per  minute, 
if  it  is  only  subjected  to  torsion.  Observe  that  if  we  double 
the  speed,  the  shaft  is  strong  enough  for  double  the  power. 
Instead  of  3 '3  we  use  2*9  for  a  mild  steel  shaft  and  4  for  cast 
iron. 

294a.  Shafts  usually  carry  pulleys,  and  are  otherwise  loaded 
as  beams,  as  by  the  pull  of  belts,  and  therefore,  for  reasons 
given  in  Art.  379,  we  take  1J  to  1^  of  the  above  size  for  mill 
shafting,  and  for  crank  shafts  and  shafts  subjected  to  shocks 
we  sometimes  add  50  per  cent,  to  the  diameter  as  given  by  (1). 
We  have  some  explanation  in  our  theory  of  Art.  263  for  this 
increase;  some  of  it  is  due  to  the  variation  in  stress,  and 
therefore  to  fatigue ;  some  of  it  is  due  to  the  fact  that  in 
crank  shafts  the  maximum  torque  is  often  double  the  average 
torque.  In  a  long  line  of  shafting,  if  the  power  is  given  off  at 
various  places  with  some  irregularity,  it  may  even  become 
evident  to  the  eye  that  the  shaft  is  perpetually  twisting  and 
untwisting,  for  of  course  the  twist  is  proportional  to  the  horse- 
power transmitted  if  the  speed  is  constant.  When  this  is  the 
case,  although  the  shaft  may  seem  to  be  strong  enough,  it  is 
weak  because  it  is  not  stiff  enough.  A  vevy  long  shaft  some 
times  gets  into  a  state  of  torsional  vibration  just  in  the  same 
way  that  the  cage-rope  of  a  coal-mine  gets  into  a  state  ol 
longitudinal  vibration.  The  nature  of  this  vibration  will 
depend  on  accidental  causes,  and  should  the  impulses  that  give 
rise  to  it  happen  to  repeat  themselves  at  proper  intervals,  the 
vibration  may  go  on  increasing  until  the  torsion  at  some  place 
may  be  sufficient  to  produce  rupture.  In  the  same  way  a 
number  of  men  walking  from  side  to  side  of  a  large  ship,  just 
taking  as  much  time  in  going  from  one  side  to  the  other  as  the 
ship  takes  to  make  a  vibration,  may  make  the  rolling  dangerous. 


352 


APPLIED    MECHANICS. 


It  is  for  this  reason  that  we  endeavour  to  make  the  period  of 
oscillation  of  a  ship  differ  greatly  from  the  probable  period  of 
waves  which  she  may  experience  (see  Art.  489). 

In  very  small  shafting  this  vibration  often  occurs,  and  it  is 
usual  to  add  vaguely  f  to  f  inch  to  the  diameters  found  by  the 
above  rule — a  sacrifice  to  the  Goddess  of  Chance. 

295.  Consider  a  little  prism,  P  B  (Fig.  188),  whose  ends  lie  in  two 
cross-sections  of  a  shaft  near  together,  o  being  the  centre  of  one  of 
the  sections,  and  o'  the  centre  of  the  other.     The  twisting  strain 
causes  B  to  move  to  B',  regarding  P  as  fixed.     (The  motion  is,  of 

course,  usually  very  much  less 

,       than  I  have  here  shown    it). 

^-j  o     There  must,  then,  be  shearing 

/»     _ -*•*"  forces  acting  on  the  ends  in 

opposite  directions.  If  a  is  the 
angle  of  twist  of  the  shaft  per 
inch  of  its  length,  then  B  o'  B"  is 
o  multiplied  by  o  o' ;  and  if  o  P 
or  o'  B  is  ?',  then  B  B'  is  r  ao  o', 
where  o  is  an  angle  measured 
in  radians.  The  shear  strain  in 
the  little  prism  is  BB'  divided 
by  P  B  or  o  o',  so  that  it  is  ra ; 
hence  the  shear  stress  is  Nra 
(see  Art.  282).  If  a  is  the  area 
of  the  end  of  the  little  prism  in 
square  inches,  the  shear  force 
acting  on  it  is  N  r  o  0,  and  as 
this  acts  in  the  direction  at 
right  angles  to  the  radius,  its 
moment  about  oo'  is  xr2aa. 

But  we  have  a  similar  moment  for  every  such  little  area  into  which 
the  cross-section  may  be  divided,  and  to  find  the  total  torque  we  must 
take  the  sum  of  all  such  terms.  Now,  N  and  a  are  the  same  every- 
where, so  that  in  taking  such  a  sum  our  only  difficulty  is  with  tho 
factors  r^a.  But  the  sum  of  all  such  terms  as  r^a  is  called  the 
moment  of  inertia  of  the  section  about  the  axis  o  o',  rihd  it  has  been 
calculated  for  us.  Thus,  if  D  is'-'the  diameter  of  a  round  shaft,  the 
moment  of  inertia  of  its  section  about  an  axis  through  its  centre 
at  right  angles  to  the  section  is  IT  D4  ^  32,  and  for  a  hollow  shaft 
whose  outside  diameter  is  D  and  inside  diameter  d,  the  moment  of 
inertia  is  ir  (D4  —  d*)  -j-  32 ;  and  hence  we  sec  that  the  twisting 
moment  T  necessary  to  produce  a  twist  of  a.  radians  per  inch  in  a 
round  shaft  of  diameter  D  is  T  —  TTN  aD4/32  ....  (1),  and  for  a 
hollow  shaft  it  is  T  =  TT  N  o  (D4  —  rf4)/32  ....  (2).  The  torsional 
rigidity  of  a  shaft  is  defined  as  A  if  o  =  T/A.  The  values  of  A  will 
be  found  in  Table  XV.,  Art.  532,  for  various  sections  of  shaft.  The 
verification  of  these  rules  is  an  excellent  laboratory  exercise. 

296.  The  strength  of  a  shaft  is  to  be  calculated  on  the  assump- 
tion that  rupture  occurs  when  the  shear  stress  Nra  mentioned 


Fig.  188. 


APPLIED    MECHANICS.  353 

ubove  exceeds  the  greatest  shearing  stress  to  which  the  material 
ought  to  be  subjected ;  and  as  the  stress  is  greatest  when  r  is  the 
outer  radius  of  the  shaft  or  \  D,  so  that  /  =  £  N  D  o,  and  as  from 
equation  (1)  (Art.  295)  we  find  that  N  a  is  32  T  -f-  it  D4,  we  know 
that/=  2°  x  32  T  +  irD4,  and  this  is  the  condition  of  strength 
of  a  cylindric  shaft.  It  is  more  compactly  put  in  the  form 

T  =  ~IQ-  for  solid  cylindric  shafts  ....  (1), 
and  in  the  same  way  we  get 

T  =  — i — -^- — —  for  hollow  cylindric  shafts  ....  (2), 
lo  D 

/  being  the  breaking  shear  stress  of  the  material  in  pounds  per 
square  inch,  x  the  twisting  moment  in  pound-inches  which  will 
cause  rupture,  D  the  outer  diameter,  and  d  the  inner  diameter  (if 
the  shaft  is  hollow)  in  inches. 

We  see,  then,  that  the  strength  of  a  solid  shaft  depends  on  the 
cube  of  its  diameter,  whereas  its  stiffness  depends  on  the  fourth 
power  of  its  diameter. 

As  to  the  practical  rule  given  in  (1)  (Art.  294),  we  saw  in  Art.  182 
that  torque  in  pound-/^,  multiplied  by  angular  velocity  in  radians 
per  minute,  divided  by  33,000,  is  the  horse-power.  As  we  use  T  in 
pound-inches,  T  =  12  x  33,000  x  n/2irn.  If  we  use  this  in  (1)  of 
this  article  and  take/=z  9,000  Ibs.  per  square  inch,  we  find  the  usual 
practical  rule  (1)  (Art.  294).  Taking  /=  4,500  for  cast  iron  and 
13,500  for  steel,  we  find  the  rules  for  these  materials  already  given. 

297.  Shafts  Subjected  to  Twisting  and  Bending. — It  is  most 
convenient  here  to  assume  that  a  student  has  read  the  larger  print 
in  the  next  chapter,  and  knows  that  if  a  round  shaft  of  diameter 
D  inches  is  subjected  to  a  bending  moment  M  pound-inches  there 
are  compressive  and  tensile  stresses  through  the  cross-section  at  the 
circumference  of  amount  p  —  32M/7T  D3  =  2aM,  say  ....  (1)  for 
a  solid  shaft,   and  32  M  D/TT  (D4  —  d4}  =  2  b  M,  say  .  .  .  .  (2)   for  a 
hollow  shaft,  if  d  is  the  internal  diameter.     We  have  just  found 
that  the  shear  stress  on  the  same  interface,  due  to  a  twisting 
moment  T,  is  /  =  a  T  and  6  T  ....  (3)  for  the  solid  and  hollow 
shafts.    The  other  stresses  across 

other  interfaces  at  the  point 
ought  now  to  be  studied,  but  it 
will  be  found  that  it  is  only 
necessary  to  determine  the  prin- 
cipal stresses  there — that  is,  the 
greatest  and  least  tensile  or  com- 
pressive stresses  which  act  across 
any  interfaces. 

298,  In  Art.  292  we  had  a 
simple  case  of  the  general  rule 
for  stresses  given  in  Art.  290, 
and  this  is  another  case  nearly 
as  simple. 


Suppose  that  across  an  inter-  j)          f 

face  A  c  at  right  angles  to  the  Fig.  189 

M 


354  APPLIED    MECHANICS. 

paper  we  know  that  there  is  a  normal  stress  (either  tensile  or 
compressive)  p  Ibs.  per  square  inch  and  a  shear  stress  /;  that  on 
planes  parallel  to  the  paper  there  is  no  stress ;  that  on  the  planes 
at  right  angles  to  A  c  and  the  paper  there  is  no  normal  stress. 
Consider  the  equilibrium  of  the  prism  A  c  D,  and  find  the  traction 
on  A  D  of  q  Ibs.  per  square  inch.  For  the  sake  of  ease  of  calculation 
we  will  assume  A  c  to  be  1  inch,  and  that  the  prism  is  1  inch  at 
right  angles  to  the  paper.  Now,  when  there  is  shear  stress  / 
across  A  c  there  is  also  an  equal  shear  stress  /  across  D  c  (Art.  282). 
What  are  the  forces  with  which  stuff  outside  acts  on  the  prism  ? 
The  resultants  of  the  forces  on  the  faces  are  : — On  A  c,  a  horizontal 
force  p  and  a  vertical  force  /,  because  A  c  is  1  square  inch  in  area  ; 
on  D  c,  a  horizontal  force  /  x  area  of  D  c  or  /.  D  c  or  /.  cot.  0, 
as  A  c  is  1 ;  on  A  D,  a  force  q  x  area  of  A  D  or  q  .  cosec.  0,  and  this 
force  and  its  direction  are  what  we  desire  to  calculate.  Given  0, 
it  would  be  easy  to  calculate  q  and  its  direction  ;  but  our  problem 
is  even  simpler.  It  is  this  :  Find  on  what  plane  A  D  there  is  only 
a  normal  stress  q  with  no  tangential  component,  and  find  q. 
Resolving  forces  horizontally,  /cot.  9  +  p  =  q  cosec.  6  x  sin.  9, 
or  /  cot.  &  +  p  =  q  .  .  .  .  (\).  Eesolving  forces  vertically, 
f=g.  cosec.  9  .  cos.  9,  or  f=  q  .  cot.  9  .  .  .  .  (2).  Hence,  as 
(2)  gives  us  cot.0— //#,  using  this  in  (1),  we  find  f'ljq  +p=q  or 
?2  -pq  — /2  ;  so  that  q  =  \p  +  V/ip2+/2 (3). 

It  is  easy  to  find  9.  There  are  two  answers,  differing  by  a 
right  angle.  A  stress  is  called  a  principal  stress  if  it  is  normal  to 
the  interface,  and  we  see  that  we  have  in  (3)  the  two  values  of  the 
principal  stress  due  to  a  combination  of  tensile  p  and  shearing 
stress  /,  such  as  we  have  supposed.  The  principal  stresses  are 
across  interfaces  which  are  at  right  angles  to  one  another. 

Example. — At  an  interface  p  =  6  tons  per  square  inch  and 
/  =  5  tons  per  square  inch,  then  q  —  3  +  5*83 ;  so  that  the 
principal  stresses  are  8*83  tons  per  square  inch  in  tension  and 
2-83  tons  per  square  inch  in  compression  at  right  angles  to  one 
another. 

Exercise. — Wrought  iron  is  not  to  receive  a  greater  tensile  or 
compressive  stress  than  5  tons  per  square  inch.  There  is  already  a 
tensile  stress  of  4  tons  per  square  inch  across  an  interface,  What 
shear  stress  may  also  cross  that  interface  ? 

Am.,  p  =  4  and  q  =  5  =  2  +  V4  +  f2  from  (3),  and  hence 
f—  2-24  tons  per  square  inch. 

Exercise. — A  round  shaft  is  in  torsion,  and  the  shear  stress 
produced  across  the  section  near  the  circumference  is  8,000  Ibs.  per 
square  inch.  At  the  same  section  the  shaft  is  subjected  to  bending, 
and  a  compressive  stress  of  6,000  Ibs.  per  square  inch  is  produced 
across  the  same  interface.  What  is  the  greatest  compressive  stress 
in  the  material  there  ? 

Ans.,  q  =  3,000  +  -v/9  x  106  +  64  x  10«  =  11,544  Ibs.  per 
square  inch. 

Exercise.— If  p  and  /  are  equal,  find  them  that  q  may  be  just 
6  tons  per  square  inch. 

Ans.,  Each  of  them  is  3 '71  tons  per  square  inch. 


APPLIED    MECHANICS. 


355 


299.  We  see,  then,  that  a  hollow  round  shaft  subjected  to  the 
twisting  moment  T  and  the  bending  moment  M  is  really  subjected, 
at  points  near  the  circumference,  to  the  maximum  compressive  or 
tensile  stress  q  where  q  =  *M  +  V  V*  M2  +  b*  T2  (see  (2)  and  (3)  of 
Art.  297),  or 


16D 


7T  (D4   - 


M  +    A/M2  +T: 


and  in  solid  shafts 


....(5). 


Hence     the    greatest    and    least    compressive    or    tensile    stress 
existing  in  a   hollow  or    solid    shaft  when    it  is   subjected    to 
T  and  M  is   exactly  the  same  as  the 
greatest    shearing    stress    in    a  shaft  ;/\ 

subjected  only  to  a  twisting  moment 

j£  -|_    -v/  M2  +  T2  .  .  .  .   (6). 

300.  An  overhung  crank  shaft  (Fig. 
190)  is  subjected  to  a  wrench.     If  the 
greatest  force  exerted  at  the  pin  A,  at 
right  angles  to  the  crank,  is  F,  the  twist- 
ing moment  is  F  .  A  c,  and  we  may  take 
the  bending  moment  as  F  .  B  c  if  B  is  the 
middle  of  the  j  ournal.  Hence  the  wrench 
is  equivalent  to  a  twisting  moment 

F  (B  c  +  A/  B  c2  +  A  c2),  or  F  .  (B  c  +  A  B), 
a  rule  which  is  perhaps  a  little  un- 
expected. 

301.  Ordinary  shafts  are  of  wrought  iron  or  mild  steel,  materials 
such  that  their  resistances  to  compression,  tension,  and  shearing 
are  not  very  different,  and  therefore  when  they  are  subjected  to  T 
and  M  we  at  once  calculate  (6),  and  say  that  the  strength  of  the 
shaft  is  the  same  as  if  it  were  subjected  to  this  twisting  moment 
only.  The  practical  rule  (Art.  294a)  is  worked  on  the  idea  of  a  twist- 
ing moment  only,  or  d  ex  \/  T.     If  there  is  also  a  bending  moment 
M,  which  is  of  the  value  k  T,  then  the  rule  becomes  evidently 


[C 


Fig.  190. 


d  =  3-3 


,  d  =  3-3 


The  extreme  values  of  k  for  many  kinds  of  shafting  have  beori 
worked  out,  and  the  consequent  change  in  the  multiplier  of  (1), 
by  means  of  shrewd  guessing  and  calculation  ;  but  to  my  mind 
it  is  a  better  recommendation  of  the  rules  given  in  Art,  2940 
that  they  are  consistent  with  theory  and  with  the  best  practice  of 
engineers. 

I  have  given  here  the  usual  theory  of  shafts  subjected  to 
twisting  and  bending  ;  it  assumes  that  the  strength  is  determined 
by  the  maximum  stress.  A  true  theory  would  take  account  of  the 
considerations  of  Art.  294a. 

If  a  tensile  stress  p  and  a  shearing  stress  /  act  on  the  same 


356 


APPLIED    MECHANICS. 


interface,  and  if  pt  and  p»  are  the  principal  stresses,  we  see  from 
(3)  of  Art.  298  that  _ 

+  f\ 


The  theory  of  Arts.  290  and  292  tells  us  that  fracture  will  take 
place  if  mlpi  —  nlpz  >  fs,  where  mlt  n1,  and  fa  are  constants  which 
ought  to  be  known  for  the  material.  If  p  is  a  compressive  stress, 
we  have  fracture  if  mpi  —  npz  >  /s. 

The  probable  values  of  m,  n,  m1,  nl  for  cast  iron  are  given  in 
Art.  292.  For  wrought  iron  and  mild  steel  it  is  possible  that  <f>  of 
Art.  292  is  0,  and  hence  m  =  J,  n  =  £,  m1  =  J,  n1  =  £.  Hence 
we  have  fracture  either  on  the  compressive  or  tensile  side  of  a 
shaft  —  that  is,  whether  p  is  compressive  or  tensile  stress  —  if 
2  \/  i.P2  +/2  >/«  •  Hence,  if  a  shaft  is  subjected  to  the  twisting 
moment  T  and  the  bending  moment  M,  we  calculate  an  equivalent 
twisting  moment  v/T2  +  M2,  and  assume  that  the  shaft  is  subjected 
to  this  alone. 

For  materials  in  general,  and  probably  even  in  the  case  of 
wrought  iron  and  mild  steel,  the  equivalent  twisting  moment 

ought  to  be  calculated  from 
h  M  +  k  x/T2  +  M2,  where 
h  and  &are  constants,depend- 
ing  upon  the  nature  of  the 
material,  which  are  not 
known  at  tho  present  time. 


Fig.  191. 


Kg.  192. 


302.  The  demonstration  of  Art.  29  6  is  found  to  agree  with  experi- 
ment, but  its  results  must  not  be  applied  except  to  shafts  which  are 
circular  in  section.  Our  assumption,  which  experience  warranted, 
was  that  when  such  a  shaft  as  A  B  (Fig.  191)  is  fixed  at  B,  and  when 


pig.  no. 


APPLIED    MECHANICS 


357 


Pig.  194. 


to  an  arm,  CD,  a  twisting  couple  is  applied,  every  straight  line 

in  a  section  remains  straight,  and  moves  through  the  same  angle 

as  evpjy  other  line.     But  it  can  be  shown  that  this  is  not  the  case 

for  a  shaft  of  any  other  than  a  circular  section.     Thus,  let  o  (Fi°-. 

192)  be  the  centre  of  gravity  of 

the  section   PQ.S,   and   let  us 

suppose   that   a   shaft   of   this 

section  is  subjected  to  the  sort 

of  strain  I  have  described.    The 

shear  strain  at  the  point  p  is 

in  the  direction  p  K,  perpen- 
dicular to  o  P.  Let  its  amount, 

which  we  know  to  be  o  p   X 

angle  of  twist,  be  represented 

by   the  length   of   P  K.     It   is 

easy  to  show  that  this  is  just 

the  same  as  a  shear  strain  p  N 

in  the  direction  P  N,  normal  to 

the  surface  of  the  shaft  at  p, 

together  with  a  shear  strain  in 

the  direction  P  T,  tangential  to 

the    shaft    at   p.       But    shear 

strain  in  any  direction  is  al- 
ways accompanied  by  a  similar  strain  in  a  plane  at  right  angles 

to  this  direction  (see  Art.  282),  so  that  since  we  have  the  shear 

p  N,  we  must  also  have  a  shear  parallel  to  the  axis  of  the  prism 

along  the  surface  at  P,  and  this  cannot  be  produced  merely  by 

a  twisting  moment.  We  must 
imagine  that  along  with  the 
twisting  moment  there  is  a 
force  distributed  over  the  sur- 
face of  the  shaft  to  produce 
the  above  effects.  The  result 
of  an  exact  investigation  (Art. 
313)  is  that  a  twisting  couple 
produces  a  greater  twist  than 
F'g-  !95-  Fig«  196.  might  appear  from  what  I  have 

said  in  Art.  295,  and  it  also 

produces  a  warping  of  the  naturally  plane  sections  of  the  shaft. 

Thus  Fig.  193  is  the  shape  assumed  by  each  section  of  an  elliptic 

shaft,  and  Fig.  194  of  a  square  shaft.     Imagine  a 

section  to  be  distinguishable,  say,  in  a  glass  shaft 

by  a  thin  layer  of  a  different  colour  from  the 

rest.     Deeper  shading  indicates  greater  distance 

from  the  observer  who  is  looking  towards  the 

fixed  end  of  the  shaft.     The  arrows  show  the 

direction  of  the  twisting  moment.  In  the  fol- 
lowing three  sections,  instead  of  the  torque  for  a 

twist  of  one  radian  being  equal  to  N  times  the 

moment  of  inertia  of  the  cross- section,  it  is  only 

•84  times  this  for  a  square  section  (Fig.  195),  "54  times  it  for  the 

section  Fig.  196,  and  -6  times  it  for  the  section  Fig.  197.     Indeed, 


Fig.  197. 


358 


APPLIED    MECHANICS. 


the  square  section  has  only  '88  times  the  torsional  rigidity  of  a 
cylindric  shaft  of  the  same  sectional  area;  Fig.  196  has  -67  times, 
and  Fig.  197  has  '73  times  the  torsional  rigidity  of  a  cylindric  shaft 
of  the  same  sectional  area. 

The  numbers  in  the  column  headed  w,  Table  XV. ,  express  the  rela- 
tive strengths  to  resist  twisting  of  the  various  sections  there  figured. 
The  torsional  rigidity  of  an  elliptic  section  whose  principal 
semi-diameters  are  a  and  b  is  N7r#3#3/(a2  +  b2).  If  M  is  the 
twisting  moment,  the  shear  stress  at  a  point  x,  y  (the  axis  of  x 
being  a)  is  2  M  ^(P#24-aV)M3*3-  This  is  greatest  at  the  end  of 
the  minor  axis,  being  2  M/7rfli2. 

The  torsional  rigidity  of  a  rectangle,  if  its  length  is  two  or 
more  times  its  breadth,  is,  with  some  accuracy,  the  same  as  that 
of  the  inscribed  ellipse  multiplied  by  the  ratio  of  their  polar 
moments  of  inertia.  The  greatest  shear  stress  occurs  at  the  middle 
of  the  longer  side  of  the  boundary,  and  is  3  M  («2  -f-  £2)/8  a362,  if  M 
is  the  twisting  moment  and  a  is  half  the  longer,  b  half  the  shorter, 
side  of  the  rectangle. 

303.  A  very  interesting  result  of  the  investigation  is  that  there 
is  always  greatest  distortion  at  that  part  of  the  surface  of  a 
shaft  where  the  surface  is  nearest  the  axis.  Thus  in  an  elliptic 
shaft  the  substance  is  most  strained  at  the  ends  of  the  shorter 
diameter  of  the  section.  Imagine  a  very  light  box  to  be  made 
so  as  to  contain  frictionless  liquid  exactly  of  the  shape  of  a 
shaft.  If  we  give  a  sudden  turn  to  the  box  about  the  axis, 
the  liquid  will  be  left  behind  if  the  box  is  circular  in  sec- 
tion, but  it  will  have  motions  relatively  to  the  box  which  can 
very  readily  be  imagined  if  the  shaft  is  not  circular  in  section. 
Now  the  actual  velocity  of 
the  liquid  at  any  place  re- 
latively to  the  box  is  in 
the  same  direction  as,  and  is 
proportional  to,  the  shear  in 
a  similar  shaft  when  it  is 
twisted.  This  has  been 
proved  by  Lord  Kelvin.  You  will  see  from 
this  that  there  is  very  little  strain  at  the  projecting  ribs  of  the  shaft, 
whose  section  is  shown  in  Fig.  196,  and  just  at  the  projecting 
angles  of  Figs.  195  and  197.  This  reminds  me  of  a  general  remark 
which  I  have  to  make,  and  which  I  must  leave  without  proof.  A 
solid  of  any  elastic  substance  cannot  experience  any  finite  stress 
or  strain  in  the  neighbourhood  of  a  project- 
ing point,  unless  acted  on  by  outside  forces 
just  at  the  point.  In  the  neighbourhood  of 
an  edge  it  may  have  strain  only  in  the  direc- 
tion of  the  edge,  and  generally  there  will  be 
exceedingly  great  strain  and  stresses  at  any 
re-entrant  edges  or  angles.  An  important 
application  of  the  last  part  of  the  statement 
is  the  well-known  practical  rule  that  every 
re-entering  edge  or  angle  ought  to  be  rounded 
to  prevent  risk  of  rupture  in  solid  pieces 


o 

Fig.  199. 


Fig.  200. 


APPLIED    MECHANICS.  359 

designed  to  bear  stress.  An  illustration  of  the  principle  is  the 
stress  at  the  centre  of  the  circular  outline  in  the  three  sections 
of  shafts  (Figs.  198,  199,  and  200).  In  Fig.  198,  at  o,  there  is 
no  stress  when  the  shaft  is  twisted;  in  Fig.  199  the  stress  may  be 
calculated ;  in  Fig.  200  the  stress  is  exceedingly  great  for  even  the 
smallest  twist  (see  Art.  302). 

EXERCISES. 

N.B. — A  most  usual  error  of  students  is  to  forget  that  moments  in 
pound-inches  are  not  numerically  the  same  as  in  pound-feet.  Beginners 
had  better  leave  the  exercises  involving  bending  moment  until  they  have 
studied  bending. 

1.  A  shaft  1  inch  in  diameter  can  safely  transmit  a  torque  of  2,400 
pound-inches.  "What  diameter  of  shaft  would  be  required  for  transmitting 
15  H.P.  at  200  revolutions  per  minute?  Ans.,  1£  inch. 

2.  Find  the  horse-power  which  may  be  transmitted  by  a  shaft  4 
inches  in  diameter  when  running  at  150  revolutions  per  minute,  if  the 
stress  due  to  twisting  be  limited  to  9,000  Ibs.  per  square  inch. 

Ans.,  269. 

3.  A  line  of  steel  shafting  is  80  feet  long ;  if  a  twisting  moment  of 
4,000  pound-inches  be  applied  at  one  end,  what  will  be  the  total  angle  of 
twist,  the  diameter  of  the  shaft  being  2^  inches  ?      What  horse-power 
will  this  transmit  at  220  revolutions  per  mimite  ?      Ans.,  4'7°  ;  14  H.P. 

4.  A  solid  wrought-iron  shaft  is  to  be  replaced  by  a  hollow  steel  shaft 
of  the  same  diameter.      If  the  material  of  the  latter   is  30  per  cent, 
stronger  than  that  of  the  former,  what  must  be  the  ratio  of  internal  to 
external  diameter  ?    What  is  the  percentage  saving  in  weight  ? 

Am.,  -693  ;  48  per  cent. 

5.  The  amount  of  twist  in  a  solid  shaft  is  to  be  limited  to  1°  for  each 
10  feet  of  length.     Find  the  diameter  for  a  twisting  moment  of  50  ton- 
inches,  the  modulus  of  torsional  rigidity  being  10,000,000  Ibs.  per  square 
inch.  Ans.,  5 '2  inches. 

6.  A  wrought-iron  shaft  is  subjected  simultaneously  to  a  bending 
moment  of   8,000  pound-inches,   and  to  a  twisting  moment  of  15,000 
pound-inches.     Find  the  twisting  moment  equivalent  to  these  two  and 
the  least  safe  diameter  of  the  shaft,  the  safe  shear  stress  being  taken  at 
8,000  Ibs.  per  square  inch.          Ans.,  25,000  pound-inches;  2-52  inches. 

7.  Find  the  diameter  of  a  shaft  for  a  winding  drum  which  works 
under  the  following  conditions:    the  load  lifted  is 1%  ton;    diameter  of 
drum,  5  feet ;  width  of  face  of  drum,  26  inches  ;  distance  from  inner  face 
of  drum  to  the  middle  of  the  bearing  of  shaft,  13  inches;  maximum 
stress,  7, 000 'Ibs.  per  square  inch.  Ans.,  5 '44  inches. 

8.  A  wrought-iron  shaft  3  inches  in  diameter,  and  making  140  revolu- 
tions per  minute,  is  supported  at  wall  brackets  16  feet  apart.     There  is  a 
pulley  on  the  shaft,  midway  between  the  bearings ;  if  the  resultant  side 
pull  due  to  the  weight  of  the  pulley  and  the  pull  of  the  belt  be  210  Ibs., 
what  is  the  greatest  horse-power  the  shaft  will  transmit  with  safety  ?    Safe 
shear  stress,  7,800  Ibs.  per  square  inch.  Ans.,  66. 

9.  A  bar  of  iron  is  at  the  same  time  under  a  direct  tensile  stress  of 
5,000  Ibs.  per  square  inch,  and  to  a  shearing  stress  of  3,500  Ibs.  per 
square  inch.     What  would  be  the  resultant  equivalent  tensile  stress  on 
the  material  ?  Ans,,  6,801  Ibs.  square  inch. 


360  APPLIED    MECHANICS. 

10.  Taking  the  safe  tensile  stress  of  wrought-iron  to  be  10,000  Ibs. 
per  square  inch,  determine  whether  it  would  be  safe  to  subject  a  piece  of 
wrought-iron  to  a  tensile  stress  of  3|  tons  per  square  inch,  together  with 
a  shear  stress  of  3  tons  per  square  inch. 

Ans.  Unsafe;  max.  stress  =  11,700  Ibs.  square  inch. 

11.  A  wrought-iron  shaft  is  subjected  to  a  twisting  moment  of  36,000 
pound-inches  and  a  bending  moment  of  18,000  pound-inches  ;  find  the 
diameter  when  the  maximum  shear  stress  is  8,000  Ibs.  per  square  inch. 
Find  also  the  twisting  moment  which  alone  would  produce  a  shear  stress 
of  the  same  numerical  value.  Ans.,  3-3";  58,200  pound-inches. 

12.  A  screw  propeller-shaft  10  inches  in  diameter  is  subjected  to  a 
twisting  moment  of  35  ton-feet,  and  to  a  bending  moment  of  10  ton-feet, 
due  to  the  weight  of  the  shaft  and  the  pitching  of  the  ship.     What  is  the 
maximum  compressive  stress  if  the  thrust  of  the  screw  be  10  tons  ? 

Ans.,  2-9  tons. 

13.  A  shaft  12  inches  diameter,  transmitting  a  twisting  moment  of 
100  ton-feet,  is  also  subject  to  a  bending  moment  of  20  ton-feet.     Find 
the  maximum  stress  induced.  Ans.,  4 '3  tons  per  square  inch. 

14.  Find  the  diameter  of  a  wrought-iron  shaft  to  transmit  90  horse- 
power at  130  revolutions  per  minute.     If  there  is  a  bending  moment 
equal  to  the  twisting  moment,  what  ought  to  be  the  diameter  ? 

.Ans.,  2-7  inches  ;  4-1  inches. 

15.  A  steam-engine  crank  is  12  inches  long,    and  the  greatest  force 
which  is  transmitted  through  the  connecting-rod  is  9,000  Ibs.     Find  the 
diameter  of  the  wrought-iron  crank- shaft,  taking  the  safe  shear  stress  at 
9,000,  the  distance  of  the  centre  of  the  crank-pin  from  the  centre  of  the 
bearing  nearest  it  being  10  inches,  measured  horizontally. 

Ans.,  5-07  inches. 

16.  A  round   shaft    3  inches  diameter,  find  the  sizes  of  equivalent 
shafts  of  square,  elliptic  and  rectangular  sections  if  the  breadth  and 
thickness  of  each  of  these  latter  are  as  1  to  2.     If  these  shafts  are  20  feet 
long,  and  they  are  transmitting  20  H.P.  at  100  revolutions,  what  is  the 
total  twist  on  each  of  them.     N  =  10,500,000. 

Ans.,  Rectangle,  2 '15  x  4-3  inches;  ellipse,  minor  axis,  2-38  inches; 
square,  2-73  inches  side.  Twist  on  circular  shaft  =  2 '07 3%  on  square 
shaft,  1-78°;  elliptical  shaft,  1-05°;  rectangular  shaft -93°. 


361 


CHAPTER    XV. 

MORE    DIFFICULT   THEORY. 

304.  MATHEMATICIANS  endeavour  to  help  engineers  (including 
in  this  term  all  men  who  apply  the  principles  of  natural  science)  by 
investigations  concerning  ideal  elastic  material  shaped  like  actual 
beams    and    shafts.      The   mathematical  analysis  is   exact  and 
difficult,  and  only  a  few  problems  have  yet  been  solved,  and  it  is 
only  by  leaving  out  terms  that  seem  insignificant  that  we  are  able 
to  apply  the  results  to  actual  problems.     The  engineer  recognises 
from  the  beginning  that  his  problems  are  too  complicated  for  any 
exact  mathematical  investigation.      He  therefore  leaves  out  his 
apparently  insignificant  terms  rather  at  the  beginning  than  the 
end  ;  but  indeed  he  leaves  them  out  in  any  part  of  his  investiga- 
tion if  they  are  likely  to  give  trouble,  for  he  recognises  from  the 
beginning  that  his  theory  is  only  to  guide  bim,  and  that  the  final 
appeal  must  be  to  experiment.     The  engineer  looks  upon  the 
phenomena  involved  in  the  loading  of  the  tie-bar  as  simple  because 
experiment  is  easy;   whereas  the  mathematician,  seeing  that  a 
lateral  contraction  accompanies  an  axial  elongation,  regards  it  as 
complicated. 

The  engineer  ought  to  study,  and  develop,  and  correct,  by 
experimental  observations,  his  usual  method  of  investigation  as 
described  in  this  book,  but  he  ought  also  to  study  the  mathema- 
tician's treatment  of  the  subject,  and  let  it  assist  him  as  he  lets 
experiment  assist  him.  The  following  very  short  sketch  needs 
much  time  for  its  proper  comprehension ;  it  is  from  the  mathe- 
matician's point  of  view.  Students  may  pursue  the  subject  in  Mr 
Love's  treatise  on  elasticity,  or  Thomson  and  Tait's  treatise  on 
natural  philosophy. 

305.  The  consideration  of  homogeneous  strain  in  general  (when 
any  portion  of  stuff  in  the  shape  of  a  sphere  is  changed  into  what  is 
called  the  strain  ellipsoid,  any  three  co-orthogonal  diameters  of  the 
sphere    becoming    conjugate    diameters  of   the    ellipsoid,   planes 
parallel  to  one  another  remain  parallel ;  all  parallel  lines  get  the 
same  fractional  changes  in  length)   is  not  so  difficult  as  it  is 
tedious.    Unfortunately  the  authors  of 

books  insist  on  its  being  studied.     For 
our  purposes  we  have  only  to  deal  with 
infinitesimal  strains,  and  this  is  easy. 
Suppose  that  a  point  #,  y,  z  is  dis- 
placed to  x  +  u,  y  +  v,  z  +  w  where 
M,  «?,  w  are  very  small,  then 
du    ,      dv  dw 


are  the  tensile  strains  of  the  stuff  in 
directions  parallel  to  x,  y,  z.  In  Fig. 
201  let  the  axis  of  a?  be  at  right  angles 

M* 


Fig.  201. 


362  APPLIED    MECHANICS. 

to  the  plane  of  the  paper.  Let  the  traces  of  two  planes,  c  B  and  c  A, 
parallel  to  o  Y  and  o  z,  and  perpendicular  to  the  paper,become  changed 
to  c/  B',  c'  A'.  The  angle  by  which  A'  c'  B'  is  less  than  the  original 
right  angle  is  evidently  a  shear  strain.  I  shall  call  the  amount  of 
it  a  ;  and  as  it  is  a  shear  of  planes  normal  to  Y,  parallel  to  z,  or  a 
shear  of  planes  normal  to  z,  parallel  to  Y,  there  is  no  great  harm  in 
calling  it  a  shear  about  the  axis  of  x.  Now  the  angle  turned 
through  by  c  A,  clockwise,  is  really  the  horizontal  motion  of  A  minus 
the  horizontal  motion  of  c,  divided  by  AC,  since  the  angle  is 
very  small.  But  this  is  dv/dz.  Let  the  student  be  quite 
sure  of  this  fact.  Similarly  the  angle,  anti- clock  wise,  turned 

through  by    c  B  is   dm/dy,  and   hence  a  —  -=^-+  -^- .  .  .  .  (2). 

az      ay 

Similarly  if  b  and  c  are  to  y  and  z  what  a  is  to  x,  then 
dm      du      du      dv 

A  student  may  keep  these  in  mind  by  means  of  the  mnemonic 

a    b  -  c  ~\ 

u    v    to  >  .  .  .  .  (2). 

x    y    z  } 

Notice  that  the  average  rotation  of  c  A  and  c  B,  clockwise,  in  Fig. 
201,  is  half  the  algebraic  sum  of  the  angles  turned  through  by  c  A 
and  c  B,  or,  calling  the  average  rotations  of  the  material  £>i  about 
the  axis  of  x,  &>2  about  the  axis  of  Y,  o>3  about  the  axis  of  z, 

'dv      dm\    .         ,  (dm      dn\    ,        ,  {du      do^ 


It  is  evident  that  when  o^  =  w2  =  <33  =  0,  or  there  is  no  rotation, 
it  means  that  the  strain  is  pure,  or  that  the  three  principal 
diameters  of  the  strain  ellipsoid,  called  the  principal  axes  of  strain, 
remain  parallel  to  their  original  directions. 

Shear  strain  involves  no  change  of  volume ;  and  if  the  edges 
parallel  to  x,  Y,  z  of  the  unit  cube  become  1  +  <?,  1  +/,  1  +  g,  the 
volumetric  dilatation  or  strain  is  e  +  f  +  g,  since  these  are  small. 
Or  if  D  is  used  for  this,  then 

du       dv       dw 

The  conditions  o>]  =  <52  =  w3  =  0  are  evidently  the  conditions 
that  there  is  a  function  <f>  (called  the  strain  potential),  such  that 

u  —  -,-? ,  v  =  -^-,  w  =  ~ ;  and  we  -see  that,  in  case  there  is  no 
dx  dy  dz 

cubical  dilatation,  (4)  becomes 

&  +  0  +  S  =  0""(5)' 

At  any  point  o  of  a  body  let  there  be  three  planes  of  reference 
meeting  at  the  three  mutually  perpendicular  axes  of  #,  of  y,  and  of  z. 
Let  tensile  stress  be  called  positive.  Across  the  plane  of  yz  (usually 


APPLIED    MECHANICS. 


363 


called  the  plane  #,  because  the  axis  of  x  is  normal  to  it)  let  the 
stress  be  resolved  into  its  three  components — XB,  parallel  to  ox; 
YZ,  parallel  to  oy ;  and  zx,  parallel  to  oz.  Across  the  planes  y 
and  z  let  the  component  stresses  be  x<</,  YI/,  zy,  and  Xz,  Y«,  z«. 
To  find  the  stress  components  F,  G,  H  in  the  three  directions  across 
any  plane  whose  direction  cosines  (or  the  direction  cosines  of  its 
normal)  are  I,  m,  n,  consider  the  equilibrium  of  the  tetrahedron 
formed  by  the  four  planes  under  the  action  of  the  tractions  acting 
from  material  outside  it.  The  area  of  the  sloping  face  being  A, 
the  areas  of  the  other  faces  are  I  A,  m  A,  and  »  A.  The  resultant 
force  on  each  area  is  stress  multiplied  by  each  area.  We  have, 
-  then,  resolving  parallel  to  a?,  F  A  =  Xx  I  A  +  Xy  m  A  +  x«  n  A 
and  three  other  equations.  Hence, 

F  =  L  x#  +  in  Xy  -f-  n  x# 

H  =  I  zx  •+  m  zy  +  nzz 

In  a  fluid  the  static  stress  is  always  normal  to  any  interface. 
Hence  F,  G,  and  H  are  in  this  case  the  components  of  a  normal  stress 
p  on  the  new  plane,  and  all  the  tangential  stresses  vanish.  Hence, 

F  =  I  P  =  I  nx    or    P  =  Xx 

so  that  the  stress  is  the  same  across  any  interface  whatsoever. 

306.  Now  consider  in  the  general  case  the  equilibrium  of  a  paral- 
lelepiped whose  corner  is  at  o  (Fig.  202),  the  co-ordinates  of  which 
are  a?,  y,  z,  and  let  the  co-ordinates 
of  the  opposite  corner  o'  be 
x  +  8a?,  y  +  8y,  z  +  Sz,  and  let 
us  consider  the  equilibrium  of  all 
the  surface  tractions  acting  on  the 
faces  from  the  outside.  The  re- 
sultants of  the  normal  forces  meet 
in  the  centre,  and  if  we  neglect 
volumetric  forces,  they  are  in 
equilibrium.  They  are  not  shown 
in  Fig.  202.  I  *y 

We  show  the  tangential  forces 
per  unit  area  with  which  outside 
stuff  acts  on  inside  on  three  of 
the  faces.  The  other  three  faces 
have  similar  forces,  the  arrows 
being  in  opposite  senses  to  these.*  Fig-  2°2- 

Hence  their  moments  are  just  the 
same  as  the  moments  of  these  about  axes  through  the  centre. 

*  It  is  easy  to  take  into  account  the  fact  that  there  may  be  volumetric 
forces,  and  that  z  z  on  the  x  plane  through  o'  is  not  equal  to  z     on  the  x 

plane  through  o,  but  is  rather  Zx  +  Sx  —  zx.     We  do  this  later,  and  it  i* 
easy  to  see  that  these  extra  terms  vanish  in  the  present  problem. 


A 


-Zx 


364 


APPLIED    MECHANICS. 


Taking  moments  of  the  figured  forces  about  the  axis  through  o 
parallel  to  a?,  we  find 

z  y  .  Sx.  8z.5y  =  Y05a3.8y.52;  so  that  z  y  =  Y  » 

This  is  Cauchy's  theorem.     We  call  each  of  them  s.     In  the  same 
way  we  have  the  other  two  relations  here  given : — 

zy=Yz==s;    z<c  =  xa  =  T;   xy  =  T«  =  u. 

We  also  give  the  names  p  to  x«    Q  to  Yy,  B  to  z*. 

s,  T,  u  are  the  shear  stresses,  and  p,  Q,  and  R  are  the  normal 
tensile  stresses  in  the  material.  Hence  our  equation  may  bf 
written : — 


.  .  .  .  (1). 


Exercise. — Across  what  interfaces  at  a  point  are  the  stresses 
normal  ?  That  is,  find  the  principal  stresses  and  their  directions. 
If  F,  G,  and  H  are  the  components  of  a  stress  normal  to  the 
plane  I  m  n,  then  B  Z  =:  F,  Bm  =  G,  BW  =  H;  and  if  we  sub- 
stitute these  in  (1),  remembering  that  I2  -f  w2  +  w2  =  1,  we  can 
find  I,  m,  n  and  B.  The  answer  tells  us  that  there  are  always 
three  directions  and  amounts  of  principal  stress. 

In  our  most  general  state  of  stress,  Fig.  203  shows  the  surface 
tractions  actin  g  on  the  element  fix  Sy  Sz  from  the  outside.  Imagine 

equal  and  opposite  forces  on  the 
other  three  faces. 

Now  let  us  consider  volumetric 
forces  and  the  rates  of  variation  of 
the  stress.  On  the  faces  meeting  at 
o  we  have  p  Q  K  s  T  and  u.  But 
the  forces  on  the  other  faces  are,  as 
regards  normal  forces, 


G 

on  Sz  .  Sx  we 


203. 


On  Sy  .  Sz  we  have  also  u  + 
dr 
dx 
dv 


and  T  +  Sx 
have  u  +  Sy 


. 
and  8 


da 
Sy-d~y> 


on  Sx 


8y  we  have  T  +  82        and  s  +  Sz 
dz 


~. 
dz 


Now  we  know 


that  the  parts  of  these  forces  p  and  p,  s  and  s,  etc.,  balance  in- 
dependently ;  therefore  we  need  only  consider  the  extra  ones  as 
shown  in  Fig.  204. 

If  the   volumetric  force  in  the  direction  of  increasing  x  is 
p  .  x  .  Sas  .  Sy  .  Sz,  then 


APPLIED    MECHANICS. 


365 


i,  x    to  .  oif .  oV  +  ov  •  o£  .  o&  -7-  -4-  5y  .  -5—  •  oa;  "5z 

dte  #V 

_I_    §2   . t    Jy   .    §<g  •  0. 

We  have   similar  equations  for  the  other  directions,  and  these 
reduce  to  the  following  equations  of  equilibrium : — 


dx 


dz 


«Y+^  +  ^  +  ^  =  0  I (2). 

dx        dy        d~ 

i    d  T   ,   d  s    ,   d 


Note  that  we  have  assumed  the  moments  due  to  the  forces 
on  the  boundary 
of  the  element  to 
balance,  in  prov- 
ing Cauchy's  the- 
orem, or  that  the 
bodily  forces  have 
no  moment.  They 
may  have  some 
moment  in  the 
matter  in  which 
magnetic 
stresses  act,  but 
in  ordinary  stress 
phenomena  we 
assume  the 
truth  of  Cauchy's 
theorem. 

If  the  mate- 
rial is  fluid,  so 
that  g  =  T  =  u  =  0, 

~   -  0 
dQ  - 

p  Y  +  dy  ~     ' 


Example. — Let 
p  x  be  called  w,  Fig-  204. 

the  force  on  unit 

volume,  say  the  weight  in  pounds  of  1  cubic  foot,  then  p  is  pressure 
in  pounds  per  square  foot  at  any  depth  in  a  fluid  under  parallel 
vertical  gravitational  force. 

If  there  is  motion  in  the  general  case,  we  must  use  x  -  T  " 


instead  of  x  merely,  where  u  is  the  displacement  in  the  direction 


366  APPLIED   MECHANICS. 

of  increasing  x  of  the  mass  p  .  5x  8y  Hz  ;  and  hence  we  have  three 
equation  like  £  +  £  +  £  +  p  x  =  ,  £  .  .  .  .  (3). 

At  any  element  of  surface  of  a  body  (I  m  n  being  the  direction 
cosines  of  its  normal),  if  there  are  surface  tractions  with  com- 
ponents F  a  H,  the  condition  (1)  is  to  be  satisfied  at  the  surface. 

Now,  taking  the  unit  cube,  and  letting  a  be  -  ,  the  reciprocal 

E 

of  Young's  modulus,  and  letting  ft  be  the  lateral  contraction  in  a 
tie-bar  of  the  material,  when  a  is  the  axial  elongation  ;  under  the 
action  of  p,  a,  and  K,  the  edges  become  lengthened  by  the  amounts 
#,  /,  and  ff,  where  e  =  p  a  —  (Q  +  R)  0,  or 

e=      pa  -  Qj8  -  R0) 
f=-T0  +  QO  -  R£      ----  (4). 
ff=.  -Pj8  -  Q/8  +  Ra) 

Compare  these  with  Art.  269. 
Solving  these  equations,  we  find,  writing  D  for  e  +  /  -f  g,  and 

recollecting  the  fact  that  a  =  _L  +  J:  =  I,  ft  =  -1  -  jL, 

SN       9K       E  GN       QK' 

P  =  AD  +  2N<M 
Q=AD  +  2N/J  ----   (5), 

R  =  AD 


where  \  stands  f  or  K  -  f  N  ;   K  being  the  modulus  of  elasticity  of 
bulk,  N  being  the  modulus  of  rigidity. 

If  s,  T,  and  u  are  the  shear  stresses,  and  a,  b,  and  c  are  the  shear 
strains  (see  Art.  305),  s  =  N  a,  T=:N#,  TT  =  NC.  ..  .  (5).     In 

that  Art.  305  we  used  -^,  etc.,  instead  of  e,  etc.,  and  so  we  can  write 
dx1 


z 

du       dv       dw 
•where  D  stands  for  T—  +  -T-  +  ^-,  and  we  also  have 

._»(*f  +  *\  ,_»(*f  +  *?Y  »-„(*  +  *!)...(•). 

\^y        rfz/  \A         ^»/  V^a;        dy) 

<Zw  =  p.^e  +  Q.  4^4-  ».^  +  s  .  da  +  T  .  db  +  u  .  dc...(6) 


was  proved  by  Lord  Kelvin  to  be  a  complete  differential  in  two 
cases  —  first,  at  constant  temperature  ;   second,  when  the  changes  of 
strain  occur  so  quickly  that  no  heat  is  gained  or  lost  by  the  stuff. 
Using  (5),  we  find  that  d  w  is  a  complete  differential  if 

2  w  =  (A  +  2ft)  (e  +/+  g?  +  p  («*  +  V  +  &  -  4fg  -  4  ge 

...  d  w  d  w  /w. 

so  that  P  =  -      ,  etc.,  u  =  —  -  ,  etc  .....  (7). 


APPLIED    MECHANICS.  867 

Kirchhoff  pointed  out  that  w  cannot  be  negative.  When  w  is  0 
the  whole  body  moves  as  a  rigid  body.  He  also  showed  that  if  tha 
six  strains  be  given  at  every  point  in  a  body  it  is  necessary  to 
impose  six  independent  conditions,  such  as  those  of  St.  Venant 
given  in  (7)  of  Art.  308.  Without  these  there  may  be  "  rigid 
body  "  displacements  as  well  as  the  strains.  He  also  showed  that 
if  the  surface  displacements  or  surface  forces  on  a  body  are  given, 
there  is  only  one  solution  possible  (Kelvin  showed  that  there  was 
also  an  unstable  solution).  Kelvin  had  shown  that  one  was  always 
possible. 

St.  Venant  showed  that  in  calculating  the  strains  due  to  surface 
loads  it  is  only  in  the  neighbourhood  of  the  place  that  the  actual 
distribution  of  the  surface  force  is  of  any  importance.  This  is 
called  the  principle  of  equipollent  loads. 

Converting  the  stresses  in  (3)  into  strains,  we  have 


with  two  other  equations,     v2  means          +         -f       .      The 


surface  conditions  become  FinZ(AD-|-2N<?)  -j-mNC-fwN*,  with 
two  others.  Translating  this,  we  have  the  surface  conditions  (1) 
becoming 


=  l\  D  +  2N 

dv 


O  =  mA  D  +  2  N  (  -3—  i  +  waij  —  l&3  j 


n\  D  +  2N 


>....  (9). 


Here  dnl  is  an  element  of  the  normal  to  the  surface. 
(8)  or  (3)  may  be  written 


i A  -r  *N )  -j *»  ~5 r  **   j     -f  p x  =  p  -=-5- ....  (iu 

'  ax  ay  az  r  at* 

with  two  other  equations. 

If  we  neglect  x,  Y,  z,  the  volumetric  forces,  differentiate 
etc.,  with  regard  to  a?,  etc.,  and  add,  we  get 

Trt_ 


(10), 


a  well-known  equation  showing  that  there  is  wave  propagation 
with  velocity  =  A  /  —  —  .     Differentiate  the  third  of  (8)  with 

respect  to  y,  and  the  second  with  regard  to  z,  and  subtract,  and 
we  get 


3G8  APPLIED    MECHANICS. 

and  two  other  equations.     These  are  equations  of  distortioual  wave 
propagation  with  a  velocity  =  A  /  N. 

307.  To  illustrate  this  method  of  study,  let  us  consider  the  prob- 
lem of  the  thick  cylinder  of  Art.  275.  There  we  considered  it  from 
the  stress  point  of  view.  We  shall  now  consider  it  from  the  strain 
point  of  view.  At  any  point  in  the  material  at  the  distance  r  from 
the  axis  there  is  a  radial  displacement,  which  we  shall  call  ?/,  as  if 
the  direction  r  were  our  old  direction  x.  There  is  displacement  at 
right  angles  to  u,  and  its  fractional  amount  —  that  is,  the  strain  —  is 
evidently  w/r,  "because  if  a  radius  r  increases  to  r  -f-  w,  the  circum- 
ference of  the  circle  increases  to  2ir  (r  +  «),  so  that  the  fractional 
increase  of  the  circumference  is  u/r.  Let  us  imagine  any  tensile 
strain  parallel  to  the  axis,  which  we  call  the  direction  of  z,  to  be 
constant.  It  is  evident,  also,  that  the  strain  is  irrotational,  and  also 
that  the  three  principal  shears  are  0.  We  make.these  statements  to 
show  that  we  are  considering  the  old  practical  problem  ;  but  the 
mathematician  puts  it  rather  in  this  way  :  he  assumes  certain 
strains  to  exist  ;  he  works  out  a  mathematical  problem,  and  he  takes 
his  chance  of  the  results  fitting  any  practical  case. 

dlL  ti 

Let    ~—  e,~=f,ff  =  g0,a  =  0,b  =  0,c^:  0. 


1.  No  bodily  forces,  and  the  problem  a  static  one—  that  is,  the 
ains  independent  of 
tion  (8)  or  (10)  becomes 


strains  independent  of  time.     Then  D  =  -^  +  -  +  ^0,  and  equa- 


(I)- 
Hence 

d?n       \du       u 

3H  +  r^-p  =  0-  "  '  (2)' 

And  we  find  on  trial  that  u  =  A  r  +  B  r  ~  *  ,  .  .  .  (3)  where  A  and  B 
are  arbitrary  constants.  Knowing  the  strains,  equation  (5)  (Art. 
306)  enables  us  to  find  the  stresses. 

p  =  2A  (A  +  N)  +  A  gQ  -  2  N  B  r2  .  .  .  .  (4), 
Q  =  2A  (\  -f  N)  +  A  <70  4-  2  N  Br2  .  .  .  .  (5), 
R  =r  2A  A  +  9  O  +  2  N)  .  .  .  .  (6). 

When  there  is  no  fluid  pressure  or  other  endlong  force  we  may 
take  it  that  R=  0  in  a  long  cylinder,  as  for  example  when  a  gun 
is  being  manufactured.  But  if  there  is  a  fluid  pressure  p,  we 

may  take  R  =   r*  Pl  .      This  affects  nothing  in   our   problems 

r°  ~ri 
except  shrinkage  or  -  -p.      We  see,  therefore,  that  the  theory 

of  Art.  275,  and  its  results  are  justified  for  all  practical 
purposes. 

It   is   good    to  remember    that  the    dilatation  is  constant. 


APPLIED    MECHANICS.  369 

If  a  cylinder  is  solid  to  the  centre  the  coefficient  B  is  0. 

Exercises.  —  (1)  A  solid  cylinder  of  4  inches  radius  is  subjected 
to  an  outside  pressure  of  5  tons  per  sq.  in.  What  are  the 
stresses  ?  Ans.,  p  and  Q  are  everywhere  -  5,  or  corapressive 
stresses  of  5  tons  per  sq.  in. 

(2)  If  this  cylinder  has  the  smallest  hole  bored  out  along  the 
axis,  and  the%  pressure  is  0  in  the  hole,  what  are  P  and  Q  ? 
Ans.,  p  and  Q  are  everywhere  -5  except  near  the  hole.  At 
the  hole  p  =  0  and  Q  =  -  10,  or  a  compressive  stress  of  10  tons 
per  sq.  in. 

2.  Rotating  Cylinder.  —  Volumetric  force  a?r  acting  radially  on 
unit  mass,  a  being  angular  velocity.  Using  this  in  (8)  or  (10)  of 
Art.  306,  instead  of  (1),  we  have 


If  we  try  u  =  Arn,  we  find  that  n  =  1,  n  =  —  I  will  satisfy  (7)  if 
A  is  an  arbitrary  constant  ;  also  n  =  3  will  satisfy  (7)  if  A  = 
—  pa2/8  (A  +  2  N)  ;  and  hence 

u  =  Ar  +  Br-1  -  paV/8  (A  +  2N)  ----  (8) 

where  A  and  B  are  arbitrary  constants.  Having  the  strains,  we 
can  calculate  the  stresses 

In  previous  editions  of  this  book  I  gave  the  stresses,  and 
applied  them  to  the  case  of  a  disc.  But  this  is  evidently  wrong, 
as  K  is  not  zero  on  the  faces.  There  is  no  theory  quite  correct, 
but  Dr.  Chree's  solution  is  correct  for  all  practical  purposes. 
He  takes 

du  u  dw  du    dw 

=  d~r'   f=T'    ff  =  ^   l  =  dz  +  -Jr' 

We  can  now  easily  write  out  the  equations.  Try  if  the  solution 
is,  a  being  Poisson's  ratio, 


(3  +  cr)  +  |«r(l  +Vr)  r  (^- 

o 


It  is  easy  to  get  the  stresses  from  these,  using  (5)  of  Art.  306. 

In  this  solution  the  planes  or  faces  z  =>  +  I  are  free  from 

stress,  and  there  is  no  tangential  stress  on  the  edge,  and  the 


370  APPLIED    MECHANICS. 

resultant  normal  traction  per  unit  length  of  the  circumference 

rl 

is  0  when  r-=.  r0  (that  is,  /  p.  dz  =  0  where  r  =  r0).      By  the 
J  -I 

principle  of  equipollent  loads  this  means  that  there  is  no  practical 
departure  from  the  real  condition  of  things  except  close  to  the 
edge.  Find  w  when  z  is  4-  I  or  - 1,  and  note  that  the  plane 
faces  become  paraboloids  of  revolution,  the  disc  being  thinnest 
at  the  centre.  If  the  disc  is  very  thin,  take  /— 0,  2  =  0,  and 

our  answers  are  p  =  ^P  (3  +  tr)  (r*  +  r*  -  r2  -  T±^)  -  -(9) 
Q  =  *  (3+,)  {^+^  +  ^-£±1^}  .  _  (10).  If 

there  is  no  central  hole    p  =&  (3  +  cr)  (r02  -  r2) (11). 

b 

Q  =  «£  (3  +  «r)  (^  -  —^  r>)  -  -  (12).     It  will  be  found  that 

having  the  very  smallest  hole  at  the  centre  just  doubles  the 
greatest  Q  which  is  at  the  centre.  These  results  (9)  to  (12)  are 
easily  obtained  by  making  x  constant  in  the  next  problem,  and 
remembering  (4)  or  (5)  of  Art.  306. 

Students  ought  to  make  examples,  taking  a  =  0-25,  cfy/8  =  1. 
Thus  take  r0=  10,  and  take  r  =  0,  or  O'Ol,  or  1,  or  2,  or  4,  or  7, 
showing  their  results  on  squared  paper. 

It  has  been  proposed  to  build  up  a  disc  of  rings  shrunk  upon 
one  another  as  a  gun  is  built,  or  of  wire  wound  on  under  tension. 
Now  it  will  be  observed  that  a  solid  disc  must  be  better  than  any 
series  of  rings,  however  put  together.  For  we  cannot  imagine 
the  rings  to  exercise  tensile  radial  forces  upon  one  another,  and 
it  is  only  a  tensional  inward  force  on  the  outermost  ring  which 
will  enable  us  to  run  it  at  a  speed  greater  than  the  critical 

speed  \/P/p,  where  P  is  the  greatest  stress  which  the  material 
will  stand. 

307.*  Disc  of  varying  thickness,  rotating.  An  approximate 
theory  is  possible  which  is  of  practical  value.  Let  x  be  the 
thickness.  It  is  easy  to  see,  if  we  study  the  forces  acting  upon 
an  element  of  the  disc  that  we  have,  if  x  does  not  vary  rapidly, 

-^(x  pr)-Q#-}-pa2r2a?=:0 (1).     Assuming  x  to  consist 

of  terms  like  rn  we  can  find  p  and  Q,  R  being  0.  A  solution  can 
also  be  found  for  a  disc  of  uniform  strength  ;  that  is,  taking 
p  =  Q  constant  everywhere.  We  have  at  once  from  (1)  the 
result 

-po2r2/2p 
x  =  xle    '  (2). 


APPLIED    MECHANICS.  370« 


In  Art.  274  we  saw  that  there  was  a  limiting  velocity 
for  a  rod  or  rope  or  rim  of  a  wheel,  but  here  we  have  a  means 
of  getting  any  velocity,  however  great.  In  the  Laval  turbine 
the  speed  of  the  rim  of  the  wheel  is  usually  more  than  1,000  feet 
per  second. 

Suppose  a?0  the  thickness  of  the  disc  at  r0  the  outside.  Let  a 
rim  of  section  a  be  outside  the  disc.  It  is  subjected  to  an 
internal  pull  of  the  amount  P  a?0  per  unit  of  length,  and  under 

these  circumstances    its  speed  may  be  ^J  tl  +  *-~)  P/P  K 
the  tensile  stress  is  P. 

Exercise.  —  It  is  necessary  to  have  the  rim  of  a  wheel  8  feet 
diameter,  14-4  sq.  inch  section  of  nickel  steel,  to  run  at  1,500  feet 
per  second  ;  let  p  be  3  x  106  Ib.  per  sq.  foot,  p  =  12,  so  that  usual 

critical  velocity  is  500  feet  per  second.    Then  1  +     ?  _ 


As  P0  =  4  and  a  =  O'l,  we  find  a?0  =  0-2,  so  that  from  (2)  we  have 
all  the  dimensions  of  the  wheel.  Laval  constructs  his  wheels  in 
this  way,  taking  care  to  have  no  hole  through  the  centre  of  his 
disc,  and  where  great  rim  velocity  is  wanted  this  method  of 
construction  ought  to  be  followed.  But  if  a  fly-wheel  is  wanted 
or  a  gyrostat  of  minimum  weight,  it  is  not  necessary. 

The  weight  of  the  disc  will  be  found  to  be  w  =  (xl—  x0) 
2  P^  7r/a2,  and  its  rim  2vr0affp.  In  the  above  case  I  find  the 
disc  to  weigh  6,907  Ib.,  and  its  rim  971  Ib.  ;  the  kinetic  energy  of 
the  whole  is  3'77  +  107  foot  Ib.  If  the  whole  mass  were  a  mere 
rim  running  at  500  feet  per  second  its  kinetic  energy  would  be 
3-06  +  10'  foot  Ib.  Therefore  for  mere  fly-wheel  purposes  there 
is  no  great  gain  in  using  this  complex  shape  of  wheel.  For  fly- 
wheel purposes  the  simple  construction  ought  to  be  adopted, 
almost  all  the  mass  in  a  thin  rim  fastened  by  arms  to  a  solid  hub. 
Let  the  section  of  each  arm  be  square,  or  elliptic,  or  circular, 

the  area  A  of  its  section  at  the  radius  r  being  A  =  c  e  ~"  ^a      '  r° 

where  c  is  a  constant.  It  is  easy  to  arrange  that  the  arms  exert 
no  pull  on  the  inside  of  the  rim,  as  the  elongation  of  the  arm 
may  just  be  equal  to  the  increased  diameter  of  the  rim,  and  the 
centrifugal  force  of  the  fastening  can  be  made  to  reduce  the 
tension  between  arm  and  rim  to  nothing.  The  tensile  stress  in 
the  arm  is  constant  everywhere  except  just  at  the  rim.  (See 
Appendix.) 

308.  Bending  or  Twisting  a  Prism.  —  A  straight  prismatic 
surface  with  plane  ends  bounds  isotropic  material.  Surface 
tractions  are  applied  to  the  ends  only.  The  axis  of  the  prism  is 
the  axis  of  z,  one  end  being  the  origin.  When  we  speak  of 
bending,  the  deflection  will  be  in  the  direction  of  x  (that  is, 
bending  will  take  place  in  the  plane  xz.) 


371  APPLIED    MECHANICS. 

The  student  will  refer  back  to  Fig.  203  to  see  the  exact  significa- 
tions of  the  stresses  P,  Q,  R,  and  s,  T,  TJ,  and  the  strains  e  =-3-, 

dx 

.      dv  dw        ,  dw      dv   ,       du      dw  dv       du 

f—-J~,g  =  -T-»  an^  «  =  -7 — h  -j-»  *  =  -j-  +  -3— »  *  =  -^ — h  -7— 
dy' y        dz  dy       dz'  dz       dx'          dx      dy ; 

also  to  the  equations  of  equilibrium  (3),  and  to  the  equations  (5) 
which  express  the  stresses  in  terms  of  the  strains,  and  also  to  the 
boundary  conditions  (1).  He  had  better  write  out  these  equations 
here.  We  do  so,  and  give  them  new  reference  numbers.  We 
assume  neither  volumetric  forces  nor  dynamical  forces  in  (3),  now 
called  (1). 


dp      dv      d? 

-fa  +  ~dy-  +  -d;  = 

rfu  do,  .dB__ 
dx  +  dy  +  ~dl  ~ 
dv  ,  ds  .  dx, 


.  .  .  .  (1), 


(2),  or 
—  0  (p 


R  =  AD+ 
8  =  Na,  T 

D  stands  f  or  e  +  f  +  g.     \  is  K  —  f  N.     Our  old 

1,1         1  1          1     ft 

~  3T+  9^  =  E'  *  =  6^~  9l»  a  = 

(Poisson's  ratio).     K  is  the  modulus  of  elasticity  of  bulk;  N  is  the 
modulus  of  rigidity,  and  the  surface  tractions  are 

F  =  ;p-ff»U  +  WT| 

G  =  Ju-j-wa  +  «s>  ....  (3). 

H  =  JT  +  ws  +  WR) 

It  is  to  be  remembered  that  the~  mathematician  assumes  a 
certain  condition  of  strain,  and  takes  his  chance  of  its  fitting  some 
particular  practical  problem.  If  it  happens  to  fit  the  conditions  of 
the  problem  exactly,  it  is  an  exact  solution  of  the  problem  ;  and, 
according  to  Art.  306,  it  is  the  only  solution.  If  it  does  not  exactly 
fit  the  conditions  (as,  for  example,  Dr.  Chree's  solution  of  the 
rotating  disc),  we  discuss  the  discrepancy  to  see  whether  it  is 
essential  or  negligible  for  practical  engineering  purposes. 

St.  Venant  studies  the  case  of  the  prism  under  the  following 
assumptions  :  p  =  Q  =  u  =  0  .  .  .  .  (4).  There  is  no  normal 
stress,  therefore,  in  the  directions  x  or  y  (that  is,  laterally  from 


APPLIED    MECHANICS. 


372 


fibre  to  fibre);  there  may  be  tangential  force  acting  from  fibre 
to  fibre  in  a  direction  parallel  to  z.     Hence  equations  (1)  give 
df        ,     ds       .     df   .    da       d& 

•as  =°-  a?=°'  s  +  ^  +  ir=°----(8)- 

The  surface  conditions  (3)  become 

IT  +  ms  =  0  ____  (6). 

He  also  assumes  that  the  origin  is  fixed,  and  that  particles  in  the 
directions  of  the  axes  of  z  and  y  just  there  are  fixed.  This  means 
that  at  the  origin 


The  following  purely  mathematical  work  is  tedious,  but  quite 
easy.  The  student  will  do  it  carefully  for  himself,  following  our 
directions.  [Inserting  (4)  in  (2),  so  as  to  calculate  the  strains, 
we  find  that 

du       dv  dw  ... 

d^-J^-       <r^-"'(8) 

where  <r  is  Poisson's  ratio,  or  )8/o,  or  |A/(A  -f  N).     Also  u  =  0 
means  that 

dw   ,    dv 

' 


di       .  ,,    ,  d*u   .      d* 

—-  =  0  means  that  -=•  •=•  +  -3 
dz  dz*       dz  . 


0, 


ds 
-^— 
dz 


.  ... 

=  0  means   that 


d*w     ,   dzv 

-Tdj+d*=Q 


....  (10), 


and  the  last  condition  of  (5)  gives 
(    dzu          d?w       (Pw 

N  \dH7dz  TJifrj       d 


The  first  and  second  of  (10)  and  (8)  give 


(11)  and  (8)  give,  after  simplification, 


Differentiate  (13)  with  respect  to  z,  and  (10)  with  respect  to 
and  y,  and  use  (8)  to  eliminate  u  and  v,  and  we  get 


Differentiate  (10)  with  respect  to  y  and  #,  add  and  use  (9), 
we  get 

T-^ST—t  r    =0....(16). 

dx  .  dy  .  ds  v     ' 


373  APPLIED    MECHANICS. 

Differentiate  (10)  with  respect  to  y  and  a?,  and  use  (8)  and  (14), 
and  we  get 


Hence  ->-  is  linear  in  z,  and  linear  in  x  and  y  separately.  Hence 

^  =  a  +  aix  +  a,y  +  *  (ft  +  /Bj*  +  /Bjy)  ....  (17). 

[a  and  )8  are  here  any  constants,  and  not  the  coefficients  of  formula 
(2)].     Now  use  (8)  with  (17),  and  also  (10),  and  we  have 

U  =  -  <f   (ax  -f 


5  «iz2  —  i  02s  +  zui  +  U0 
where  ut  and  u0  are  functions  of  y.     Similarly, 

f>=  —  ff  (ay  -f  o^y  +  %  o^2  +  )8zy  +  fray*  +  ^  /82y2z)  - 


where  v0  and  vt  are  functions  of  x.    Now  try  these  in  (9),  and  find 
values  of  w0,  uv  v0,  vv  and  get 


w?  =  z  (a  +  a^x  +  «ay)  +  i  22  (j3  +  fax  +  /82y)  +  </>  (a?,  y). 
Using  this  in  (13),  we  see  that  we  must  solve 


To  get  the  complete  solution,  we  find  the  general  solution  when 
the  right  hand  side  is  zero,  and  add  to  it  a  particular  solution. 
For  the  particular  solution,  try 

<f>  =  A  x*  +  B  y1  -f-  c  xyz  -f 
and  we  find  it  to  be 


This,   then,  is  what  we   must    add  to  the  general  solution   of 
-T-J  +  -=-^  =  0,  taking  care  that  the  value  of  ^  so  found  enables 

the  whole  value  of  w  to  satisfy  the  boundary  conditions.     We 
may  now  write  out  wt 

w  =  z  (a  +  aim  +  <**y)  +  £z2  (|8  -f-  fta?  -f-  £#)  -  {^  j8  (^  +  y2) 

+ 


309.  Boundary  Condition. — Use  (2),  and  find  from  the  above 
values  of  v,  v,  w  values  of  s,  T,  and  B.  Use  them  in  (6).  Integrate 
both  sides  of  the  equation  round  the  boundary  of  a  cross-section,  and 


APPLIED    MECHANICS. 


374 


transform  the  line  integrals  to  surface  integrals  over  the  section. 
This  gives  us  )8  =  0.     Hence  our  complete  answers  are : — 

u  =   -  a  (ax  +  %  Ojo;2  +  o^xy  +  %  fax2z  + 

v  —   -  <r  (ay  +  a-^xy  +  %  a^y2  +  faxyz 

w  =  z  (a.  +  a-p  +  a^y)  +  %  z2  (fax  +  fay)  +  <f>  - 


where  d>  satisfies  -7     +  -=-     =  0  at  all  points  of  any  cross-section, 
dx2      dy2 

and  the  condition 


)32 


m<ra;2     ---- 


(19) 


at  all  points  of  the  cylindric  Jboundary.    The  stresses  at  any  point 
are  p  =  0,  Q  =  0,  u  =  0. 


K  =  E  [o  +  o^  +  a^y  +  z  (fax  + 

By  giving  any  values  to  a,  a1}  o2,  j80,  ft,  02  we  get  all  the  possible 
solutions.  It  must  be  remembered  that  a  point  x  y  z  has  the  new 
position  x  +  u,  y  +  v,  w  +  z. 

Notice  that  if  /30,  fa,  fa  are  0,  then  $>  =  0,  because  s  =  N  -=? 
and  T  =  N  -i,  and  each  of  these  is  0  at  the  boundary,  and  the  only 

conditions  to  be  satisfied  by  d>  are  that  l^-  +  m^  =  0  at  the 

d£          dy 

boundary.  Hence  <£  may  be  a  constant,  but  it  is  0  where 
x,  y,  z  =  0,  and  hence  0  =  0  everywhere. 

310.  Example  1.  —  Let  all  the  arbitrary  constants  vanish  except  a. 
We  must  remember  that  o  is  any  arbitrary  constant,  and  is  not  the 
o  of  Art.  289.     Then  u  =  -  trctx,  v  =  -  a-ay,  w  =  az,  R  =  E  o,  and 
the  other  stresses  vanish.     We  have,  therefore,  a  tie  bar. 

311.  Example  2.  —  Bending.  —  Let  all  vanish  butoj,  then  u  =  - 
\  ttj  (z2  +  ax2  -  (ry2),  v  =  -  a^xy,  w  =  a\xz.    Hence  P  =  0,  Q  =  0, 
B  =  E  a,:r,  s  =  0,  T  =  0,  u  =  0.     The  shape  of  the  centre  line  is 
obtained  by  putting  x  =  0,  y  =  0,  and  we  have  u  =  -  £  ai«2,  v  =  0, 


375  APPLIED    MECHANICS. 

w  =  0.  So  that  it  becomes  the  arc  of  a  parabola,  or  nearly  a  circle 
of  curvature  av  lying  in  the  plane  x,  z.  The  total  force  parallel  to 

z  across  the  section  =1   I  E  a-^x  .  dx  .  dy  =  0,  because  x  is  measured 

from  the  centre  of  area.  Hence  the  tractions  at  a  section  are  a 
torque  merely,  called  here  a  bending  moment,  whose  amount  is 
evidently  -  E  01  ij  where  IA  is  the  moment  of  inertia  of  the  section 
about  the  axis  of  y  in  the  section.  Since  w  =  aixz,  and  we  make 
z  =  z0  a  constant,  we  have  the  displacement  at  all  points  in  a 
section.  The  section  remains  plane,  and  turns  through  the 
angle  z0ai. 

Change  of  Shape  of  Section. — Take  a  portion  of  the  section 
of  rectangular  shape.  The  boundaries,  having  been  x  =  ±  a,  y  =  ±  b, 
have  become  x  =  ±  «  -  \a-\a  (a2  -  y2),  y  =  ±  b  -  cra^x.  So  that 
the  boundary  consists  now  of  two  straight  lines  and  two  arcs  of 
parabolas.  We  note  that  the  upper  and  lower  surfaces  of  the 
beam  are  anticlastic  surfaces,  the  principal  curvatures  being  in  the 
ratio  ff.  Note  (Art.  330)  the  result  of  the  ordinary  theory.  It  is 
then  evident  that — at  all  events,  when  the  bending  moment  is 
constant — the  hypothesis  on  which  our  ordinary  theory  is  based, 
namely,  that  a  plane  cross-section  remains  plane,  agrees  with  the 
more  fundamentally  correct  theory.  But  we  have  now  to  notice 
that  this  case  of  the  St.  Venant  theory  only  contemplates  small 
displacements,  and  may  not  be  applied  to  cases  where  the  displace- 
ments are  large.  The  ordinary  theory  is  more  useful,  therefore, 
for  it  makes  the  shape  of  the  centre  line  to  be  circular,  as  it 
obviously  must  be ;  and  no  attention  need  be  paid  to  the  fact  that 
this  case  of  the  St.  Venant  theory  makes  it  the  arc  of  a  parabola. 

312.  Example  3.— Let  all  the  constants  except  £„  vanish,  and  let 
j80  be  called  -  r.  Then  u  =  -  rzy,  v  =  rzx,  w  =  <f>  (x,  y),  where 
T  is  a  constant  and  w  is  a  function  of  x  and  y  only.  We  find  that 


and 


/d<b  \  /d<b  \ 

p  =  <j  =  R  =  0,  s  =  N  I  -^  +  rx},  T  =  N  I  -^-  -  ry],  u  =  0, 

at  the  boundary  l-~  +  m-~  =  r  (ly  -  mx}.     Everywhere  in  the 
section  -^\  +  -j-f  =  0.     The  total  tangential  force  parallel  to  x  is 

I   I  T  .  dx    dy.    On  writing  this  out,  it  will  be  found  transformable 


and  into  N   I   j  Ix  (-^  -  ryj  +  mx  (-^  +  rx\  (  ds   round 


the 
boundary,  and  this  is  0  by  the  boundary  condition      Hence  the 


APPLIED    MECHANICS.  376 

resultant  force  parallel  to  x  is  0,  and  the  resultant  force  parallel 
to  y  is  0.    So  that  the  forces  over  the  section  are  a  torque 

N  1    I    )  X  \d     +  T/  ~  y  \d     ~  T^)  !  dX  '  ^'  wnicl1  k  e(lua^  to 

f*  C* 
TNI  +  N  I   I  (x-jr  ~  y  f\  dx  .  dy  =  qnri,  say,  where  i  is  the 

moment  of  inertia  of  the  cross-section  about  its  centre. 

This  case,  then,  is  that  of  a  prism  subjected  merely  to  a 
twisting  moment  q  N  T  i.  When  the  section  is  circular,  q  is  1,  and 
T  is  the  angle  of  twist  per  unit  length  of  the  prism.  When  the 
section  is  not  circular,  q  is  not  1,  and  is  called  St.  Venant's  torsion 
factor.  To  solve  the  problem,  then,  when  the  shape  of  section  is 

given,  we  must  find  <f>  such  that  ~  +  ~  =  0  all  over  the  section, 

with  the  boundary  condition  l-^-  +  m-r^  =  r  (ly  -  mx).     To  do 
this,  it  is  well,  as  in  many  other  such  problems,  to  find,  first,  a 

tefi 

d$ 


conjugate  function  ^,  such  that          +     -    =  0,  and  -     =      ,  and 

ax*        dy*  ax       ay 


=  -  -.     So  that  the  boundary  condition  becomes 

.dty          dti 

l~  -  m^-—  T  (ly  -  mx). 
dy          dx 


Now  we  see  by  trial  that  if  ^  =  £  T  (x2  +  y2)  +  c,  the  boundary 

d\b  ddj 

condition  is  satisfied  f  or  -?-  =  r  y,  and  -~  =  T  x.     It  is  evident, 

then,  that  if  we  can  find  a  solution  of  -j\  +  ~  =  0,  subject  to 

the  condition  if/  =  \  r  (x*  '+  y*}  +  c  at  all  points  of  the  boundary, 
we  have  the  correct  answer. 

Example.—  Prism  of  elliptic  section,  ^  +-^  =  !>  g™g  the 

shape  of  the  boundary.  It  is  found  by  guessing  that  ^  =  A  (a:2  -  y*) 
is  a  solution,  if  A  is  properly  evaluated  to  .satisfy  the  boundary 
condition.  So  that  A  (x2  -  y1}  =  \T  (x2  +  y*}  +  const,  at  the 

i2 
boundary—  that  is,  where  y2  =  ^  (a2  -  x2).    We  substitute  this, 

and  find  that  A  is  £T  (a2  -  £2)/(a2  +  P).  So  that  ^  =  A  (x*  -  yz), 
and  </>  =  -  2,Axy. 

Evidently  the  curves  in  the  section  along  which  $  is  constant 
are  rectangular  hyperbolas,  with  the  principal  diameters  as  axes. 
The  total  twisting  couple  is 


377  APPLIED    MECHANICS. 

The  shears  are  now  easily  calculated. 

8  =  N  ( -  2  A  #  +  T  #),  T  =  N  ( -  2  Ay  - 


2  -  i2\  /    2  a2 


8  is  greatest  when  x  =  a,  then 

also  s2  +  T  2  =  f*  = 


a2  +  i2' 
The  maximum  stress  anywhere  in  the  section  is  the  value 

of  T  when  y  =  b,  viz.  N  r  -^  -  r§.     For  /  is  a  maximum  when 

44#2  +  a4*/2  is  greatest.  Now,  whatever  be  the  value  of  x  for 
which  the  maximum  occurs,  for  that  value  of  x  the  maximum  will 
be  when  y  is  greatest  —  namely,  on  the  boundary.  Put  x  =  a  cos.0, 
y  =  b  sin.0,  then  b*a?  +  a4*/2  becomes  aW  [b2  +  (a2  —  P)  sin.20], 
which  is  obviously  greatest  when  9  —  90°.  Hence  the  stress  is  a 
maximum  at  the  extremities  of  the  minor  axis. 

Example.  —  Torsion  of  a  Shaft  whose  Section  is  an  Equilateral 
Triangle.  —  Let  3  a  be  the  vertical  height  of  the  triangle,  then  the 
equation  of  the  boundary  is  (x  -  a)  (x  -  y  V  3  +  2  a)  (x  +  y  V  3  + 
20)  =  0,  or  z2  -  3xy2  +  3a  (x2  +  y2)  -  4a3  =  0.  The  function  ^  = 
A  (tf8  -  3  xy*}  satisfies  the  differential  equation,  and  A  (x3  -  3  xy2) 

-  i  r  (x2  +  y2)  will  be  constant  over  the  boundary  if  A  =  -  —  . 
Hence  *  =  -  (*»  -  3^2),  and  *  =  -  (y»  -  3afy)  ....  (1). 


From  this  it  is  easy  to  find  the  strains  and  stresses.  It  will  be 
found  that  the  shear  is  0  in  the  corners,  and  is  a  maximum  at  the 
middle  of  the  sides. 

313.  Approximate  Formulae.—  St.  Venant  worked  out  </>  and  the 

strains  and  stresses  for  a  number  of  sections,  and  he  found  that  if 
the  section  of  a  shaft  is  not  too  different  in  any  two  of  its 
dimensions  across  the  centre,  the  torsional  rigidity  (or  twisting 

64N 

moment  per  unit  angle  of  twist)  is  A  =  m  —  where  s  is  the  area 

of  the  section  and  i  its  moment  of  inertia  about  its  centre  of 
gravity  ;  N  is  the  modulus  of  rigidity,  and  m  is  a  number  which 
does  not  greatly  differ  in  different  cases,  m  =  -02533  for  an 
ellipse;  and  in  the  sections  examined  its  lowest  value  was  -023, 

and  its  highest  was  *026.  Consequently,  if  we  take  A  =  j^  s4N/i  in 
all  such  cases,  there  is  no  great  error. 


APPLIED    MECHANICS.  378 

314.  Non-uniform  Flexure.— In  Art.  309  let  all  the  constants 
be  zero  except  0V     The  displacements  are 


where  </>  satisfies  -^\  +  —  *  =  0  at  all  points  of  a  normal  section  ; 
and  at  the  boundary  the  condition 

id£  +  ™^  =  &  Ei  ^2  +  (2  +  *)  (^  +  i  fo8) 

The  stresses  at  any  point  of  a  section  z  =  constant,  are 


s  =  N  f  -    -  (2  +  <r)  £i#*      parallel  to  the  axis  y, 


T  =  N  ^  -40!  (<rz2  +  2  y2)  Y  parallel  to  the  axis  a?, 
,  parallel  to  the  axis  z. 


The  total  force  parallel  to  x  (really  the  shearing  force  x  at  the 
section)  turns  out  to  be  E  i  0lt  and  the  resultant  total  force  parallel 


f  f 
to  y  (call  it  a  shearing  force  Y)  is  E  £j  I   I  xy  .  dx  .  dy,  and  this  is  0 

if  the  axes  are  principal  axes  ;  and  we  shall  assume  them  to  be  so. 
The  resultant  stress  parallel  to  z  vanishes.  The  couple  about  the 
axis  of  y  is  -  Z^EI.  The  couple  about  the  axis  oi  2  may  be 
written  out.  By  a  combination  of  this  and  previous  solutions,  we 
have  the  case  of  any  twisting  and  bending  couples  applied  to  a 
prism.  If  we  merely  take  the  case  of  a  prism  being  bent,  being 
fixed  at  one  end,  and  loaded  at  the  other  with  a  load  w,  we  must 
make  our  twist  0  ;  and  hence  we  must  use  the  solution  in  which  a^ 
was  constant,  together  with  this  in  which  fa  is  constant,  so  that 
the  two  twists  may  just  neutralise  each  other.  We  get  this 
condition,  we  find,  by  taking  aa  and  j8j,  such  that  ttj  +  fij  =  0.  In 
this  case  we  have  a  total  tangential  force  parallel  to  x  of  the 


amount  w  =  x  =  Efri,  and  a  couple  EjSji  (/  -  z),  or  w  (I  -  z), 
which  we  usually  call  a  bending  moment,  due  to  a  load  w  at  the 
distance  I  from  the  fixed  end  of  the  beam,  cti  =  -  &J  =  -  vrl/E  i. 

The   equation   to  the  centre  line  is  x  —  —  (£z2J  -  ^z3).     We 


see, 
auch 


therefore,  that  the  ordinary  theory  is  right  as  to  ihe  shape  of 
L  a  beam.     The  curvature  «i  is  the  bending  moment  at  the 


379 


APPLIED    MECHANICS. 


fixed  end  divided  by  E  i,  the  flexural  rigidity.     The  displacements 
are,  if  there  is  no  twist, 

~w 


(I  -  «)  *y, 


Notice  from  «  and  v  that  the  change  of  shape  of  the  section  is 
as  given  in  Art.  311.  The  strain  g  —  —  =  —  (z  -  T)  x  (that 
is,  the  tensile  strain  at  any  point  of  the  section  is  proportional 

to  *).     d~  =  e  =  —<r  (I  -  z)  x,  f'=^-=  —  (l-  z)  x.    We  may 
dx  EI  dy      EI  v 

write  out  the  shear  strains  a  and  b  in  terms  of  <J>,  etc.     c  =  0. 

Now  in  cases  that  have  been  considered  in  which  x  and  y  are 
small  compared  with  I  -  z,  it  is  found  that  <f>  is  of  the  third  degree 
in  x  and  y.  So  that  a  and  b  are  small  compared  with  e,  /,  and  g. 
For  the  same  reason,  the  term  2  xyz  is  not  important  in  w  above. 
And  the  engineer's  theory  based  on  the  assumption  that  a  plane 
section  remains  plane,  may  be  taken  as  correct.*  When  x  and  y 
are  not  small  compared  with  I  -  z,  we  must  find  <£,  and  this  is 
difficult.  As  w  contains  </>,  although  the  tensile  strain  is  pro- 
portional to  x,  the  plane  section  does  not  remain  plane  if  beams  are 
short.  St.  Venant's  solution  assumes  that  w  is  distributed  in  a 
particular  manner  over  the  end  section.  But  by  the  principle  of 
equipollent  loads  the  actual  distribution  of  w  is  of  very  little 
consequence,  except  close  to  the  end  section  itself,  and  hence  is  of 
no  practical  importance  except  in  beams  that  are  not  long  in 
comparison  with  the  0 

values  of  x  and  y  in 
their  sections. 

316.  Vertical  loads 
are  often  applied  to 
beams  on  their  hori- 
zontal top  surfaces. 
We  know  from  the 
principle  of  equipollent 
loads  that  the  actual 
distribution  has  little 
effect  except  in  the 
neighbourhood  of  the 
surfaces  to  which  the 
loads  are  applied.  We 
can  obtain  a  fairly 
clear  notion  of  the  Fig.  205. 

effect  by  thinking  of 
the  load  on  a  plane  surface  bounding  an    infinite  elastic  solid. 

*  See  Appendix. 


APPLIED    MECHANICS.  380 

M.  Boussinesq  has  given  the  solution.*  The  following  brief 
memorandum  may  be  useful,  oo  being  the  plane  surface 
bounding  the  infinite  isotropic  solid.  M  is  a  point  within, 
situated  at  a  distance  MN  =  z  below  the  surface.  K  is  any 
element  of  the  surface,  situated  at  the  distance  K  M  —  r 
from  the  point  M,  and  subject  to  a  given  exterior  surface 


pressure  Kp  =  p  per  unit  area,  having  the  component  JLp1  =  p1 

he  pressure  which  a  plane   ele- 
ment   E  E1    taken    through    M    parallel   to  the  surface   o  o   will 


per  unit  area  along  KM.  The  pressure  which  a  plane 
ment  E  E1  taken  through  M  parallel  to  the  surface  o  o 
support  per  unit  area  in  consequence  of  the  pressure  p  will  be 

found  directed  along  K  M  produced,  and  is  equal  to  M  r  =  „  *. 
If,  as  a  particular  case,  the  pressure  K  p  =  p  be  normal,  then 
p1  =  p  cos.  NMK  =p-,  and  MP  =  3pzzj2irr*.  If  we  want  the 

vertical  component  of  M  p,  we  have  3^/2  iri*. 

Generally.  —  Let  w\  be  the  normal  force  per  unit  area  at  any 
point  on  the  plane  bounding  surface  at  the  point  x  =  z1,  y  =  y1, 
z  =  0,  the  axes  of  x  and  y  being  in  the  plane,  and  the  axis  of  z 
being  the  normal  to  the  plane  drawn  into  the  material. 


Let      = 


=  I   1  MI  .  rdx1  .  dylt  and  x  =  I   I  w^  log.  (z  +  r)  dxl  .  dyl, 

where  r  is  the  distance  from  xlylQ  to  xyz.     Then  u,  v,  and  w  being 
the  displacements  at  #,  y,  z, 

1          d  1       ffl 


__ 

4  IT  (\  +  N)  dx       4irN<fe 
rfx          1       d 


_    __        __ 
(A.  +  N)  dy       4  v  N  dz  .  dy' 

, 


_    __  _ 

4ir  (\  +  N)   dz         4irN  *   dz2        4fl-N  (A  +  N) 

From  w,  v,  and  w  all  the  strains  and  stresses  may  be  calculated  at 
any  point. 

*  The  interested  student  may  refer  to  a  paper  read  by  Prof.  Carus  Wilson 
before  the  Physical  Society  of  London  (June,  1891)  on  "  The  Influence  of 
Surface  Loading  on  the  Flexure  of  Beams,"  and  to  another  by  Dr.  Chreo. 


381 


6 


CHAPTER  XVI 

BENDING. 

316  IN  Fig.  206,  c  D  is  a  beam  carrying  a  weight  w.  We 
know  that  the  beam  transmits  the  weight  to  the  walls,  and  that 
in  doing  so  the  beam  is  kept  in  a  strained  condition ;  we  must 
consider  what  is  the  state  of  strain  in  the  beam.  To  observe 
this  it  will  be  well  to  take  a  beam  which  is  very  visibly  strained, 
a  beam  of  indiarubber.  A  B  is  its  appearance  when  lying  on 
the  table;  draw  upon 
it  a  number  of  parallel 
lines  in  chalk  or  pencil, 
a  b,  c  d,  ef,  etc.  Now 
if  we  support  the  beam 
at  its  two  ends,  and  load 
it,  we  find  that  the  lines 
a  6,  c  d,  etc.,  remain 
straight,  but  they  are  no 
longer  parallel.  We  find 
the  distance  a  c'  to  be 
less  than  a  c,  but  b'  d'  is  greater  than  b  d.  In  fact,  a'  c'  is 
compressed,  b'  d'  is  extended.  We  find  also  that  along  the 
line  B'  F'  there  is  neither  compression  nor  extension.  E'  p' 
remains  of  its  old  length,  although  it  is  no  longer  straight. 
If  we  consider  each  cross  section 
of  such  a  beam  we  see  that  the 
upper  part  of  it  is  in  compression, 
the  lower  part  of  it  is  in  exten- 
sion, and  there  is  a  straight  line 
in  the  middle  where  there  is 
neither  compression  nor  extension. 
This  line  is  called  the  neutral  axis 
of  the  cross  section,  and  all  these 
axes  lie  in  a  surface  called  the  neutral  surface  of  which  E'  F'  is 
an  edge  view.  Fig.  207  is  a  magnified  drawing  of  the  small 
portion  of  the  beam  between  two  cross  sections,  c  e  f  d 
shows  its  original  shape,  c'  e'  f  d'  its  shape  when  strained. 
Evidently  there  is  more  compression  at  c'  tf'  than  at  I'  m. 


Fig.  206. 


382  APPLIED    MECHANICS. 

The  compression  becomes  less  and  less  as  we  come  nearer  H'  j', 
then  the  extension  begins,  and  becomes  greater  and  greater  as 
we  get  farther  away  from  H'  j'  until  we  get  to  d'f',  where  it  is 
greatest.  If  the  material  is  likely  to  break  in  compression 
it  will  be  most  likely  to  break  at  c'  e'.  If  it  is  likely  to  break 
in  tension  it  will  be  most  likely  to  break  at  d'f. 

317.  If  we  know  the  compression  or  extension  at  any  place, 
we  can  calculate  what  it  is  at  any  other  place,  for  the  strain  is 
evidently  proportional  to  the  distance  from  the  middle.  Thus  if 
at  e'  there  is  a  compressive  strain  of  -002,  that  is,  there  is  a 
compression  of  '002  foot  for  every  foot  in  length,  then  at  ra', 
half-way  between  j'  and  e',  there  is  only  a  strain  of  '001.  There 
is  the  same  strain  at  i'  the  same  distance  below  j',  but  it  is  now 
an  extension.  The  material  resists  being  strained  in  this  way, 
and  the  pushing  and  pulling  forces  which  it  exerts  at  the  section 
e'f,  Fig.  207,  are  just  the  forces  required  to  balance  all  the 
other  forces  acting  on  the  part  e'  D  T/'. 

As  e'  D  Tf  is  a  body  kept  at  rest  by  forces,  and  which  is  no 


w,  wa  L 

jl 1 1 


-5  H 
O 


Pig.  208. 


longer  altering  in  shape,  it  is  to  be  regarded  as  a  rigid  body.* 
Now  what  is  the  condition  under  which  it  is  kept  at  rest  ? 

318.  The  beams  used  by  us  are  almost  never  deformed  so 
much  as  the  beam  shown  in  Fig.  206,  and  indeed  our  theories 
are  only  true  on  the  assumption  of  exceedingly  small  changes  of 
shape.  Let,  then,  the  part  e'  D  Tf  be  drawn  less  deformed  as 
E  D  T  P  in  Fig.  208,  and  consider  its  equilibrium.  We  had 
better  consider  more  weights  than  one,  loading  it. 

The  forces  wp  W2,  and  W3  represent  loads,  and  P  is  the 

*  In  boobs  on  mechanics  you  may  have  read  much  about  rigid  bodies  and 
the  laws  of  their  equilibrium,  and  you  may  have  thought  that  such  bodies 
had  no  existence ;  but  you  must  remember  that  we  can  regard  a  quantity  of 
water,  or  a  piece  of  steel  spring,  or  a  rope,  as  a  rigid  body  for  the  time  being, 
If  it  is  being  acted  on  by  forces,  and  is  no  longer  changing  its  shape. 


APPLIED    MECHANICS.  383 

supporting  force  at  the  end  T  of  the  beam.  In  Art.  99  we  saw 
that  if  all  the  loads  on  a  structure  are  given,  the  supporting 
forces  may  be  calculated. 

319.  We  are  now  considering  only  horizontal  beams  on 
which  the  loads  are  only  vertical  and  the  supporting  forces 
vertical.  Students  had  better  work  again  here 
a  few  exercises,  to  find  the  supporting  forces 
when  the  loads  are  given.  Either  neglect  the 
weight  of  E  D  T  F  itself  or  imagine  it  represented 
by  w2.  Suppose,  then,  p  to  be  known. 

The  molecular  forces  acting  on  the  surface 
E  Fv(by  the  material  to  the  left  upon  the 
material  to  the  right  of  E  F)  balance  all  the 
external  forces  acting  upon  E  D  T  F,  namely,  wx,  w2,  w3, 
and  P.  Art.  98  gave  the  following  three  as  the  conditions 
of  equilibrium. 

I.  The  upward  tangential  resultant  of  the  molecular  forces, 
which  I  shall  call  s,  the  shearing  force  at  the  section,  is  equal 
to  wx  -f  W2  +  w3  —  P.     The  student  ought  now  to  work  a  number 
of  exercises.     Given  loads,  find  supporting  forces ;  find  s  at 
many  sections ;  show  all  answers  in  one  diagram  and  call  it  the 
diagram  of  shearing  force.     Observe  that  we  assert  nothing  as 
to  how  this  shearing  force  is  distributed  over  the  section.     We 
shall  find  in  Art.  369  that  it  is  most  intensely  dis- 
tributed about  the  middle  parts  near  the  neutral 

axis  o  o. 

II.  As  the  loads  are  all  vertical  there  is  no 
horizontal  component  in  the  resultant  of  the  mole- 
cular forces;  in  fact,  all  the  pushing  forces  on 
E  F  balance  all   the  pulling   forces.     Figs.  209, 
210  show  E  F  magnified,  its  actual  shape  and  side 
elevation  also.      At  H,  a   point   in  E  F   at   the 
distance  y  from  o  0  the  neutral  axis,  the  pushing 
force  per  square  inch  or  the  compressive   stress 
being  proportional  to  y,  let  us  call  it  c  y,  when 

c  is  some  constant ;  if  a  point  is  at  H'  (Fig.  209),  Kg  2iQ 
we  shall  call  o  H'  a  negative  value  of  y,  so  that  a 
pulling  force  is  regarded  as  the  negative  of  a  pushing  force.  Now 
at  H  let  there  be  an  exceedingly  small  portion  of  area  a,  the 
force  on  this  is  c  y  a,  and  we  must  have  the  sum  of  all  such 
terms  as  c  y  a  for  the  whole  area  to  be  zero.  This  is  only  another 
way  of  saying  that  the  pushing  and  pulling  forces  are  equal. 


384 


APPLIED    MECHANICS. 


All  the  terms  c  y  a  have  the  same  multiplier  c ;  hence  what 
we  state  is  that  if  every  little  portion  of  area  a  be  multi- 
plied by  its  i/,  the  sum  is  zero.  When  this  is  so,  Art.  109  tells 
us  that  o  o,  the  neutral  axis,  must  pass  through  the  centre  of 
gravity  of  the  area.  This  is  why  we  are  always  so  anxious  to 
find  the  centre  of  gravity  of  the  section  of  a  beam.  The  rules 
for  finding  the  centre  of  gravity  of  the  area  are  given  in 
Art.  111.  We  are  now  about  to  find  the  value  of  c.  Notice 
that  if  we  know  c  we  know  the  stress  at  any  point  in  the 
section,  and  we  particularly  want  to  know  c.  o  E  or  c.  o  F,  the 
greatest  stress. 

III.  The  moments  of  all  the  molecular  forces  about  any  axis 
balance  the  moments  of  wl5  W2,  W3  and  p  about  the  same  axis. 
Now  as  these  are  all  vertical  forces,  if  we  choose  an  axis  at 
right  angles  to  the  plane  of  bending  (the  plane  of  the  paper)  in 
the  plane  E  F,  notice  that  about  any  such  axis  these  moments 
will  be  the  same  ;  hence  we  speak  of  the  moments  of  W1?  W2,  W3 
and  P  about  any  such  axis  as  the  bending  moment  M  about  the 
section  E  F.  When  it  tends  to  make  the  beam  convex  upwards 
I  call  it  positive.  Hence  M=  -  PTF  +  wa'i  E  +  W2'J E  +  W3'LK 

The  student  ought  to  practise  the  calculation  of  M  for  many 
sections  of  a  beam  and  then  show  his  answers  in  a  diagram 


Fig.  211. 

We  have  an  easy  graphical  method  of  drawing  such  a 
diagram  (see  Art.  349).  We  call  it  a  diagram  of  bending 
moment. 

320.  Very  well,  then,  the  sum  of  the  moments  of  all  such 
molecular  forces  as  c  y  a  must  be  equal  to  M.  Take  the  moments 
about  o  o,  the  neutral  axis ;  the  moment  of  cy  a'\$  cya  x  y 
or  c  2/2  a.  Now,  when  every  little  portion  a  of  an  area  is  multi- 
plied by  the  square  of  its  distance  from  a  line  and  the  sum 
taken,  it  is  called  the  moment  of  Inertia  I.  of  the  whole  area 
about  the  line,  and  it  is  easy  to  find  I.  for  any  area  about  the  axis 
O  O.  Part  of  Chap.  VII.  is  devoted  to  this  subject  of  moments 
of  inertia.  We  have,  then,  c  I  —  M,  or  c  =  M/I ;  and  hence  the 
compressive  stress  p  at  points  y  inches  from  the  neutral  axis  is 


APPLIED    MECHANICS. 


385 


p  =  M  y/I .  .  .  (1).  As  y  is  greatest  at  points  like  E  or  P,  we  have 
the  greatest  compressive  and  tensile  stresses  at  these  points. 
In  fact,  the  greatest  compressive  stress  in  the  section  is  at  E, 
and  its  amount  is  o  E.  M/I  .  .  .  (2) ;  the  greatest  tensile  stress  is 
at  P,  and  its  amount  is  o  F.  M/I . .  .  (3),  and  these  two  expressions 
give  us  the  great  laws  of  strength  of  beams.  If  we  know  the 
stresses  which  the  material  will  stand,  we  know  whether  the 
section  E  F  will  withstand  the  bending  moment  M. 

321.  If  the  material  is  cast  iron  it  is  advisable  to  have  o  E= 
about  4|  times  o  P,  because  cast  iron  will  stand  about  4J  times 
as  much  compressive  as  tensile  stress.    Hence 
the  usual  economical  cast-iron  sections  are  as 
shown  in  Fig.  211,  with  centres  of  gravity 
near  the  bottom  boundaries  of  the  sections. 
Whereas  in  wrought  iron  and  mild  steel  and 
other  materials  the  resistance  to  compressive 
stress  is  much  the  same  as  the  resistance  to 
tensile  stress,  and  consequently  o  E  is  made 
equal    to     o     P,     and    the 
M        usual    economical    sections 
are  as  shown  in  Fig.  221. 

322.  In  the  model,  Fig. 
212,  which  shows  a  beam 
fixed  at  one  end  and  loaded 
at  the  other,  part  of  the 
material  has  been  removed, 
ind  instead  of  it  we  have 
inserted  a  chain  or  link  A, 
which  is  only  capable  of 
which  is  only  capable  of 


Pig.  212. 


exerting 
exerting 


pull,    and    a  rod  B, 

push.  It  is  found  that  forces  acting  merely 
horizontally  on  M  N  o  p  are  not  sufficient  to  keep  it  at  rest ; 
we  also  need  an  upward  force  at  M  F,  which  is  equal  to  the 
weight  w,  together  with  the  weight  of  M  N  o  F  itself.  We  see  then 
that  at  such  a  section  as  M  P  of  a  beam  we  need  pulling  and 
•  pushing  forces,  but  also  to  satisfy  the  first  condition  given  above 
we  need  the  shearing  force  at  M  P.  In  fact,  an  upward  force  w' 
must  be  exerted  at  M  P  equal  to  the  weight  w  and  also  to  the 
weight  of  M  N  o  P.  At  M  P  the  bending  moment  is  w  x  o  P, 
together  with  the  weight  of  M  N  o  F  x  the  distance  of  its  centre 
of  gravity  from  M  F.  This  is  to  be  balanced  by  the  pull  in  the 
chain  A  or  the  push  in  the  rod  B,  for  these  are  equal,  multiplied 


386  APPLIED    MECHANICS. 

by  the  distance  between  their  lines  of  action.  If  a  beam  is 
long,  the  shearing  force  exerted  by  the  material  at  a  section  of 
the  beam  is  usually  not  so  important  to  consider  as  the  pushing 
and  pulling  forces,  and  in  many  cases  it  is  neglected.  When  a 
beam  is  very  short  the  shearing  force  becomes  more  important 
to  consider. 

323.  We  shall  now  take  a  casein  which  there  is  only  bending 
moment  to  be  balanced  by  the  material  at  a  section.  Let  A  B 
(Fig.  2 13)  be  a  strip  of  wood  or  metal  originally  straight,  whose 
weight  we  shall  neglect.  Fix  or  solder  to  the  ends  stout  pieces 
of  metal,  and  by  means  of  cords  and  weights,  or  in  any  other 


Pig,  218.  Fig.  214. 

way,  exert  couples  on  these  ends  as  shown.  Consider  now  the 
equilibrium  of  any  portion  say,  c  D  B  (Fig.  214).  At  the  sec- 
tion c  D  we  know  that  pulling  and  pushing  forces  must  be 
exerted  by  the  material  which  exists  at  the  left  of  c  D  on  the 
material  which  exists  at  the  right  of  c  D,  and  the  moments  of 
these  just  balance  the  moment  of  the  forces  F  and  F,  and  this  is 
evidently  the  same  at  any  section  of  the  strip.  The  bending 
moment  at  any  section  is  then  the  moment  of  the  couple  or 
torque  M  acting  at  either  end.  Magnifying  the  section  E  F,  as 
in  Fig.  210,  and  representing  the  amounts  of  the  pulling  and 
pushing  stresses  by  arrows,  we  see  as  before  that  as  the  sum 
of  all  the  forces  one  way  must  be  equal  to  the  sum  of  all  the 
forces  acting  the  other  way,  and  as  the  stress  at  each  place  is 
proportional  to  distance  from  o,  the  part  where  there  is  no 
stress  or  the  neutral  axis  is  as  before,  a  line  through  o 
at  right  angles  to  the  paper,  and  this  must  pass  through 
the  centre  of  gravity  of  the  section.  We  see  also  that  the 
stresses  at  all  points  of  the  section  are  given  by  (1),  and  if  we 
particularly  desire  to  know  the  greatest  stresses  they  are 
given  us  by  (2)  and  (3). 

324.  Unsymmetrical  Bending.— If  the  bending  moment  M  in  a 
section  is  about  an  axis  o  x1  through  o  the  centre  of  gravity  of  the 
section,  and  if  this  is  not  a  principal  axis  (see  Art.  114),  but  makes 


APPLIED   MECHANICS. 


387 


the  angle  x1  o  x  =  o  with  a  principal  axis  o  x  (about  which  tha 
moment  of  inertia  is  ij),  the  other  being  OY  (about  which  the 
moment  of  inertia  is  LJ),  then  M  may  be  resolved  into  M  cos.  a  about 
o  x,  and  M  sin.  a  about  o  Y.  ,The  stresses  and  strains  due  to  these 
separately  have  now  to  be  combined 
to  obtain  the  actual  stresses  and  strains 
due  to  M.  Thus  at  a  point  whose  co- 
ordinates referred  to  the  principal  axes 
are  x  and  y,  we  have  the  total  stress 


f= 


M  cos.  a 


y       M  sn.  a 

~~ 


If  /  is  put  equal  to  0,  we  find  the 
position  of  the  neutral  line.  It  evi- 
dently is  like  o  Q,  and  makes  the  angle 
/3  =  Q  o  x  with  o  x,  such  that  tan. 

£  =  -1  tan.  a.  It  is  now  easy  to  find 

x? 
the  points  which  are  at  the  greatest 

distance  from  o  Q,  and  these  are  the 
points  at  which  there  is  greatest  stress. 


X' 


Fig.  215, 


Exercise. — In  a  beam  of  rectangular  section,  if  the  axis  about  which 
the  bending  moment  takes  place  is  at  right  angles  to  a  diagonal,  show 
that  the  greatest  stress  is  at  corners,  and  is  of  the  amount  6  M  -~  bd  D  if  b, 
d,  and  D  are  the  breadth,  depth,  and  diagonal  of  the  rectangle. 

325.  The  line  which  passes  through  the  centre  of  gravity 
of  every  cross  section,  being  neither  extended  nor  compressed, 
is  of  the  original  length  of  the  strip.  When  the  beam  is  bent 
as  in  the  figure,  A  B  becomes  longer  than  this,  and  a  b  shorter, 
yet  their  ends  are  in  the  same  planes  A  a  and  B  b.  Thus  the 
strip  may  be  considered  as  a  bundle  of  fibres  lying  in  arcs  of 
circles  which  have  the  same  centre  and  subtend  the  same  angle 
at  that  centre.  If  we  know  their  relative  lengths  we  can  tell 
where  the  centre  of  the  circle  is.  Now  we  know  the  stress  per 
square  inch  on  a  certain  fibre,  and  we  know  its  original  length, 
hence  we  can  calculate  its  present  length  (see -Arts.  241  and  265), 
and  its  length  is  to  the  length  of  the  neutral  fibre  as  its  radius 
is  to  that  of  the  neutral  fibre.  In  this  way  we  find  that  the 
radius  of  the  neutral  fibre  is  numerically  equal  to  the  modulus 
of  elasticity  of  the  material  multiplied  by  the  moment  of  inertia 
of  the  cross  section,  and  divided  by  the  bending  moment  at  the 

1        M 

section,  or  the  curvature  -  =  —  ...  (4). 
r      E  I 

To  put  it  in  another  way  : — The  curvature  in  Fig.  207  being 


388  APPLIED    MECHANICS. 

angular  change  per  unit  length,  is  the  angle  between  c'  d  and 
e'jf  if  H'  j'  is  unity.  But  this  angle  is,  strain  at  y  -4-  y,  or 
p  -T-  Ey  .  .  .  (5)  and  by  (1)  this  becomes  (4).  The  form  (5)  is 
sometimes  useful  in  indicating  the  greatest  curvature  which 
may  be  given  to  a  beam  without  hurt,  by  making  p  the  greatest 
stress  which  the  material  will  stand,  y  being  the  greatest  distance 
of  any  point  in  the  section  from  the  neutral  axis  on  the  com- 
pression or  tension  side.  In  the  same  way  it  is  easy  to  see  that 

if  a  beam  was  already  bent,  having  a  curvature  — ,  the  change 

^*o 

1  1  M 

of  curvature  --—  =  -...  (6). 

TQ  H-    1 

Example. — A  straight  strip  of  tempered  steel,  07  inch 
broad,  (H  inch  thick  (this  represents  the  depth  of  a  beam),  is 
subjected  to  a  bending  moment  of  100  pound  inches  :  find  its 
radius  of  curvature.  Answer : — The  moment  of  inertia  of  the 
section  is  07  x  -1  x  -1  x  '1-4-12,  or  -0000583.  The  modulus 
of  elasticity  of  steel  is,  say  36,000,000,  and  36,000,000  x 
•0000583  -^  100  gives  21  inches  for  the  radius  of  curvature  of 

the  bent  strip.  The  curvature  is  ^T.  If  the  strip  was  already 
bent  before  the  load  was  applied  and  had  a  radius  of  curvature 
50  inches,  then  the  change  of  curvature  —  —  ¥77— o?j  '>  hence 

r  =  14 -8,  or  if  the  new  bending  takes  place  in  an  opposite  way 
to  the  old  so  that  the  new  curvature  is  called  negative  for  our 

purpose,  _  —  _  =  _-  so  that  r  =  -  36  -2  inches. 

OU  T         A\. 

Exercise. — A  beam  is  supported  horizontally  on  two  points,  one  under 
each  end ;  c  is  a  point  of  the  "beam  one-fourth  of  its  length  from  one  of 
the  points  of  support.  Compare  the  curvature  at  c,  supposing  the  beam 
to  be  uniformly  loaded,  with  what  it  would  be  if  the  beam  were  without 
weight  and  the  load  concentrated  at  the  middle  point,  the  total  load  in 
both  cases  being  the  same.  Ans.,  As  3  to  4. 

326.  When  a  beam  is  loaded  in  any  way,  we  know  now  how 
to  find  the  bending  moment  at  any  place,  and  if  we  know  the 
modulus  of  elasticity  of  the  material,  and  the  moment  of  inertia 
of  the  section,  we  can  find  the  curvature  of  the  beam.  We 
may  draw  a  bent  beam,  then,  in  the  same  way  as  we  draw  the 
springs  of  Fig.  216,  but  the  beam  is  so  little  curved  usually  that 


APPLIED    MECHANICS. 


389 


we  have  difficulty  in  getting  compasses  long  enough.  In  this  case 
it  is  usual  to  diminish  all  the  radii  in  some  large  proportion, 
remembering  that  the  deflection  of  the  beam  as  you  draw  it  is 
increased  in  this  proportion.  For  a  beam  fixed  at  one  end  and 
loaded  at  the  other  we  get  a  curve  just  like  the  portion  s  T  in 
Fig.  216  c,  a  being  the  fixed  end  and  T  the  loaded  end. 

The  following   method    is    not    perhaps    so    instructive    for 
beginners,  but  it  is  the  quicker  and  more  accurate  method.    The 


390  APPLIED   MECHANICS. 

theory  is  given  in  Art.  358.  Suppose  A  c  D  E  B  is  a  diagram  showing 
the  value  of  M  4-  E  i  at  every  place,  E  being  Young's  modulus,  and 
i  the  moment  of  inertia  at  each  section.  Treat  it  as  if  it  were  a 
diagram  of  loading,  as  described  in  Art.  350,  and  proceed  as  if  you 
wanted  to  find  M  ;  in  truth  what  you  will  find  is  a  diagram  which 
shows  everywhere  the  deflection  of  the  beam.  That  is,  if  you  get 
a  b  of  Fig.  235  horizontal,  you  will  have  found  the  shape  of  the 
beam  turned  upside  down.  The  scale  will  be : — If  the  beam  is 

drawn  to  a  scale  of  1  inch  represented  by  n  inches,  and  if  —  is 
drawn  to  a  scale  of  unity  in  pounds  and  inches  represented  by 

M 

ij  inches,  and  if  an  area  of  1  square  inch  on  the  diagram  of  - — 

in  Fig.  237  is  represented  on  the  diagram  of  Fig.  236  by  v  inches, 
then  a  deflection  of  the  beam  of  1  inch  is  represented  by  a  distance 
of  y  ri?  v/o  n  inches,  o  H  is  to  be  in  inches. 

327.  Elastic  Curve. — If  we  take  a  straight  uniform  strip  of 
steel  and  subject  it  to  two  equal  and  opposite  forces  in  the  same 
straight  line,  the  strip  will  assume  one  of  the  forms  shown  in 
Fig.  216,  which  all  go  under  a  common  name — the  elastic  curve. 
Now  consider  the  part  p  B,  Fig.  21 6  a.  Neglecting  its  own 
weight,  it  is  acted  on  by  a  force  F  at  B,  and  at  P  there  must  be 
an  equal  and  opposite  force  to  produce  balance.  There  is  a 
force  at  P  tending  to  compress  the  steel,  but  what  is  of  more 
importance  is  the  fact  that  F  produces  a  bending  moment  at  p, 
and  the  amount  of  it  is  the  force  x  the  distance  p  K.  Now  our 
strip  is  everywhere  of  the  same  material  and  section,  and  the 
only  thing  that  can  alter  is  its  curvature.  This  curvature  at 
any  place  we  know  to  be  greater  when  the  bending  moment  is 
greater,  and  less  when  the  bending  moment  is  less  ;  in  fact,  the 
radius  of  curvature  is  inversely  proportional  to  the  bending 
moment,  and  this  really  comes  to  the  fact  that  the  radius  oi 
curvature  at  any  place  P  is  inversely  proportional  to  the 
distance  p  K  ;  or  if  the  distance  p  K  of  any  point  from  the  line 
of  force  is  called  x,  as  r  =  E  I/M  or  E  I/F.JC,  and  as  E  i  and  F  are 
constant,  r  x  is  constant. 

We  can  obtain  the  shapes  shown  in  Fig.  216  in  two  ways : 
first,  by  taking  a  straight  strip  of  steel  and  performing  the 
operation ;  secondly,  by  drawing  the  curves  ift  a  series  of  area 
of  circles.  Suppose  we  have  calculated,  as  in  the  above  example, 
that  the  modulus  of  elasticity  of  the  material  multiplied  by  the 
moment  of  inertia  of  its  cross-section  is,  say,  200  in  inches  and 
Ibs.,  and  suppose  we  know  that  the  force  acting  at  B  is  10  Ibs., 
then  we  know  that  the  radius  of  curvature  at  p  is  equal  to  200 


APPLIED    MECHANICS.  391 

divided  by  the  bending  moment  at  p,  which  is  10  x  p  K.  In  Pact, 
the  radius  of  curvature  at  P  is  equal  to  20  divided  by  p  K  or  x. 
Choose  now  in  Fig.  217  the  point  c  as  the  middle  point  in  the 
strip.  Suppose  c  D  or  the  value  of  x  for  c  to  be  4  inches,  then 
the  radius  here  is  5  inches.  Take  c  o  =  5  inches,  and  with  o 
as  centre  describe  a  small  arc,  c  E.  Join 
E  o  and  produce  it.  Now  at  E  measure 
E  F,  the  value  of  x  for  E,  and  suppose  you 
find  it  34  inches;  divide  20  by  34,  and 
we  get  5 '88  inches,  and  set  this  new  radius 
off  from  E  to  o'.  Take  o'  as  a  new  centre, 
and  describe  the  short  arc  E  G  of  any  con- 
venient small  length,  and  in  this  way  proceed 
until  the  curve  is  finished.  This  is  not  a  very  Fig.  217. 

accurate  method  of  drawing  the  curve  unless 
the  arcs  are  very  short,  and  small  errors  are  apt  to  have 
magnified  evil  effects,  but  I  know  of  no  better  exercise  to 
impress  upon  you  the  connection  between  radius  of  curvature 
of  a  strip  and  the  bending  moment  which  produces  it.  You  are 
therefore  supposed  to  have  actually  drawn  one  such  curve  at 
least  before  proceeding  with  your  study  of  this 
subject.  There  is  a  way  of  diminishing  errors,  easy 
to  discover  for  yourself  if  you  are  interested. 

328.  Parts  of  these  curves  happen  to  be  the  shape 
taken  by  liquids,  because  of  their  capillary  action, 
between  two  parallel  solid  faces.      They  are  also  the 
shapes  of  the  arches  which  are  best  fitted  to  with- 
stand fluid  pressure.     Thus,  for  instance,  in  Fig.  218 
the  curve  from  M  to  N  is  of  the  shape  of  the  curve 

Fig.  216  e,  from  M  to  N,  the  free  water  level  being  the  line  A  B ; 
and  in  Fig.  219  the  middle  line  of  the  joints  of  the  arch  M  to  N 
is  the  same  curve  inverted.  The  water,  whose  pressure  it 
resists,  has  as  free  water  level  the  line  A  B  in  the  relative  posi- 
tion of  the  line  of  force  to  the  elastic  curve. 

329.  When  a  strip  of  elastic  material  is  bent,  it  not  only 
alters  its  shape  in  the  well-known  way,  but  it  alters  the  form  of 
its  cross-section.     On  the  convex  side  of  the  strip  the  breadth 
becomes  concave,  and  on  the  concave   side  of  the  strip  the 
breadth  becomes  convex.     It  is  very  easy  to  try  this  for  your- 
self on  a  broad  strip  of  steel  or  a  bar  of.  india-rubber.     These 
saddle  shapes  of  the  surfaces  are  due  to  the  fact  that  when  each 
fibre  is  pulled  it  gets  thinner  as  well  as  longer  (see  Art.  265), 

N* 


Fig.  218. 


392 


APPLIED   MECHANICS. 


IS 


and  when  it  is  pushed  it  gets  broader  as  well  as  shorter,  and  it 
very  curious  that  this  action  should  not  interfere  perceptibly 

with  the  laws  of  bending  as  I 

have  given  them  to  you.  In 
all  probability  it  does  so  in- 
terfere, however,  in  the  case 
of  a  very  broad  thin  strip  of 
material  greatly  curved. 

330.  In  any  section  of  a 
beam,  consider  a  rectangular 
portion  of  area  of  breadth  b 

Fig.  219.  (parallel  to  the  neutral  line) 

and  depth  d.      It  evidently 

undergoes  a  change  of  shape  of  outline  whose  character  does  not 
depend  upon  what  portion  of  the  section  it  is  in.      But,   for 
ease  of  calculation,  let  us  take  it  to  be  symmetrical  above  and 
below  the  neutral  line  oo  (Fig.  220).     The 
A          F        R          curvature  of  the  beam  is  M/EI.     The  com- 
pressive  stress  at  a  distance  y  from  the  neutral 
line  is/=  My/i.      This  produces  compressive 
strain,  but  it  also  produces  a  lateral  thickening 
strain  pf  (see  Art.  265).      Hence  the  breadth 
b  of  the  rectangle  at  any  place  y  becomes 
broadened  by  the  amount  b  M  y)8/i,  and  hence 
the  straight  sides  A  c  and  B  D  alter  to  the 
straight  sides  E  H  and  G  L.   Also  the  dimensions 
parallel  to  A  o  and  B  o  on  the  A  B  side  of  o  o 
lengthen  equally,  and  on  the  c  D  side  they 
shorten  equally.   Consequently  A  B  and  c  D  be- 
come arcs  of  circles  with    the  same  centre, 
is  £  d  at  A  B,  and  -  \  d  at  c  D.  Consequently 
,  HL  =  b  (1  -  M  ^dp/i).     And  if  r  is  the 

'  +  **  =  ;  +  |*^A   and  hence  J  =  JL, 


Fig.  220. 


EG 


a  (i  +  M 

radius  of   O'NO', 


or 

The  student  will  recollect  that  we  used  a  in  Art.  265  to 
represent  — ;  and  £  is  the  smaller  number,  such  that  j3/a  was  called 

Poisson's  ratio  <r.    We  see  now  that  the  curvature  of  o'  N  o'  is  <r 

times  the  curvature  of  the  neutral  line  of  the  beam.     In  the  above 

case  the  curvature  of  either  the  top  or  bottom  surface,  or  of  the 

neutral  surface,  is  anticlastic  or  saddle-shaped. 

Exercise. — One  side  of  a  plate  of  metal  is  at  01°c  and  the 

other  is  at  02°c.     The  plate  when  cold  is  plane.     What  is  now 

its  curvature,  and  what  is  the  greatest  stress  in  the  material 

if  curvature  be  prevented?     Ans.— a2.  (Bl  —  B^  2/2/2.      (This 

measures   the   specific   curvature   of    the  surface,    being   the 

product  of  the  two  principal  radii  of  curvature  at  any  point 

of  the  surface.)     Greatest  stress  =-Ea   (02— 02).     (See  Art.  5.) 


393 


CHAPTER   XVII. 

STRENGTH    AT   ANY    SECTION    OP   A   BEAM. 

331.  OBSERVE  that  in  any  section  of  a  beam,  the  stress  being 
greatest  at  places  furthest  away  from  o  o  (Fig.  220),  these  are 
places  where  the  material  is  most  useful  in  resisting  bending.  In 
some  cases,  as  in  railway  girders,  economy  of  material  and  its 
weight  are  so  important  that  there  is  very  little  area  of  section  ex- 
cept at  E  and  F,  where  there  are  two  booms  or  flanges  each  of  area 


Fig.  221. 

A  :  one  where  nearly  all  the  compressive  forces  act  (/A  is  the  total 
compressive  force  if/'  is  the  compressive  stress),  and  one  where 
nearly  all  the  tensile  forces  act  (being  equal  to  the  same  /  A), 
and  the  sum  of  their  moments  about  o  (or  any  parallel  line  at 
right  angles  to  the  plane  of  bending),  being  f  A  d  if  d  is  the. 
distance  from  E,  the  centre  of  one  boom,  to  F,  the  centre  of  the 
other  (and  called  the  depth  of  the  beam),  is  equal  to  the  bending 
moment.  In  plate  girders  there  is  a  thin  web,  or  perhaps  two 
(Fig.  221),  holding  the  booms  or  flanges  in  their  proper  posi- 
tions ;  in  open-work  girders  there  are  diagonal  braces  to  do 
this.  The  function  of  the  web  or  diagonal  braces  is  to  resist 
the  shearing  forces,  and  we  have  good  reason  to  know  (see 
Art.  334)  that  the  booms  or  flanges  need  be  proportioned  only 
to  withstand  the  bending  moment.  In  Fig.  100  we  had  a 


394 


APPLIED    MECHANICS. 


Ill  f  ] 


girder  with  few  diagonal  pieces.  Now  we  already  know  that 
for  absolute  certainty  in  calculating  the  forces  exerted  by  the 
pieces  it  was  necessary  to  imagine  pin  joints  not  merely  between 
the  braces  and  the  booms,  but  between  one  piece  of  boom  and 
the  next.  Evidently  this  is  nothing  more  than  assuming  that 
each  piece  can  only  exert  a  pulling 
or  pushing  force,  and  has  no  shear 
in  a  cross  section.  It  will  be  found 
in  Art.  369  that  there  is  practically 
no  shear  in  the  cross  section  of  the 
flanges  of  even  a  plate  girder,  and 
that  the  above  easy  practical  rule 
(fAd  =  ia.)  used  so  generally  by  en- 
gineers is  quite  legitimate. 

332.  When  the  web  is  a  plate  it 
is  hardly  worth  while  to  calculate 
whether  it  will  resist  the  shearing 
force  ;  we  always  find  that  the  web 
which  is  thick  enough  (never  less 
than  f  inch)  to  let  the  girder  be 
handled,  and  to  let  other  girders  be 
fastened  to  it  and  pieces  to  prevent 
buckling,  is  ever  so  much  larger  to 
resist  shearing  than  is  merely  ne- 
cessary for  this  purpose.  Plate 
girders  are  used  up  to  75  and  even  100-feet  span  for  railway 
bridges.  For  great  spans,  as  in  the  Menai  tube  of  Fig.  222, 
this  form  is  no  longer  thought  to  be  economical. 

In  open-work  girders  it  is 
necessary  to  make  the  calcula- 
tion, and  it  is  done  in  the 
following  way.  In  built-up 
structures,  if  iron  and  timber  0. 
are  used,  it  is  well  to  use 
timber  for  the  struts  (see 
Art.  372),  because  it  is  late-  i 


1  1 


Fig.  222. 


Fig.  223. 


rally  large,  and  iron  for  the 

fcies.     In  structures   in  which 

different  materials  are   used,    expansion  is   different   in    the 

different  parts;  and  there  ought  to   be  no   redundant  parts. 

Thus   if   four   pieces   form  a   parallelogram   and   it  has  two 

diagonal  pieces,  of  the  six  one  is  redundant.      If  all  expand 


APPLIED   MECHANICS.  395 

equally  there  is  no  great  harm,  but  if  one  expands  more  than 
another  great  weakness  may  result. 

Exercise. — A  hinged  square  of  four  pieces  of  iron,  each  6  feet  long,  has 
two  diagonal  pieces,  one  of  brass  and  the  other  of  iron.  The  iron  bars 
are  of  equal  cross  section,  and  three  times  tiat  of  the  brass  one.  There  is 
no  initial  strain  in  them  at  0°  0.  What  are  the  forces  in  the  pieces 
at  30°  C.  ? 

333.  Redundant  parts  are  often  useful  in  stiffening  a  struc- 
ture when  the  loads  are  apt  to  be  very  different  at  different  times. 
When  it  is  necessary  to  reject  redundant  parts  we  need  no 
rules ;  common-sense  will  tell  us,  for  example,  that  if  the  bars 
are  long  and  slender  we  ought  to  reject  the  struts  rather  than 
the  ties.  When  there  are  no  redundant  bars,  the  student 
follows  the  graphical  method  of  Chap.  VIII.,  or  the  correspond- 
ing analytical  method  which  gives  the  same  answer.  When  there 
are  redundant  bars  he 
must  make  certain  as-  , 

sumptions.  Thus,  if  a 
table  with  fairly  uni- 
formly distributed  load 
upon  it  has  20  legs,  we 
usually  assume  that  each 
leg  has  a  twentieth  of 
the  load  if  they  are  all 
equal  in  length  and  the 
floor  equally  yielding 
everywhere.  Probably  Fig.  224. 

this  is  wrong,  and  if  we 

had  sufficient  information  we  might  see  reasons  from  theory  01 
common-sense  to  suspect  more  load  on  some  of  the  legs  than  on 
others.  But  we  can  do  no  better.  If  in  Fig.  224  B  c  and  D  E 
are  the  booms  in  compression  and  tension,  if  B  D  is  A  inches,  the 
vertical  distance  between  the  centres  of  area,  and  if  the  sectional 
area  of  each  is  A,  then,  as  we  have  seen,  f  A  h  =  the  bending 
moment  at  the  section  B  D.  We  are  supposed  to  know  s,  the 
total  shearing  force  at  the  section.  This  is  the  force  with 
which  the  material  to  the  left  of  B  D  acts  upwards  on  the 
material  to  the  right  of  B  D.  If  the  push  or  pull  in  a  diagonal 
piece  is  p,  the  upward  component  of  this  is  P  sin.  0  if  0  is  its 
inclination  to  the  horizon,  and  the  total  shearing  force  is  resisted 
by  the  vertical  components  in  the  diagonal  pieces.  Thus  if 
all  the  diagonal  bars  are  of  iron,  and  if  they  are  equally  inclined 


396  APPLIED   MECHANICS. 

at  45°  to  the  vertical,  if  the  force  in  each  of  them  is  p,  the 
upward  component  of  each  is  p  cos.  45°  or  P/v/jJ";  if  there  are  n  of 
them  crossing  the  vertical  plane  B  D  then  n  P/-V/S  ^s  ^ne  whole 
shearing  force  s.  If  s  is  known,  p  may  be  calculated.  When  a 
figure  is  drawn  it  is  easy  to  see  which  piece  exerts  a  push  and 
which  exerts  a  pull. 

Exercise. — In  Fig.  224  there  are  four  diagonal  pieces  crossing 
the  section  at  45°.  If  the  bending  moment  at  BD  is  3  x  108 
pound-inches  and  B  D  is  5  feet,  find  the  area  of  the  boom  at  B  or 
at  D.  If  the  shearing  force  at  B  D  is  4  x  105  pounds  find  the 
probable  push  or  pull  in  each  of  the  four  diagonal  pieces  and 
distinguish  struts  from  ties.  Ans.,  5  x  106-r/;  7'07  x  10*. 

If  instead  of  Fig.  224  we  have  a  section  of  a  Warren  girder 
with  only  one  diagonal  piece  crossing  the  section  at  an  angle  of 
60°  with  the  horizontal,  what  is  the  push  or  pull  in  it?  Ans., 
4-62  x  105. 

jExercise. — If  only  one  diagonal  piece  at  45°  meets  a  boom 
and  a  vertical  piece,  show  that  the  force  in  the  vertical  piece  is 
equal  to  the  total  shearing  force  in  the  girder  section  at  the 
place. 

As  the  student  must  by  this  time  have  drawn  diagrams  of 
shearing  forces,  he  knows  that  when  the  load  of  a  beam  is 
uniformly  distributed  over  it  the  shearing  force  is  greatest  at 
the  left  end,  diminishes  to  zero  at  the  middle,  and  increases 
positively  to  the  right-hand  end  of  the  beam.  The  diagonal 
pieces  are  therefore  large  at  the  ends  of  a  beam  and  small  in  the 
middle,  and  especially  in  large  girders,  for  much  of  the  load  of 
any  large  girder  must  be  distributed  uniformly. 

334.  We  see  that  when,  as  in  large  girders,  we  think  greatly 
of  economy  and  we  know  our  loads  and  that  they  are  vertical,  we 
have  flanges  or  booms  whose  sizes  may  be  calculated  with  a  fair 
amount  of  correctness.  Even  in  smaller  beams  to  carry  vertical 
loads,  it  is  convenient  to  look  at  what  occurs  at  a  section  from 
this  point  of  view ;  that  the  flanges  resist  the  bending  moment 
and  the  web  the  shearing  stress.  Taking,  in  Fig.  223  for 
example,  h  the  vertical  distance  in  inches  from  the  middle  of  the 
top  flange  to  the  middle  of  the  bottom ;  Ac  the  area  of  the  top 
flange,  At  the  area  of  the  bottom  one,  in  wrought  iron  we 
make  Ac  =  At,  because  fc  is  much  the  same  as/"t,  but  in  cast 
irony*c  is  4J  times  /t,  and  hence  we  make  4£  Ac=  At;  or  the 
bottom  flange,  which  is  in  tension,  4J  times  the  area  of 


APPLIED    MECHANICS.  397 

the  top  one.  The  bending  moment  is  /c  AC  or  ft  At  multi- 
plied by  h. 

Exercise. — The  top  and  bottom  flanges  of  a  rolled  section 
of  wrought  iron  are  8"  x  f  ".  The  web  is  of  same  thickness. 
The  height  over  all  is  12".  What  is  the  bending  moment  when 
the  greatest  tensile  stress  is  10,000  Ibs.  per  square  inch  ? 

Work  this  in  two  ways.  I.  The  tensile  or  compressive  force 
in  each  flange  is  8  x  f  x  10,000=50,000  Ibs.  The  value  of  h  is 
12"  -  jj"  or  llf.  Hence  the  bending  moment  is  llf  x  50,000, 
or  569,000  pound-inches.  II.  The  moment  of  inertia  of  the 

*    i  8x123    7fx(10f)3 

section  about  its  neutral  axis  is  — ^ — - — TT> — ^-  =  000. 

U  1J 

This  divided  by  6  is  z,  the  strength  modulus,  and  10,000  z,  or 
647,000,  is  the  bending  moment. 

335.  The  above  example  gives  a  fairly  good  idea,  of  the  error 
in  adapting  the  usual  practical  rule  for  a  railway  girder  where 
there  is  almost  no  web  to  a  rolled  girder  where  there 
is  a  web.  The  web  is  here  distinctly 
of  importance  in  resisting  bending 
moment.  Some  engineers,  instead 
of  taking  h  from  centre  to  centre  of 
flange,  take  it  the  height  over  all ; 
but  even  if  we  take  ^=12"  in  the 
above  we  still  have  an  answer  0_ 
which  is  about  7  per  cent,  too  small. 
The  error  is  on  the  safe  side.  r 

Nevertheless,  it  is  always  better         |_ 


I 

5" 

I 

_P 


to   keep    to    the    correct   rule    of 

Art.  320  in  girders  which  have  an  Fig.  225. 

important  web,  and  in  all  mechanical 

engineering   calculations  we    keep    to   the    correct    rule.       I 

therefore  give  here  a  list  of  the  values  of  I  and  of  z  for  various 

forms  of  section. 

Exercise. — Show  that  the  centre  of  gravity  o  of  the  area  in 
Fig.  225  is  2  inches  above  the  bottom.  Take  it  as  x  inches. 
The  middle  of  the  bottom  rectangle  is  x  -  \  from  o  and  the 
middle  of  the  top  one  is  3J-&  from  o.  Hence  (x— f)  5  = 
(3J  —  x)  5  or  a?  =2  inches.  The  moment  of  inertia -of  the  top 

5  x  I** 

rectangle  about  axis  o  o  is  +  5  (ly)2  =  2l£.     The  moment 


398 


APPLIED    MECHANICS. 


of  inertia  of  the  bottom  rectangle  about  axis  o  o  is     '  9    +  5 

(]J)2=llf  ;  and  the  sum  isi  =  33-J-.    zl  for  the  top  is  I  -h  4  =  8  J- 
z2  for  the  bottom  is  I  -=-  2  =  16f. 

TABLE  VI. 


Moment  of  Inertia  of  Section. 


12 


Strength  Modulus  of  Section. 


/2s» 
112" 


BD8  - 


APPLIED    MECHANICS. 


399 


Moment  of  Inertia  of  Section. 


z  =  i/y. 
Strength  Modulus  of  Section. 


(B  n2  -  bh'2)'2  -  4  B  H  bh  (H  - 
12  (BH  -  bh) 


36  (ft  + 


+  ja 

' 


(BH2- 


6  (BH2  + 


, 
8  =  24 


12  (2  &  +  A,) 
b  +  ^bb,  + 
12   A  +  2 


2!  = 


400 


APPLIED    MECHANICS. 


Moment  of  Inertia  of  Section. 


z  =  ify 
Strength  Modulus  of  Section. 


+  V)  + 


(I3  T      3\~l 

**•!       _     7i         II          •»•  •  -  E—  .  -^ 
/       «^^^^^ 


)  +  «3  (**'  -  A.')] 


V 


(Parabolic  segment) 


I 

* 


p4~ 


32' 


32V 


~ 
32 < 


BD»  - 

=  3T' 
8 


APPLIED    MECHANICS.  401 

Exercise.  —  Find  the  greatest  load  that  may  be  uniformly  distributed 
on  a  cast-iron  girder  having  top  and  bottom  flanges  united  by  a  web,  of 
the  following  dimensions:  —  Width  of  upper  flange,  3  inches;  of  lower 
flange,  9  inches;  total  depth,  12  inches;  thickness  of  each  flange  and  of 
the  web  being  1  inch;  distance  between  the  points  of  supports,  10  feet; 
when  the  greatest  admissible  stress  in  the  compression  flange  is  6  tons  per 
square  inch,  and  that  in  the  tension  flange  1^  tons  per  square  inch. 

Ans.,  9  '8  tons. 

336.  Proportion  of  Depth  to  Length  in  Railway  Girders.  —  It  is 
usually  assumed  that  maximum  economy  of  weight  in  booms  and 
diagonal  pieces  leads  to  a  most  economical  ratio  of  depth  d  to 
length  I,  but  we  must  confess  that  we  feel  dissatisfied  with  the  easy 
mathematical  statements  sometimes  deduced  on  this  subject  from 
incomplete  data.  Take  it  that  in  girders  of  the  same  style  the 
diagonal  pieces  make  some  known  angle  with  the  horizontal.  Let 
us  take  45°,  for  example.  Then  each  bar  of  length  d_*/  2  and 
cross-section  a  withstands  a  shearing  force  s  =  of  x  A/  2,  where  / 
is  the  shear  stress,  and  has  a  weight  d*/  2  .  a  ('28),  or  the  weight 

is  d  A/  2  (-28)  j—  ^..     The  weight  of  the  corresponding  pieces  of 

boom  is  2c?A  (-28),  where  Kfd  =  M,  where  /  is  also  the  tensile 
stress.     Total  weight  of  a  bay  is  therefore 


or  the  weight  in  pounds  per  inch  run  of  the  girder  is 

....  „. 


We  see,  therefore,  that  if  the  inclination  of  the  diagonal  pieces  is 
fixed,  the  greater  d  is  the  better  ;  and  there  is  no  most  economical 
depth  of  a  girder  derivable  —  at  all  events,  from  these  simple 
considerations.  It  is  true,  however,  that  making  the  depth  great, 
if  the  inclination  is  constant,  means  that  we  are  increasing  the 
length  of  the  -unsupported  part  of  the  compression  boom  and  struts, 
and  they  may  need  more  lateral  bracing  ;  or,  as  it  is  better  to  put 
it,  the  cost  of  the  compression  members  per  pound  will  increase. 
Again,  the  weight  of  the  platform  will  also  increase.  These  are, 
however,  questions  of  a  different  kind,  'difficult  to  settle  by 
elementary  mathematical  equations  when  systems  are  so  different. 
Writing  down  such  general  expressions  as  we  may,  there  is 
evidence  that  there  is  greater  economy  in  weight,  whatever  there 
may  or  may  not  be  in  cost,  in  letting  the  depth  get  less  where  the 
bending  moment  is  less,  instead  of  keeping  it  constant. 

The  most  important  matter,  how  the  natural  period  of  vibration 
of  a  bridge  ought  to  come  in,  seems  never  to  be  brought  forward  in 
these  calculations.  Professor  Milne  finds  that  the  horizontal  trans- 
verse deflection  is  the  most  serious  motion  of  a  railway  bridge.  It 
begins  when  the  train  is  perhaps  200  feet  or  more  away  ;  it 
becomes  accentuated  with  every  passing  carriage,  and  when  the 


402  APPLIED    MECHANICS. 

whole  train  has  passed  there  is  the  natural  vibration  of  the  bridge, 
which  continues  for  some  time.  These  vibrations  are  due  to  lurch- 
ing of  carriages  and  impact  of  wheel-flanges  against  the  rails 
Sometimes  a  light  waggon  seems  to  produce  much  larger  effects 
than  the  heavy  locomotive,  and  there  is  some  speed  of  train  with 
which  the  vibration  is  much  more  serious  than  with  quicker  ^  or 
slower  speeds.  Bridges  have  not  yet  been  studied  from  this  point 
of  view,  and  engineers  must  for  the  present  rely  upon  their  large 
factors  of  safety.  We  are  sure  that  more  attention  ought  to  be 
paid  to  these  lateral  vibrations,  which  seem  to  be  greatly  accen- 
tuated when  gusts  of  wind  are  acting  laterally. 

The  Board  of  Trade  rule  is  5  tons  per  square  inch  on  wrought 
iron  and  6^  on  steel.  This  is  not  sufficiently  safe.  It  is  well  to 
say  5  in  tension  on  iron,  4  in  compression,  and  steel,  say,  |  greater. 
In  some  large  bridges  it  is  estimated  that  stresses  due  to  wind  are 
greater  than  those  due  to  rolling  loads. 

The  force  acting  on  the  rails  in  the  direction  of  their  length  is 
sometimes  as  much  as  £  th  of  the  weight  of  a  train  if  the  train  is 
quickly  stopped. 

As  for  the  vertical  motion,  its  consideration  will  probably  lead 
to  some  rule  connecting  maximum  deflection  yl  under  static  load, 
and  length  of  girder  I.  There  is  a  vague  sort  of  understanding  that 
y\  shall  be  something  between  ^oth  and  y^jth  of  I.  If  beams  are 
of  uniform  strength  and  depth,  d,  the  curvature  is  constant,  being 
2  //Ed  (see  Art.  362),  /  being  the  greatest  stress  in  e  very  ^  section; 
and  hence  in  girders  supported  at  the  ends  the  deflection  y\  is 


equal  to  Z2//4rfE.     If,  now,  we  take  yx  to  be  f/1,200, 


or     =  ~      Taking  E//as  3,000,      =  10  .....  (3). 


It  is  not  quite  fair  to  say  that  the  calculation  of  the  probable 
loading  of  railway  bridges  is  more  scientific  in  America  than  in 
Britain.  British  engineers  differ  much  more  in  their  assumed 
loads  than  American,  but  in  neither  case  can  it  be  said  that  there 
is  a  scientific  basis  for  the  rules  in  use.  It  is  very  important  that 
the  student  should  know  this,  because  there  is  sometimes  a  pre- 
tence of  accuracy  of  treatment  in  bridge  calculations  which  is 
quite  misleading.  The  following  rules  are  more  common  than 
others,  and  may  well  be  used  in  academic  problems. 

In  ordinary  railway  bridges  the  greatest  possible  rolling  load 
may  be  taken  as  if  it  were  a  static  load  of  wl  tons  per  foot-run  for 
a  double  line,  where  w1  =  3^  +  176/1,  if  I  is  the  span  in  feet. 
This  is  really  on  some  such  assumption  as  that  a  rolling  load  must 
be  multiplied  by  If  to  convert  it  into  the  equivalent  static  load. 
The  diminution  with  length  is  due  to  the  fact  that  the  engine 
weight  is  more  intense  per  foot  than  other  parts  of  the  train.  The 
weight  of  the  platform  may  be  taken  as  w^  =  0-7  +  -0072  I  tons 
per  foot  of  the  span  for  a  double  line.  w2  increases  with  the  span 
because  of  greater  wind-bracing  and  the  greater  width  of  larger 
spans  necessary  for  lateral  stability. 


APPLIED    MECHANICS.  403 

Half  of  every  term  may  be  taken  for  a  single  line,  and  used 
even  as  low  as  for  15 -feet  spans.  A  railway  girder  is  usually 
built  with  so  much  negative  deflection  or  "  camber,"  as  it  is  called, 
that  it  will  become  just  about  level  when  loaded. 

From  (2)  of  Art.  336,  taking  M  as  proportional  to  t^Z2,  if  w  is 
the  total  load  per  foot  run,  and  s  as  proportional  to  wl,  the  whole 

weight,  WB,  of  the  girders  per  foot  run  is  equal  to  (  a  —  +  bio  \  I, 

where  a  and  b  are  two  constants.  If  we  take  it,  as  is  usual,  that 
Ijd  is  nearly  constant  (say  10,  as  above  mentioned),  this  becomes 
tv3  =  lw/c,  where  c  is  some  constant.  Now, 


Hence  w3  =      c  or  — -,  (wt  +  w2)  .  .  .  .  (4), 

T~ 

/         I  \ 

and  w  =  (wi  +  w2)  /  (  1 )• 

V        c  / 

Whatever  may  be  the  worth  of  this  reasoning,  it  gives  us  a  rule 
which  is  taken  to  hold  in  all  railway  girders.  The  value  of  c  is 
about  1,000  feet  for  plate  web  girders,  1,200  for  lattice  girders  of 
ordinary  construction,  and  1,400  feet  for  bow-string  and  well- 
designed  lattice  girders. 

337.  Small  pistons  for  cylinders  up  to  20  inches  diameter  are 
packed  so  as  to  be  steam-tight  in  the  following  way.  The  method  is 
quite  wrong.  Suppose  the  cylinder  is  to  be  15  inches  diameter ;  there 
are  two  or  more  grooves  turned  in  the  piston  block,  about  ^  inch 
or  more  broad  and  J  inch  deep,  to  receive  cast-iron  rings.  Many 
such  rings  may  be  made  at  the  same  time.  A  hollow  cylinder  of 
cast  iron  is  turned  up  about  15 J  inches  outside  and  14|  inches 
inside,  and  many  rings  are  cut  from  it.  Each  ring  has  a  piece  cut 
out,  so  that  when  it  is  sprung  into  place  in  the  piston  groove  it 
may  be  jammed  into  the  cylinder,  its  ends  coming  now  close 
together.  The  cylinder  keeps  it  smaller  than  its  unstrained  size, 
and  the  pressure  all  round  is  assumed  to  be  uniform. 

Exercise  for  Students. — Prove  that  the  pressure  is  not  uniform 
all  round.* 

*  To  show  that  the  pressure  cannot  be  uniform  all  round  in  the  usual  method 
of  manufacture.  The  ring  unloaded  is  of  a  circular  shape ;  loaded,  it  is  circular 
and  of  smaller  radius.  The  bending  moment  must  therefore  be  constant. 
Let  us  try  what  distribution  of  pressure  p  will  produce  a  constant  bending 
moment ;  that  is,  using  the  symbols  which  follow  above,  if  M0  is  the  bending 


moment  at  the  end  where  0  =  0,  MO  -f-  r2  V     p  sin.  $  .  ety  =  M,  a  constant. 

Jo 

Differentiating  with  regard  to  0,  we  find  p  sin.  9  =  0  ;  and  as  this  is  true  for 
all  values  of  0,  we  must  have  p  =  0  everywhere.  That  is,  the  shape  can  only 
be  maintained  by  couples  applied  at  the  two  ends.  It  is  quite  a  usual  thing 
to  see  workmen  beating  such  a  ring  out  of  shape  near  the  joint,  after  it  has 
been  manufactured,  because  it  is  known  that,  so  far  from  pressing  uniformly 
all  round,  it  does  not  even  fit  the  cylinder,  but  concentrates  its  pressure  at 
certain  points. 


404 


APPLIED    MECHANICS. 


B 


The  following  method  of  making  a  piston  ring  produces  uni- 
form pressure  all  round,  even  if  the  section  of  the  ring  varies  very 
greatly.  A  ring  is  cast  which  is  larger  than  what  is  required  ;  a 

piece  is  cut  from  it,  and  the  two 
free  ends  are  brought  together 
hya  clamp.  This  clamp  pulls 
on  hoth  ends,  and  the  more 
nearly  the  resultant  pull  is  tan- 
gential to  the  mean  cylindric 
surface  at  the  end  the  more 
nearly  perfect  will  the  ring  be. 
The  ring  is  now  turned  up  to 
the  size  of  the  cylinder  and 
finished.  It  is  undamped,  and 
may  be  sprung  into  place. 

To  prove  that  the  pressure 
must  be  uniform  all  round.  In 
Fig.  226  let  BPA  be  part  of  a 
piston  ring  constrained  to  be  of 
the  size  of  the  cylinder,  of 
radius  r.  It  may  either  be  kept 
in  its  present  shape  by  the  equal 
and  opposite  forces  r  at  its  ends, 
or  ty"  pressures  p  Ibs.  per  inch 
of  its  length.  Let  us  suppose  at 
first  that  p  may  not  be  the  same  all  round,  being  some  function 
of  the  angle  A  o  p  =  6.  Draw  p  Q  perpendicular  to  o  A.  F  with- 
out p  produced  bending  moments  at  all  the  sections  of  the 
ring ;  the  pressures  p  without  F  must  produce  the  same  bending 
moments.  Let  A  o  R  =  <j>,  where  p  and  R  are  any  two  places  on 
the  ring,  and  A  is  one  end.  Bending  moment  at  p  due  to  F  is 

M  =  F  .  A  Q,   OF 

M  =  Fr  (1    —  COS.0)   ....  (1). 

The  pressure  at  n  on  the  element  r  .  8<J>  is  pr$$,  and  the  bending 
moment  due  to  this  at  P  IB  prz  .  50  .  sin  (0  —  <f>),  and  we  require 


r. 


pr*  sin.(0  —  <J>)  .  dq>  =  F  r  (1  —  cos.0)  ....  (2). 


In  the  integral^;  is  a  function  of  <f>.   Now  we  know  that  in  general 


(3). 


and  hence  equation  (2)  is 


re 

r*  \     p  sin.<J>  .  d<j>  =  F  r  (1  -  cos.0)  ---- 
Jo 


Differentiating  with  regard  to  0,  we  have  r^sin.0  =  Fr  sin.  6,  or 
p  =  F/r,  a  constant  ....  (4), 
•  See  Appendix. 


APPLIED    MECHANICS.  405 

It  will  be  observed  that  the  pressure  p  is  uniform  all  the  way 
round,  even  if  the  ring  varies  greatly  in  section.  It  is  quite  true 
that  the  proof  assumes  the  thickness  to  be  everywhere  incon- 
siderable. If,  however,  we  assume  a  uniform  thickness,  it  is  easy 
to  see  that  if  the  resultant  force  F  exercised  by  the  clamp  acts  on 
each  end  exactly  at  the  mean  radius,  there  is  absolutely  a  constant 
pressure  per  inch  all  the  way  round.  In  Japan,  twenty  years  ago, 
Professor  R.  H.  Smith  told  me  that  the  above  proposition  was 
true.  I  worked  out  the  proof  very  easily.  I  do  not  think  that  it 
has  been  published  before. 

The  usual  method  of  making  the  rings  is  very  much  more 
mischievous  than  it  may  appear  to  be  on  a  hurried  examination 
It  seems  to  me  that  if  the  correct  method  is  adopted,  care  being 
taken  as  to  the  proper  method  of  applying  the  clamp,  piston  rings 
may  be  made  in  this  way  for  the  very  largest  cylinders,  and  it  is 
evident  that  there  must  be  a  very  great  reduction  in  the  cost  of 
large  pistons  in  consequence. 

338.  Curvature. — The  curvature  of  a  circle  is  the  reciprocal  of  its 
radius ;  and  of  any  curve,  it  is  the  curvature  of  the  circle  which 
best  agrees  with  the  curve.  The  curvature  of  a  curve  is  better 
given  as  "  the  angular  change  (in  radians)  of  the  direction  of  the 
curve  per  unit  length."  Now  draw  a  very  flat  curve,  with  very 

little  slope  t.     Observe  that  the  change  in  i  or  ~  in  going  from  a 

dx 

point  P  to  a  point    Q    is    almost    exactly  a   change    of    angle 
change  in  -^  is  really  a  change  in  i,  the  tangent  of  an  angle ;  but 

when  an  angle  is  very  small,  the  angle,  its  sine  and  its  tangent, 

~1  dy 

ore  all  equal     .     Hence,  the  increase  in  -^  from  p  to  Q,  divided  by 

the  length  of  the  curve  P  Q,  is  the  average  curvature  from  p  to  Q  ; 
and  as  P  Q  is  less  and  less,  we  get  more  and  more  nearly  the 
curvature  at  P.  But  the  curve  being  very  flat,  the  length  of  the 

arc  P  u  is  really  Sx,  and  the  change  in  ~  divided  by  5#,  as  Sx  gets 

less  and  less,  is  the  rate  of  change  of  jL  with  regard  to  aet  and  the 

d2y  •    d*y 

symbol  for  this  is  ^-|.   •  Hence  we  may  take  *4  as  the  curvature 

of  a  curve  at  any  place,  when  it  is  everywhere  nearly  horizontal. 
If  the  beam  was  not  straight  originally,  and  if  y?  was  its  small 

deflection  from  straightness  at  any  point,  then  — f  was  its  original 
curvature.  We  may  generalise  the  following  work  for  beams 
not  straight  to  begin  with  by  using  -^  (y  -  y'}  instead  of  -£ 

everywhere. 

It  is  easy  to  show  that  a  beam  of  uniform  strength — that  is,  a 
beam  in  which  the  maximum  stress  /  (if  compressive,  positive ;  U 


406  APPLIED    MECHANICS. 

tensile,  negative)  in  every  section  is  the  same—  has  the  same 
curvature  everywhere,  if  its  depth  is  constant. 

If  d  is  the  depth,  the  condition  for  constant  strength  is  that 

M  M 

-  .  i  d  =  +  /,  a  constant.  But  -  =  E  x  curvature ;  hence  curva- 

tie  =  ?/, 
E.cT 

Example. — In  a  "beam  of  constant  strength,  if  d  =  =-,  then 

a  +  ox 

g  =  ?/(«  +  te).  Integrating,  we  find  ^  .  ^  =  c  +  ax  +  J  te», 
and  again,  £-  .  y  =  e  +  ex  +  f  ax"2  +  £ .  bx*,  where  e  and  c  must 
be  determined  by  some  given  conditions.  Thus,  if  the  beam  is 
fixed  at  the  end,  where  x  —  0,  and  -j-  =  0  there,  and  also  y  =  0 

there,  then  c  =  0  and  e  =  0. 

339.  In  a  beam  originally  straight,  we  know  now  that  if  x  is 
distance  measured  from  any  place  along  the  beam  to  a  section,  and 
if  y  is  the  deflection  of  the  beam  at  the  section  and  I  is  the  moment 
of  inertia  of  the  section,  then 

d2y  _  M          m 
*  d?  ~  El  •  '    •  *  * 

where  M  is  the  bending  moment  at  the  section,  and  E  is  Young's 
modulus  for  the  material. 

We  give  to  ~2  the  sign  which  will  make  it  positive  if  M 

is  positive.  If  M  would  make  a  beam  convex  upwards  and  y 
is  measured  downwards,  then  (1)  is  correct.  Again,  (1)  would 


•1-x— --»- 


Fig.  227. 

be  right  if  M  would  make  a  beam  concave  upwards   and  y  is 
measured  upwards. 

Example  1. — Uniform  beam  of  length  I  fixed  at  one  end, 
loaded  with  weight  w  at  the  other.  Let  x  be  the  distance  of 
a  section  from  the  fixed  end  of  the  beam.  Then  M  =  w  (I  -  x) ;  so 
that  (1)  becomes 


APPLIED    MECHANICS.  407 

Integrating,  we  have,  as  E  and  i  are  constants, 


+  c. 


From  this  we  can  calculate  the  slope  everywhere. 

To  find  c  we  must  know  the  slope  at  some  one  place.     Now, 
we  know  that  there  is  no   slope  at  the  fixed  end,  and  hence 

~  =  0,  where  x  —  0  ;  hence  c  =  0.     Integrating  again, 


To  find  c,  we  know  that  y  =  0  when  x  =  0,  and  hence  c  =  0  ; 
so  that  we  have  for  the  shape  of  the  beam  —  that  is,  the  equation 
giving  us  y,  the  deflection  for  any  point  of  the  beam  — 


We  usually  want  to  know  y  when  x  =  I,  and  this  value  of  tj 
is  called  D,  the  maximum  deflection  of  the  beam  ;  so  that 


Example  2.— A  beam  of  length  I  loaded  with  w  at  the  middle 
and  supported  at  the 
ends.  Observe  that 
if  half  of  this  beam 
in  its  loaded  condition 
has  a  casting  of  cement 
made  round  it  so  that 
it  is  ligidly  held,  the 
other  half  is  simply 


W 


a  beam  of  length  £  I,  **•  228. 

fixed  at  one  end  and 

loaded    at    the   other    with   \    w,    and,    according    to    the    last 

example,  its  maximum  deflection  is 


The  student  ought  to  make  a  sketch  to  illustrate  this  method  of 
solving  the  problem. 

Example  3. — Beam  fixed  at  one  end  with  load  w  per  unit 
length  spread  over  it.  The  load  on  the  part  pQiswxpQorto 
(I  -  x).  The  resultant  of  the  load  acts  at  midway  between  p  and 
Q,  so,  multiplying  by  £  (I  -  x},  we  find  M  at  p,  or 

M  =  \w  (I  -  x)* (6). 

Using  this  in  (1),  we  have  -E~  ^ \  =  P  -  2lx  +  #2.     Integrating, 
we  have  -  -     -  =  Px  -  lx2  +  %*P  +  e.     This  gives  us  the  slope 


408  APPLIED    MECHANICS. 

everywhere.     Now  -j-  =  0  where  x  =  0,  because  the  beam  is  fixed 
there.     Hence  c  =  0.     Again  integrating, 


Q 


Fig.  229. 

and  as  y  =  0  where  #  =  0,  c  =  0,  and  hence  the  shape  of  the 
beam  is 


y  is  greatest  at  the    end  where  x  =  lt  so  that  the  maximum 
deflection  is 


if  w  =  wl,  the  whole  load  on  the  beam. 

Example  4.  —  Beam  of  length  I  loaded  uniformly  with  w  per 
unit  length,  supported  at  the  ends.  Each  of  the  supporting 
forces  is  half  the  total  load.  The  moment  about  p  of  ^wl,  at  the 
distance  p  Q,  is  against  the  hands  of  a  watch,  and  I  call  this 
direction  negative  ;  the  moment  of  the  load  w  (\l  -  x)  at  the 
average  distance  J  P  Q,  is  therefore  positive,  and  hence  the  bending 
moment  at  p  is 


or  -  s  -givF  -  \W3?-  \  .  .  .  .  (9), 
so  that,  from  (1),  EI  -j—  =  -  j  ^wt2  -  \w&  \ .     Integrating,  wo 

F*  Q 


have  B  i  -  =  -  |  w?x  +  $  wx*  +  c,  a  formula  which  enables  us  to 
find  the  slope  everywhere,  c  is  determined  by  our  knowledge 
that  ~~  '=.  0  where  x  =  0,  and  hence  c  =  0.  Integrating  again, 


APPLIED    MECHANICS.  409 

+  -£fWX*  +  c,   and    c  =  0,   because   y  =  0   where 
=  0.     Hence  the  shape  of  the  beam  is 


y  is  greatest  where  x  =  %l,  and  is  what  is  usually  called  the 
maximum  deflection  D  of  the  beam,  or  i>  =  ^-r  —  if  w  =  Iw,  the 

Ou4:  E  I 

total  load. 

Example  5.  —  In  any  beam,  whether  supported  at  the  ends  or 
not,  if  w  is  constant,  integrating  (4)  of  Art.  357,  we  find 

~  =  b  +  wx,  and  M  =  a  +  bx  +  ^wx^  ----  (5). 

In  any  problem  we  have  data  to  determine  a  and  b.  Take  the  case 
of  a  uniform  beam  uniformly  loaded,  and  merely  supported  at  the 
ends.  Measure  y  upwards  from  the  middle,  and  x  from  the 
middle.  Then  M  =  0  where  x  =  \l  and  -  %l,  0  =  a  +  \bl  +  |tt>P, 
and  0  =  a  -  %bl  +  $wP.  Hence  *  =  0,  a  =  -  ^wP,  and  (5) 
becomes 

M  =  -  $wP  +  \wx*  ____  (6), 

which  is  exactly  what  we  used  in  Example  4,  where  we  afterwards 
divided  M  by  E  i,  and  integrated  twice  to  find  y. 

With  regard  to  the  following  important  practical  problem  :  —  When 
the  sizes  of  an  angle  iron  are  given  to  find  the  least  radius  of  gyration  of 
its  section  about  a  line  through  its  centre  of  gravity,  I  give  students  a 
number  of  angle  irons  every  year  ;  for  each  of  them  they  find  the  least 
radius  of  gyration.  From  the  tabulation  of  these  results  I  hope  to  get  an 
empirical  formula.  I  had  hoped  to  be  able  to  publish  this  now,  but  I 
have  not  yet  obtained  a  sufficient  number  of  results.  (See  Appendix.) 


410 


CHAPTER    XVIII. 

SOME   WELL-KNOWN   EULES    ABOUT    BEAMS. 

340.  WHEN  beams  have  the  same  section  everywhere  we 
look  for  the  place  where  the  bending  moment  is  greatest,  as 
that  is  the  place  where  fracture  tends  most  to  take  place,  and 
we  find  the  cross  section  to  withstand  this  greatest  bending 
moment.     We  shall  now  consider  such  a  uniform  beam  loaded 
and  supported  in  various  ways. 

Thus,  the  load  may  be  hung  from  one  end  of  the  beam,  the 
other  end  being  rigidly  fixed,  say  by  being  built  into  a  wall. 
When  we  say  that  the  end  of  a  beam  is  fixed,  we  mean  that  it 
is  rigidly  held  in  position,  whereas  when  we  say  that  a  beam  is 
supported  at  its  ends,  we  mean  that  it  is  merely  held  up  there. 
In  Table  VIII.  six  ways  are  shown  in  which  the  same  length  of 
beam  is  supposed  to  be  loaded.  The  total  load  is  supposed  to 
be  the  same  in  every  case,  and  the  length  from  A  to  B  is 
supposed  to  be  the  same.  Then,  we  see  that  when  the  beam  is 
fixed  at  both  ends,  and  the  load  spread  over  it,  it  is  twelve  times 
as  strong  as  when  one  end  is  fixed,  and  the  whole  load  hung  from 
the  other  end.  This  means  that  if,  with  the  beam  fixed  at  one 
end,  a  load  of  one  ton,  hung  at  the  other  end,  breaks  the  beam, 
then,  when  fixed  at  both  ends,  and  the  load  spread  uniformly 
over  it,  the  same  sized  beam  will  carry  12  tons.  Hence,  if 
experiments  are  made  on  the  strength  of  the  beam  when  loaded 
in  any  of  these  ways  we  know  what  its  strength  ought  to  be  when 
loaded  in  any  of  the  other  ways.  Now  a  great  many  experi- 
ments have  been  made  upon  beams  of  rectangular  section, 
supported  at  both  ends  and  loaded  in  the  middle,  the  third  case 
given  in  the  Table  ;  and  from  these  experiments  we  know  how 
to  find  the  load  which  such  a  beam  will  carry.  Having  found 
this,  we  know  that  when  loaded  and  supported  in  a  different  way, 
the  beam  will  carry  more  or  less  according  to  the  numbers  in 
the  column  headed  "Strength." 

341.  If   M   is   the  bending  moment   and   z   the   strength 
modulus  of  the  section,  and  f  the  stress  which  the  material 
will  stand,  M  =  z/.  .  .  .  (1). 

Let  us  take  as  an  example  beams  of  rectangular  section, 


APPLIED    MECHANICS. 


411 


breadth  b,  depth  d;  the  strength  modulus  is  b  c?2/(5>  so  that 

M  =fb  d*/Q (2). 

Now  our  theory  is  based  on  the  idea  of  perfect  elasticity  ; 
we  cannot,  therefore,  assume  that  M  is  the  bending  moment 
which  will  break  a  beam  if  y*is  the  ultimate  tensile  or  compres- 
sive  stress,  because  our  theory  cannot  hold  beyond  the  elastic 
limit,  but  we  find  by  experiment  that  the  breaking,  bending 
moment  is  proportional  to  b  6?.  Thus  if  rectangular  beams  of 
the  same  material,  but  of  different  lengths  (I  feet),  breadths  (b 
inches),  and  depth  (d  inches)  are  supported  at  the  ends  and 
loaded  in  the  middle,  we  find  that  the  breaking  load  w  Ib.  follows 
with  fair  accuracy  the  rule — w  =  c  b  dz/l  (3)  where  c  is  given  in 
the  following  table  and  stands  for  1/18  of  our  old/  It  is  well 
to  try  to  remember  that  doubling  the  breadth  of  a  beam  doubles 
its  strength,  but  doubling  its  depth  gives  four  times  the  strength. 

TABLE    VII. 
BEAMS  SUPPORTED  AT  THE  ENDS  AND  LOADED  IN  THE  MIDDLE. 


Nature  of  Material. 

c  to  Calculate  Strength. 

e  to  Calculate  Deflection. 

Teak 

820 

•00018 

Oak 

450  to  600 

•00044  to  -00020 

English  Oak 

557 

•0003 

Ash. 

675 

•00026 

Beech       .  -J.* 

518 

•00031 

Pitch  Pine 

544 

•00035 

Red  Pine 

450 

•00023 

Fir. 

370 

•0005  to  -0002 

Larch      .        f 

284 

•00041 

Deal 

COO 

•00023 

Elm          .   '7 

337 

•00061 

Cast  Iron 

2,540 

•000024 

Wrought  Iron 

3,470 

•000016 

Hammered  Steel 

6,400 

•000013 

Marble     . 

150 

Good  Sandstone 

50  to  80 

... 

The  numbers  given  in  this  table  are  merely  the  average 
values  found  by  various  experimenters.  You  may  wish,  how- 
ever, to  find  for  yourself  whether  they  are  correct  or  not.  You 
are  designing  a  beam  of  pitch-pine,  say ;  then  take  a  rod  of 
pitch-pine,  1  foot  long,  1  inch  broad,  1  inch  deep  ;  support  it  at 
the  ends,  and  load  in  the  middle  till  it  breaks :  the  Table  says 
that  the  load  will  be  544  Ibs.,  but  you  may  find  it  to  be  more 


412  APPLIED   MECHANICS. 

or  less  than  this.  Remember  also  that  it  is  near  the  middle 
that  your  beam  is  likely  to  break ;  this,  then,  ought  to  be  the 
soundest  and  most  evenly  grained  part  of  the  timber  if  possible, 
and  the  specimens  which  you  try  ought  to  be  as  nearly  as 
possible  the  same  kind  of  timber. 

If  your  beam  is  loaded  or  supported  in  any  of  the  other  five 
ways  described  in  Table  VIIL,  you  will  multiply  the  breaking 
load  which  you  have  found  by  the  number  called  strength  in 
Table  VIIL  The  reason  is  obvious. 

342.  In  certain  standard  cases  we  like  to  state  algebraically 
the  amount  of  bending  moment  at  any  section  of  a  beam.  We 
shall  do  this  in  the  six  standard  cases,  so  well  known  to  all  carpen- 
ters, shown  in  pages  413  to  415. 

I.  Beam  of  length  I  inches  fixed  at  one  end  A,  loaded  only 
with  w  at  the  other  end  B.     At  a  section  x  inches  from  B  the 
bending  moment  is  evidently  M  =  w  x.     The  shearing  force  is 
s  =  -  w.     The  diagram,  Case  I.,  shows  M.     Notice  that  M  is 
greatest  at  A,  and  is  then  w  I. 

II.  Beam  fixed  at  A  (Case  II. ,),  load  w  Ib.  per  inch  of  its 
length  or  total  load  w  spread  uniformly.     Note  that  the  load  on 
p  B,  if  P  is  a  section  x  inches  from  B,  would  be  w  x,  and  the 
resultant  of  this  acts  at  the  distance  J  x  from  the  section,  so 
that  its  moment  is  ^  x  x  w  x  or  ^  w  #2.     It  is  greatest  at  A,  being 
there  J  w  ft.     s  the  shearing  force  at  p  is  —  w  x.     Note  that 

w  I  =w,  so  that  the  bending  moment  anywhere  is  J—  as2  and  is 

J  w  I  at  the  end  A.  The  shearing  force  at  the  end  A  is 
numerically  greater  than  anywhere  else,  being  —  w. 

III.  Beam  of  length  I,  load  w  in  middle  supported  at  the 
ends.     The  supporting  forces  are  each  \  w.    At  a  section  x  inches 

from  either  end  the 
bending    moment    is 
—   J  w  cc,  being  -  j 
&^    wl    at   the    middle, 
and      the      shearing 
force  s  is  -  J  w  from 
A  to  the  middle,  when 
it   suddenly  changes 
Fte-  231.  and  becomes  +  J  w 

from  the  middle  to  B. 

IV.  Beam  of  length  I  supported  at  the  ends,  load  w  Ib.  per 
inch  of   its   length   or   total    load   w   spread    uniformly    (see 


"A 


APPLIED   MECHANICS  413 

Fig.  231).  The  supporting  forces  at  A  and  B  are  each  \  w 
or  \  w  1.  At  P  if  o  P  =  x,  the  bending  moment  is  the  moment 
of  the  supporting  force  about  p  minus  the  moment  of  the  load 
on  P  B,  or,  since  o  B  =  \  I  and  p  B  =  J  I  —  x, 


and  this  simplifies  to  M  =  -  -J-  w  (Z2  —  4  x2).  It  is  numerically 
greatest  at  the  middle  where  x  is  o,  being  there  —  ^  w  I2  or 
—  J  w  L  The  shearing  force  at  p  is  J  w  I  —  w  (  J  l—x),  or  w  x. 

The  student  ought  himself  to  draw  all  the  diagrams  of  M 
(bending  movement)  and  s  (shearing  force). 

The  diagrams  of  M  for  cases  V.  and  VI.  are  shown  in  the 
figures.  (Case  VI.  will  be  worked  out  in  Art.  360).  They  are 
simply  the  diagrams  of  III.  and  IV.  with  the  average  value  of 
M  subtracted  from  every  ordinate  —  that  is,  the  whole  diagrams 
are  lowered  by  this  amount.  The  diagrams  of  S  are  exactly 
the  same  whether  a  beam  is  merely  supported  at  the  ends  or  is 
fixed,  if  the  loading  is  symmetrical  (see  Art.  362)  —  that  is,  the 
fixing  does  not  alter  the  actual  supporting  forces  at  the  ends. 

We  have,  in  fact,  the  following  rule  for  finding  the  bending 
moment  diagram  for  a  uniform  beam  symmetrically  loaded, 
fixed  at  the  ends.  Find  the  diagram  of  bending  moment  as  if 
the  beam  were  merely  supported  at  the  ends  :  raise  it  by  a 
distance  equal  to  its  average  height.  We  now  have  the  diagram 
of  the  bending  moment  when  the  ends  of  the  beam  are  fixed. 

The  shearing  force  diagram  is  not  altered  by  fixing  the  ends. 

If  for  any  two  kinds  of  loading  we  have  the  diagrams  of  M 
and  s,  then  for  the  two  kinds  of  loading  applied  at  the  same 
time  we  simply  add  algebraically  the  ordinates  of  the  separate 
diagrams. 


414 


APPLIED   MECHANICS. 


6-3  b 


I' 


OQ 


APPLIED    MECHANICS. 


415 


| 

to 

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A 

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CD 

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CQ 

v               tfi'/ 

n 

x 

N                           /' 

1 

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o 

/ 

^ 

to 
a 

X 

*/ 

1 

1 

/ 

K 

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PQ 

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M 

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f| 

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< 

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M 
M 
h-t 

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Nature  of  Support  and 
Loading. 

Beam  supported  at  both 
ends. 
Length  I  =  A  B. 

T,nar1  -or  in  rnirlrllp 

' 

j  n 

; 

j  " 

5 

[« 

/)  ^>  ^    pj    d  H|M 

2  d  ^  -^  n 

J                 QQ 

'a 

Beam  supported  at  both 
ends. 
Length  1  =  A  B. 

1 
=3 

g 

Greatest  bending  mo- 
ment occurs  in  the 
middle,  and  is  -  |w  L 
Shearing  force  is  ^  w 
at  B,  -  i  w  at  A,  o  at 

;j,n« 

J 

3 

3- 

a 

416 


APPLIED   MECHANICa 


1 

's 


% 

PQ 
3 

H 


/s 


~      i 

s  £ 

u      * 


APPLIED    MECHANICS.  417 

343.  At  the  Imperial  College  of  Engineering,  in  Japan,  wo 
had   a   testing-machine   with    which    I    made   a   great   many 
experiments  with  my  students.     It  increased  the  load  on  a 
beam  at  a  uniform  rate,  and  registered  the  load  and  deflection 
of  the  beam  at  every  instant — that  is,  it  drew  a  curve,  each 
point  of  which  showed  the  deflection  and  the  load  which  pro- 
duced  it.      Mr.    George    Cawley,    instructor    in    mechanical 
engineering  at  the  college,  lithographed   a  number  of  these 
curves,  taken  by  himself ;  and  although  the  experiments  were 
made  on  Japanese  wood,  so  that  the  actual  amounts  of  load  and 
deflection  are  not  of  general  interest,   yet  the  shapes  of  the 
curves  are  so  interesting  as  to  be  worthy  of  publication.     With 
only  one  exception,  two  beams  were  broken  and  two  curves 
taken  for  each  kind  of  wood.     The  mean  of  these  two  curves 
has  been  given  in  Fig.  232  — that  is,  a  curve  lying  between  the 
two.     The  specimens  were  all  free  from  knots.     They  were  all 
28  inches  long  and  If  inch  square.     The  distance  ow  repre- 
sents one  ton,  and  the  distance  o  D  represents  a  deflection  of  2 
inches,  so  that  the  scale  of  the  diagram  is  known.     The  load 
was  in  each  case  added  to  at  a  uniform  rate,  beginning  with  o, 
and  the  rate  at  which  it  increased  was  one  ton  in  two  minutes, 
and  we  see  from  the  figure  that  practically  only  in  three  cases 
did  the  breaking  of  the  beam  take  more  than  two  minutes. 
The  end  of  each  curve  shows  where  the  specimen  broke ;  it  is 
easy  to  see  where  the  curve  ceases  to  be  a  straight  line — that 
is,  where  the  law,  "  Deflection  is  proportional  to  load,"  ceases 
to  be  true ;  and  this  point  is  therefore  the  elastic  limit.     In 
some  cases  the  load  corresponding  to  the  elastic  limit  is  less 
than    half    the   breaking    load,    and    in    some    cases    greater 
than   this,    but   usually    it   may   be   seen   that   it    is    about 
one-half. 

344.  What  about  beams  that  are  not  rectangular  in  section? 
Suppose  we  have  a  beam  of  the  same  section  everywhere,  whose 
strength  and  stiffness  we  know,  and  suppose  we  want  to  know 
the  strength  and  stiffness  of  another  beam  which  has  the  same 
form  of  section — that  is,  suppose  the  new  section  is  such  that 
all  the  old  lateral  dimensions  are  increased  in  a  certain  ratio — 
then  the  strength  and  stiffness  increase  in  this  ratio  ;  if  all  the 
old  vertical  dimensions  are  increased  in  a  certain  ratio,  then  the 
strength  increases  as  the  square  of  this  ratio,  and  the  stiffness 
increases  as  the  cube  of  this  ratio.     The  effect  of  change  of 
length  is  just  the  same  as  it  was  with  rectangular  beams,  and 


418 


APPLIED   MECHANICS. 


.  —Students  will  do  well  to  calculate  for  each  of  these  materials  (1) 
Young's  modulus,  (2)  c  and  e  of  Table  VII.,  and  insert  in  an  extra  page  in 
their  copy  of  this  book. 


APPLIED   MECHANICS.  419 

we  know  the  effect  produced  by  different  methods  of  supporting 
and  loading  the  beam  from  Table  VIII. 

From  Arts.  341  and  342  it  is  evident  that  the  load  which 
a  beam  will  carry  without  breaking  is  proportional  to  the 
strength  modulus  of  its  section  divided  by  the  length  of  the 
beam.  The  deflection  of  the  beam  is  proportional  to  the  load 
multiplied  by  the  cube  of  the  length,  divided  by  the  moment 
of  inertia  of  the  cross  section, 

345.  Beams  of  uniform  strength  are  those  in  which  the 
nature  of  the  loading  is  exactly  known,  and  every  section  is  made 
just  of  such  a  shape  and  size  as  to  be  equally  ready  to  break 
with  all  the  other  sections.  There  is  no  difficulty  in  making  z, 
the  strength  modulus,  exactly  proportional  to  M.  Thus  taking 
up  the  four  well-known  cases  already  described ;  let  us 
design  beams  of  rectangular  section  everywhere  of  breadth  b 
or  depth  d,  or  of  circular  section  of  diameter  D,  which  shall 
be  of  uniform  strength.  Note  that  for  a  rectangular  section 
z  or  bd2,  and  for  a  circular  section  z  a  D3.  In  the  case  of  the 
circular  section,  the  plan  and  elevation  of  the  beam  are  of  the 
same  shape. 

Case  1. — M  =  w#,  so  that  bd2  oc  x.  Keep  b  constant,  the 
elevation  showing  d  is  a  parabola.  Keep  d  constant,  the  plan 
showing  b  is  a  triangle.  D3  or  #,  so  that  plan  and  elevation  are 
what  is  sometimes  called  the  cubic  parabola ;  anyhow,  it  is  easy  to 
draw. 

Case  2. — M  oc  x2,  so  that  bd2  oc  x2.  Keep  b  constant,  the 
elevation  showing  d  is  a  triangle.  Keep  d  constant,  the  plan 
showing  b  is  a  parabola.  D3  cc  x2  ;  the  plan  and  elevation  are 
easily  drawn. 

Case  3. — From  the  middle  to  each  end,  this  beam  is  the  same  as 
the  beam  of  Case  1. 

Case  4.— M  a  (P  -  4  x2),  so  that  bd2  a  (P  -  4  x2).  Keep  b 
constant,  the  elevation  showing  d  is  an  ellipse.  Keep  d  con- 
stant, the  plan  showing  b  is  two  parabolas.  D3  oc  (j2  -  4  x2),  so 
that  the  plan  and  elevation  are  easily  drawn. 

We  cannot  treat  Cases  5  and  6  in  the  same  way,  because 
we  only  know  M  in  these  cases  on  the  assumption  that  the  beams 
are  of  the  same  section  everywhere  (see  Art.  362). 

EXERCISES. 

Rolled  girder  section,  or  two  equal  flanges  and  web,  like  A, 
Fig.  223.  In  Table  VI.  we  see  that  the  moment  of  inertia  is 

_  (b  -  6i)  dj  |  -f-  12, 


and  the  strength  modulus  z  is  i  divided  by  £  d,  if  d  is  the  depth 
over  all,  d\  the  depth  between  the  flanges,  b  breadth  of  either  flange, 
bi  the  thickness  of  web. 


420  APPLIED    MECHANICS. 

1.  Let  the  student  calculate  i  and  z  in  all  the  cases  of  the  following 
Table,  page  421. 

2.  Taking  /  =  20  tons  per  square  inch  for  iron,  and  30  tons  for  steel, 
find  fz  in  each  case.     This  is  the  greatest  bending  moment  in  inch-tons 
which  each  section  will  stand. 

3.  Show  that  a  beam  I  feet  long,  supported  at  the  ends  and  loaded  in 
the  middle,  will  break  when  the  load  in  tons  is  fz  -~  3 1,     Calculate  this 
for  I  =  10  feet  in  each  case  of  the  table. 

4.  Beams  10  feet  long.     For  one  ton  at  the  middle  the  deflection  in 

2  240  x  1203 
inches  is  D  =  •£%  .  ^ ^ ,  or  2*688  -5-  i.     Find  this  in  each  case. 

I  take  E  the  same  for  iron  and  steel — namely,  30  x  106  Ibs,  per  square 
inch. 

5.  An  iron  beam  of  the  rolled  girder  section  of  the  table,  18  inches 
deep,  25  feet  long,  fixed  at   the  ends.     What  is  the  breaking  load  if 
spread  uniformly  ? 

Ans.,  The  load  for  a  10-foot  beam  in  the  table  is  89  tons  ;  for  a  25-foot 

89 

beam,  supported  at  the  ends,  and  loaded  in  the  middle,  it  is  —  x  10,  or 

zo 

35-6  tons.     Table  VIII.  shows  that  fixing  at  the  ends  and  loading  all  over 
allows  us  three  times  as  much  breaking  load,  or  106-8  tons. 

6.  What  is  the  mid-deflection  of  the  beam  of  Exercise  5,  fixed  at  the 
ends,  when  the  load  spread  all  over  is  20  tons  ? 

Am.,  For  a  10-foot  beam  it  would  be  -00223  x  -125,  according  to  the 
"deflection"  column  of  Table  VIII.;  and  for  a  25-foot  beam  we  multiply 
by  253  -?•  103,  and  the  answer  is  '0044  inch. 

7.  Compare  the  strength  to  resist  bending  of  a  wrought-iron  I  section 
when  it  is  placed  like  this :    I,  and  like  this :  M.     The  flanges  of  the 
beam  are  each  6  inches  wide  and  1  inch  thick,  and  the  web  is  f  inch 
thick,  and  measures  8  inches  between  the  flanges.  Am.,  4-57  ;  1. 

8.  What  is  the  greatest  stress  in  a  bar  which  is  subject  to  a  bending 
moment  of  4,000  inch-pounds    (1)  if  the  section  is  a  circle  of  £  inch 
radius;  (2)  if  of  I  form,  2  inches  deep  and  1  inch  wide,  the  web  and 
flanges  each  being  f  inch  thick.  Am.,  5-4  tons;  3-1  tons. 

9.  The  dimensions  of  the  section  of  a  cast-iron  girder  are  the  follow- 
ing:—top  flange,  4  by  1£  inches;  bottom  flange,  12  by  If  inches;  web, 
16  by   1£  inches.      Determine  the  position  of  the  neutral  axis,   and 
calculate  the  moment  of  inertia  of  the  section.     Find,  also,  the  moment 
of  resistance,  the  greatest  permissible  tensile  and  compressive  stresses 
being  2£  and  7£  tons  per  square  inch  respectively.     If  the  girder  be  20 
feet  long,  and  is  supported  at  its  two  ends,  find  also  the  greatest  safe  load 
which  it  will  carry  when  uniformly  distributed  along  its  length. 

Ans.,  7|  inch  from  bottom ;  2,280'5  inch-units;  800  ton-inches ;  26|  tons. 

10.  Find  the  moment  of  resistance  to  bending  of  a  beam  of  wrought 
iron  which  has  a  section  like  that  of  the  third  figure  in  Table  VI.,  taking 
b  =  4  inches ;    depth,  5  inches ;   the  thickness  of  metal  everywhere,  £ 
inch ;   and  /  =  20  tons  per  square  inch.     What  is  the  greatest  load, 
placed  at  the  centre,  which  a  10-foot  beam  of  this  section  will  stand  when 
supported  at  both  ends?  Ans.,  60-45  inch-tons;  2  tons. 

11.  In  a  wrought-iron  girder,  supported  at  both  ends,  the  section  is 
that  shown  in  the  third  figure  of  Table  VI.     K  =  8  inches,  b  =  7  inches, 
d  =  10  inches,  D  =  12  inches.     The  length  is  24  feet.     Find  the  greatest 


APPLIED    MECHANICS. 


421 


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422  APPLIED    MECHANICS. 

uniformly  distributed  load  which  the  girder  will  safely  bear,  taking 
f  —  9,000  Ibs.  per  square  inch.  Calculate  the  deflection  at  the  middle  of 
the  beam.  Ans.,  12*29  tons;  0-447  inches. 

12.  A  steel  plate  girder,  70  feet  span,  7  feet  deep,  has  a  uniform  load 
of  1  ton  per  foot,  a  detached  load  of  10  tons  at  the  middle,  and  loads  of  5 
tons  each  at  15  feet  on  each  side  of  the  centre.     Find  the  diagram  of 
bending  moment,  and  state  its  amount  at  the  detached  loads.     Calculate 
the  best  areas  of  cross-section  of  the  booms  at  the  middle.     Take  6  tons 
per  square  inch  in  compression  and  7  tons  in  tension  as  safe  stresses. 

Ans.,  21-13  square  inches;  18-1  square  inches. 

13.  A  hollow  tube  of  wrought  iron  20  feet  long,  3  inches  outside  and 
2^  inches  inside  diameter.     Find  its  weight.     What  is  its  deflection  with 
its  own  weight  ?    What  further  weight  on  its  middle  will  it  carry  safely 
if/  =  4£  tons  per  square  inch  ?        Ans.,  145-2  Ibs.  ;  '44  inch  ;  158  Ibs. 

14.  The  supporting  forces  of  a  35-foot  beam  are  20  and  15  tons. 
There  is  a  load  uniformly  spread  of  10  tons,  and  one  detached  load. 
Find  the  detached  load  and  its  position,  and  find  the  bending  moment  at 
the  detached  load,  and  also  at  the  middle. 

Ans.,  25  tons  at  21  feet  from  one  end;  252  foot-tons;  2  18'7  foot-tons. 

15.  A  beam  70  feet  long,  with  weights  of  5,  2,  4,  6,  and  5  tons  at 
distances  10,  20,  30,  35,  and  60  feet  from  one  end.     Find  the  bending 
moment  at  each  of  these  places,  and  find  the  supporting  forces. 

Ans.,  117-1,  184-3,  231-4,  234,  and  102-9  foot-tons;  11-7  tons;  10'3 
tons. 

16.  If  we  take  it  that  /  really  means  some  kind  of  ultimate  stress 
which  ought  to  be  considered  in  the  formula  for  a  rectangular  section  : 

Greatest  bending  moment  =  /  .  —  ^-, 

6 

we  know  that  in  a  beam  of  length  I  inches,  the  breaking  load  at  the 
middle  being  w,  the  beam  being  supported  at  the  ends,  the  breaking 

bending  moment  must  be  wJ/4,  and  hence  we  have  w£/4  =/-g->  or 

w  =  |/  .  **!.    ln  the  beams  of  Table  VII.,  b  =»  1  inch,  d  =  1  inch,  I  =  12 

inches.     Calculate  /  for  each  of  the  materials  mentioned  in  Table  VII. 

17.  A  beam  12  inches  long,  1  inch  broad,  1  inch  deep,  has  a  deflection 
D  for  a  load  of  1  Ib.     The  values  of  D  are  given  in  Table  VII.  for  many 
materials.    They  are  there  called  e.     Calculate  the  Young's  modulus  in 

each  case.    Here  D  —  —      becomes 


wP  ^  1  x  12»  x  12  _  432 

48i*         48  x  1  x  e    *'~    e   ' 

Thus,  for  English  oak  e  =  -0003,  or  3  x  10~4,  and  hence  E  =  1,440,000. 
Divide,  therefore,  432  by  every  number  in  the  "deflection"  column  of 
Table  VII.  to  find  Young's  modulus. 

18.  A  bar  of  wrought  iron   3  inches  broad  and  \\  inch  thick  is 
supported  in  a  horizontal  position  at  two  points  2^  feet  apart.     What 
deflection  at  the  middle  will  be  caused  by  placing  there  a  load  of  15 
cwts.  ?  Ans.,  0-OG4  inch. 

19.  A  straight  bar  of  wrought  iron  1  inch  x  1  inch  in  section  is 


APPLIED    MECHANICS.  423 

loaded  as  a  tie  bar  with  5  tons.  It  is  found  that  the  portion  between  two 
points  on  it,  4  feet  apart,  elongates  -019  inch.  What  is  the  value  of  E? 
If  the  bar  be  subject  to  a  bending  moment  of  1,800  inch-pounds,  what 
would  be  the  radius  of  curvature?  Find  also  the  greatest  stress  and 
deflection  if  the  bar  be  supported  at  points  4  feet  apart  and  loaded  with 
120  Ibs.  at  the  middle.  Am.,  28'3  X  106 ;  109  ft. ;  8,640  Ibs. ;  0-0123  inch. 

20.  A  square  bar  f  inch  x  f  inch  is  subjected  to  a  bending  moment  of 
350  inch-pounds.     What  is  the  greatest  stress  in  the  bar,  and  the  radius 
of  the  circle  into  which  it  is  bent,  E  being  taken  at  2,000,000  Ibs.  per 
square  inch  ?  Arts.,  4,978  Ibs. ;  12-5  ft. 

21.  A  bar  of  deal  6  feet  long,  2  inches  broad,  and  3  inches   deep, 
supported  at  the  ends,  is  broken  by  a  weight  of  1,200  Ibs.  suspended  at 
the  centre.     What  uniformly  distributed  load  would  a  beam  of  the  same 
length  and  material  bear  if  the  depth  were  4  inches  and  breadth  3  inches  ? 

Ans.,  6,400  Ibs. 

22.  Sketch  the  diagrams  of  shearing  force  and  bending  moment  for  a 
uniform  beam,  supported  at  both  ends,  30  feet  long,  weighing  10  Ibs.  per 
foot  run,  and  carrying  a  load  of  1  ton  at  a  point  12  feet  from  one  end. 
Give  the  numerical  values  of  the  shearing  force  and  bending  moment  for 
the  middle  transverse  section  of  the  beam. 

Am.,  896  Ibs. ;  174,780  inch  Ibs. 

23.  A  beam  42  feet  span  supports  five  wheels  of  a  locomotive.     The 
fore  wheel  is  1  foot  from  the  left  end,  and  the  distances  between  the 
wheels,  in  order,  are  5,  8,  10,  and  7  feet,  and  the  loads  transmitted  to  the 
beam,  in  order,  are  5,  5,  11,  12,  and  9  tons.    Find  the  maximum  bending 
moment.  Ans.,  3,612  inch  tons. 

346.  In  designing  railway  girders  the  theory  of  bending  is 
found  to  lead  to  useful  rules ;  but  in  machine  design  some  judg- 
ment is  necessary  in  applying  rules,  and  want  of  judgment  is 
conspicuous  sometimes  in  the  writers  of  books  on  this  subject. 
What  we  would  urge  upon  students  is  the  necessity  for  a  great 
respect  being  shown  to  what  are  sometimes  called  "rule-of- 
thumb-proportions " :    great    respect,    tempered   by    criticism. 
When  all  the  people  of  an  old  trade  have  made  a  pedestal  or 
hanger  of  much  the  same  proportions,  we  must  remember  that 
these  proportions   have  been  reached  at  considerable  cost  in 
trials  and  failures. 

347.  The  Teeth  of  Wheels.— When  toothed  wheels  drive 
each  other,  their  teeth  tend  to  break  like  little  beams  fixed  at 
one  end.     It  is  usual  in  considering  their  strength  to  regard  the 
pressure  between  two  teeth  as  acting  at  a  corner,  because  this 
may  accidentally  occur,  and  it  is  the  most  trying  condition. 
There  are  usually  two  pairs  of  teeth  in  contact  at  once,  so  we 
consider  that  only  half  the  total  horse-power  has  ever  to  be 
transmitted  by  one  pair  of  teeth.     This  transmitted  horse-power, 
multiplied  by  33,000,  divided  by  the  circumferential  velocity  of 


424  APPLIED    MECHANICS. 

the  wheel  in  feet  per  minute,  is,  of  course,  the  force  in  pounds 
which  each  tooth  has  to  withstand  (see  Art.  41).  Imagine  the 
tooth  to  tend  to  break  at  a  section  a  L  c  d  b,  Fig.  233,  making  45° 
with  the  depth  a  H,  just  as  we  know 
it  would  break  if  the  corner  were 
struck  smartly  with  a  hammer.  This 
consideration  leads  to  the  rule  that 
the  square  of  the  pitch  is  propor- 
tional to  the  force,  divided  by  the 
greatest  safe  stress  per  square  inch 
to  which  the  material  may  be  sub- 
jected. If  teeth  are  so  carefully 
trimmed  up  that  we  imagine  the  load 
always  to  be  distributed  over  the 
Fig.  233.  whole  breadth,  it  is  easy  to  see  that 

the  pitch  ought  to  be  proportional 
to  the  force  per  inch  breadth  of  wheel.  Taking  the  first 
case,  it  is  easy  to  see  that  the  practical  rule  becomes 

p  =  C  \/HJV 

where  H  is  the  horse-power  transmitted  at  a  velocity  of  v  feet 
per  minute.  If  p  is  the  pitch  in  inches,  c  may  be  taken  as  7  for 
well-made  cast-iron  wheels  working  without  shock ;  c  is  9  for 
ordinary  mill  wheels ;  c  is  11  in  wheels  subjected  to  shocks ;  c 
is  also  usually  taken  as  11  in  mortise  wheels.  At  high  speeds 
c  is  usually  taken  greater  as  shocks  are  probably  greater. 

Exercise. — A  spur  wheel  making  100  revolutions  per  minute 
has  60  cogs  of  3  inch  pitch  (the  circumference  is  therefore  180 
inches,  or  15  feet,  so  that  v=  1,500  feet  per  minute);  what 
power  will  it  transmit  safely?  Here  vp^/c-  is  1,500  x  9  -r-c2,  so 
that  the  horse-power  is — 

275  in  the  most  favourable  case, 

167  in  mills, 

112  when  there  is  much  shock. 

When  teeth  are  well  trimmed  we  may  take  it  that  the  above 
numbers  fore  ought  to  be  diminished  by  25  percent,  if  the  breadth 
of  the  wheel  is  four  times  the  pitch.  In  very  broad  bevel  wheels 
the  above  velocity  is  to  be  taken  as  the  average  velocity.  In 
ordinary  or  narrow  bevel  wheels  it  is  the  velocity  of  the  inner 
parts  of  the  teeth. 


APPLIED    MECHANICS.  425 

348.  Similar  Structures  Similarly  Loaded. — If  a  girder  is 
loaded  mainly  by  its  own  weight,  then  any  other  girder  made  to 
the  same  drawings  but  on  a  different  scale  would  be  a  similar 
structure  similarly  loaded ;  and  this  is  the  name  given  to  all 
structures  made  from  the  same  drawings  but  to  different  scales, 
if  their  loads  are  in  the  same  proportions  to  the  weights  of  the 
structures  themselves.  It  will  be  found  that  in  all  such  cases  the 
stress  at  similar  places  is  proportional  to  the  size  of  the  structure 
—that  is,  the  weakness  of  the  structure  is  in  direct  proportion 
to  its  size. 

This  is  easily  seen  if  we  imagine  the  structure  to  be  such  a 
simple  one  as  a  rod,  A,  Fig.  234,  carrying  a  weighty  ball,  w.     If 
there  is  another  such  arrangement,  of  twice  the  size  in  every 
direction,  the  area  of  cross  section  of  the  rod  would 
be  four  times  as  great ;  but  the  load  to  be  carried 
would  be  eight  times  as  great,  and  therefore  the 
stress  per  square  inch  at  a  section  would  be  twice 
as  great— that  is,  the  larger  rod  and  ball  would  be 
twice  as  weak.      As  the  stress  would  be  twice  as 
great  and  the  length  of  the  rod  twice  as  great,  the 
extension  would  be  four  times  as  great.     The  exten- 
sion of  the  rod  per  foot  in  length  would  only  be 
twice  as  great.     In  the  same  way  a  beam  of  cast 
iron,  1  inch  square  and  1  foot  long,  is  1,700  times 
too  light  to  break  with  its  own  weight,  whereas  a 
beam  of  cast  iron  whose  length,  breadth,  and  depth 
are  in  the  same  proportion,  if  1,700  feet  long  and       Fis-  2S4* 
1,700  inches  square  in  section,  would  break  with 
its  own  weight.      The  deflection  of  similar   beams  similarly 
loaded  is  proportional  to  the  square  of  their  dimensions ;  but 
the  deflection  per  foot  of  length  is  only  proportional  to  their 
dimensions. 

Imagine  similar  beams  of  the  same  material  whose  similar 
dimensions  are  as  1  to  s  and  the  loads  are  as  1  to  s3 ;  evidently 
bending  moments  will  be  as  1  to  s4;  moments  of  inertia  of 
cross  sections  will  be  as  1  to  s4 ;  curvatures  at  similar  places 
will  be  equal ;  deflections  will  be  as  1  to  s2 ;  stresses  will  be  as 
1  to  s. 

If  we  dare  take  the  loading  of  similar  ships  as  1  to  sn  where 
n  is  something  between  2  and  3,  bending  moments  will  be  as 
1  to  sn  + l  •  curvatures  will  be  as  1  to  sn~3 ;  deflections  will  be 
as  1  to  a"'1 ;  stresses  will  be  as  1  to  su~^ 


426  APPLIED   MECHANICS. 

In  machines,  accelerating  forces  are  proportional  to  s  X 
masses  x  N2  if  N  is  the  number  of  revolutions  per  second  or  per 
minute.  Hence  in  similar  machines  the  forces  are  proportional 
to  s4  N2,  bending  moments  are  proportional  to  s5  N2,  curvatures 
to  s  N2,  deflections  to  N2  s3,  stresses  to  s2  N2.  If,  therefore,  the 
stresses  due  to  mere  accelerations  in  machines  are  to  be  the 
same,  s  N  must  remain  constant — that  is,  if  the  size  of  a  machine 
is  doubled  its  speed  must  be  halved. 

348a.  Strength  of  Ships.— Let  A  c  represent  the  length  of  a 
ship.  Imagine  the  ship  divided  by  sections  at  equidistant 
points  in  A  c.  Let  the  ordinatea  of  the  curve  w  represent 
the  weights  of  these  portions,  and  the  ordinates  of  the  curve  B 
represent  the  buoyancy  or  weights  of  water  displaced  by  them. 
The  two  areas  must  be  equal,  and  their  centres  of  gravity  lie  on  the 
same  ordinate.  The  ship  is  "  water-borne  "  at  D  and  E.  We  take 
the  vertical  distances  between  A  B  D1  and  A  w  D1,  and  between  c  B  EI 
and  c  w  E1,  as  representing  the  downward  load  per  foot  or  per  inch 


on  a  beam  A  D,  and  on  c  E  and  the  vertical  distances  between  D1  B  B1 
and  D1  w  E1  as  upward  load  per  foot  or  per  inch  on  the  part  of  the 
beam  DB.  If  we  set  off  these  distances,  therefore,  as  ordinates 
from  a  line  A  c  (Fig.  234A),  we  have  the  curve  of  positive  and 
negative  loading  of  a  beam.  "We  can  use  them  to  obtain  the  shear- 
ing force  and  bending  moment  at  every  section  of  the  beam  or  ship. 
Even  when  lying  level,  it  is  evident  that  the  loads  on  a  ship, 
regarded  as  a  beam,  are  in  other  directions  besides  the  vertical. 
Shipbuilders  must  also  take  account  of  the  loading  due  to  the 
inertia  of  the  parts  of  the  ship  in  her  various  kinds  of  motion, 
When  these  loading  forces  are  known  it  is  not  difficult  to  calculate 
the  strength.  Besides  the  strength  as  a  whole,  it  is  important  to  see 
how  each  part  communicates  the  load  on  it  to  the  general  system. 
The  complete  problem  is  therefore  a  complicated  one,  which,  how- 
ever, can  all  be  worked  out  according  to  the  principles  given  in  this 
"book.  The  general  student  is  supposed  to  know  the  above  simple 
method  of  obtaining  loading  and,  therefore,  bending  moment  and 
shearing  force  at  each  section  of  the  ship  as  if  it  were  a  beam. 


427 


Fig.  235. 


CHAPTER    XIX. 

DIAGRAMS    OF    BENDING    MOMENT   AND   SHEARING    FORCE. 

349.  WHEN  a  beam  is  loaded  in  any  way  whatever,  it  is  easy  to 
obtain  by  a  graphical  method  the  diagram  of  bending  moment ; 
in  fact,  in  finding  tlio  supporting  forces  one  finds  the  diagram 
of  bending  moment.  Let  A  B  (Fig.  235)  represent  the  length  of 
a  beam  which  has  three  ver- 
tical loads— 1,  2,  3.  To  find 
the  vertical  supporting  forces 
at  A  and  B,  draw  the  unclosed 
force  polygon,  K  L  (Fig.  236) — 
before  the  student  arrives  at 
this  part  of  the  book  he  will 
probably  have  drawn  other 
force  polygons  where  all  the 
sides  were  really  in  the  same 
straight  line — 1,  2,  and  3  (Fig. 
236),  representing  in  direction 
and  magnitude  the  three  loads 
of  Fig.  235.  Choose  any  point 
0.  Join  o  K,  o  1  2,  o  2,  3,  and 
o  L  (o  2,  3,  means  the  line  join- 
ing o  with  the  point  where  the 
sides  2  and  3  meet).  Now  draw 
the  link  polygon  (Fig.  235), 
beginning  at  any  point  a,  in  the  vertical  from  A,  and  ending 
in  the  point  b.  Now  a  b  is  the  side  closing  the  link  polygon. 
Draw  o  N  (Fig.  236)  parallel  to  a  b  (Fig.  235).  Then  L  N  is 
the  amount  of  the  supporting  force  at  B,  and  N  K  is  the 
amount  of  the  supporting  force  at  A.  Also  draw  any  vertical 
line,  s  T  (Fig.  235).  Then  the  length  s  T,  intercepted  by  the 
sides  of  the  link  polygon,  represents  the  bending  moment  of  the 
beam  at  any  point  p  on  some  scale  which  it  is  easy  to  find. 

To  prove  this  :  Draw  o  H  horizontally.  The  moment  at  any 
point  P  due  to  the  supporting  force,  N  K  at  A,  is  N  K  x  A  P  ;  and 
this  is  equal  to  o  H  x  FT;  for,  by  similar  triangles, 


Fig.  236. 


and  therefore 


«T  :  TP  :  :  ON  :  NK, 
AP:TP::OH:NK. 


428  APPLIED    MECHANICS. 

This  second  proportion  gives  N  K  .  A  p  equal  to  o  H  .  T  F.  In  the 
same  way  the  moment  at  p  due  to  the  force  1  is  o  H  .  p  s  ;  and 
hence  the  true  moment  at  P,  being  the  difference  of  these,  is 
o  H  .  s  T.  Let  the  student  prove  as  an  exercise  that  if  the  beam  is 
drawn  to  a  scale  of  1  inch  represented  by  x  inches,  and  if  the  loads 
are  drawn  to  a  scale  of  1  pound  represented  by  y  inches,  then  s  T 
is  the  bending  moment  at  p,  on  a  scale  such  that  1  pound-inch  is 

represented  by  -—  inches,  o  H  being  measured  in  inches. 

O  H 

If  the  load  is  not  concentrated  at  a  number  of  points,  it  is 
usual  to  imagine  it  divided  into  a  number  of  loads,  each  of  which 
acts  at  one  point.  The  diagram  of  bending  moment  is  drawn  in 
the  way  which  I  have  just  described,  and  then  for  the  polygon 
with  its  straight  sides  we  substitute  a  curve  which  touches  all  the 
sides  of  the  polygon. 

After  you  have  found  a  diagram  of  bending  moment,  if  you 
wish  to  see  the  effect  of  additional  loads,  draw  a  diagram  for  these 
loads  as  if  they  acted  alone,  but  take  care  that  the  horizontal 
distance  o  H  is  the  same  as  before.  Add  together  the  ordinates  of 
your  two  diagrams  to  get  your  new  diagram  of  bending  moment 
for  all  the  loads. 

350.  It  is  very  unfortunate  that  this  subject  is  usually 
taken  up  from  an  academic  standpoint.  Students  discuss  much 
theory  sufficiently  well  for  examination  purposes,  but  they  do  not 
understand  the  most  elementary  things.  They  seldom  state  on 
a  drawing  what  is  the  actual  bending  moment  in  pound  or  ton- 
inches  at  any  section.  Nor,  indeed,  do  they  ever  seem  to 
use  these  drawings  for  practical  purposes.  Instead  of  one 
diagram  covering  a  large  sheet  of  paper,  we  have  the  usual 
diagram  looking  like  a  mere  book  illustration  in  size. 

A  student  will  find  that  unless  he  works  a  number  of 
exercises  graphically  he  cannot  comprehend  this  subject.  Let 
him  take  such  an  exercise  as  this.  A  beam  A  B  is  loaded  with 
5  tons  at  c,  7  tons  at  D,  2  tons  at  E  where  A  c  is  8  feet,  A  D  is 
13  feet,  A  E  is  18  feet,  A  B  is  24  feet.  It  is  also  loaded  with  1 
ton  per  foot  from  A  to  c,  0'5  tons  per  foot  from  c  to  D,  0-7  tons 
per  foot  from  D  to  E,  and  O'S  tons  per  foot  from  E  to  B.  It  is 
worth  while  making  a  diagram  for  the  distributed  load,  its 
ordinate  showing  the  amount  of  load  per  foot  or  per  inch. 

Take  also  a  case  with  no 
detached  loads  but  only  a 
diagram  showing  w  Ib.  per 
inch  as  shown  in  Fig.  237. 
We  divide  the  area  up  into  a 
convenient  number  of  .parts, 
Fig.  237.  c  F  K  A,  F  I  K  H,  &c.  Find  the 


APPLIED    MECHANICS.  429 

centre  of  gravity  of  u  L  K  and  let  the  arrow  there  represent 
the  area  I  J  L  K-^-that  is,  the  load  on  the  portion  K  L.  The 
student  ought  to  state  to  what  scale  his  area  represents 
load.  Thus  we  represent  the  whole  by  a  number  of  detached 
loads;  we  draw  the  diagram  of  bending  moment  which  is  a 
polygon,  but  it  is  to  be  noticed  that  lines  F  H,  I  K,  J  L,  &c.,  pro- 
duced, meet  the  polygon  at  points  which  are  on  the  true  bending 
moment  diagram  for  the  distributed  load,  and  so  the  curve  is 
easily  drawn  touching  the  polygon  at  these  points. 

351.  It  is  proved  in  Art.  339  that  if  D  is  the  greatest  distance 
moved  by  any  point  in  a  "beam,  and  this  is  called  the  beam's  deflec- 
tion, and  if  the  cross-section  is  the  same  everywhere,  w  the  load, 
L  the  length,  i  the  moment  of  inertia  of  the  section,  and  E  the 
modulus  of  elasticity, 

W  L3 

D  =  - —  for  a  beam  fixed  at  one  end  and  loaded  at  the  other. 

O  £  I 

D  =  -    -  —  for  a  beam  fixed  at  one  end  and  loaded  uniformly. 

_    1  w  L3  for  a  beam  supported  at  the   ends  and  loaded  in 

=  4S~    E~T  the  middle. 

5   w  L3  for  a  beam  supported   at  the  ends  and  loaded 

~  384  ¥7  uniformly. 

The  third  of  these  formulae  is  the  one  most  needed.  It  is  by 
means  of  this  formula  that  the  modulus  of  elasticity  is  generally 
determined.  Thus  in  careful  experiments  with  an  iron  beam 
1  inch  broad,  1  inch  deep,  carried  on  supports  24  inches  asunder, 
suppose  we  find  that  a  load  of  2,000  Ibs.  produces  a  deflection  of 

1      V      1       V      1       V       1 

one- quarter  of  an  inch.     Now,  i  for  the  beam  is   — 

1 

or  — .     The  third  formula  given  above  becomes 

^  2,000  x  24  x  24  x  24 
"16  3s  x  ^        ~~' 

and  from  this  we  find  that  E  is  27,648,000  Ibs.  per  square  inch. 

Again,  taking  the  third  of  the  cases  shown,  I  find  that  560  Ibs. 
produced  a  deflection  of  0-22  inch  in  a  beam  of  wood  24  inches 
long,  If  inch  square,  supported  at  the  ends.  Here 

i    =  1-75  x  1-75  x  1-75  x  1-75  ^  12,  or  -781, 
,  _    1   560  x  24  x  24  x  24 

~  16  3E  x  -78"!          ' 

from  which  we  find  that  E  is  938,656  Ibs.  per  square  inch. 

Again,  from  Table  VII.,  we  see  that  a  beam  of  teak  12  inches 
long,  1  inch  broad,  1  inch  deep,  gets  a  deflection  of  -00018  inch  for 
a  load  of  1  Ib.  Here  the  moment  of  inertia  of  the  cross-section 

is  1  and  -00018  =  I   1  X  ^  *x  ^  12,  from  which  we  find 
-    that  E  for  teak  is  2,400,000  Ibs.  per  square  inch. 


430 


APPLIED   MECHANICS. 


352.  Take  a  small  belfm,  A  u,  Fig.  238,  supported  at  the  ends, 
and  load  it  in  the  middle.  Measure  carefully  the  deflection  or 
lowering  of  the  middle  point.  This  is  called  the  deflection  of 
such  a  beam.  Now  this  distance  will  usually  be  small,  and  so 
we  had  better  magnify  it  by  letting  the  string  c  w  pass  over 
the  little  axle  E,  which  carries  a  long  pointer.  This  pointer  will 
show  on  the  scale  p  K  a  magnification  of  the  deflection.  We 


Fig.  238. 

shall  find  that  the  more  load  we  place  at  c  the  greater  is  the 
deflection ;  arid  in  fact  the  deflection  is  proportional  to  the  load, 
until  our  loads  become  great  enough  to  produce  permanent  set, 
when  (Art.  244)  the  deflections  increase  more  rapidly  than  the 
load.  If  now  we  use  a  beam  of  the  same  material  but  of  double 
the  breadth,  then  for  the  same  load  we  shall  get  one-half  the  old 
deflection.  If  we  use  a  beam  of  double  the  depth,  then  for  the 
same  load  we  shall  get  only  one-eighth  of  the  old  deflection. 
Also,  if  we  double  the  length  of  our  beam,  using  the  same  load, 
we  shall  get  eight  times  the  old  deflection.  A  very  instructive 
series  of  experiments  may  be  made  very  easily  in  this  subject, 
and  we  shall  not  thoroughly  understand  the  matter  unless  we 
make  a  few  such  experiments.  It  is  found  that  a  beam  of  pitch 


APPLIED   MECHANICS.  431 

pine,  1  foot  long,  1  inch  broad,  and  1  inch  deep,  supported  at 
its  two  ends  and  loaded  in  the  middle,  is  deflected  -00035  inch 
by  a  load  of  1  Ib.  This  explains  the  numbers  given  in  Table  VII. 
It  is  found  that  if  the  same  beam  is  fixed  at  one  end  and  loaded 
at  the  other  (first  case  of  Table  VIII.),  the  deflection  is  16  times 
as  great,  whereas  if  the  beam  is  fixed  at  both  ends  and  the  load 
is  spread  uniformly  (last  case  of  Table  VIII.),  the  deflection  is 
only  -125,  or  one-eighth  as  great.  This  explains  the  "deflec- 
tion "  column  of  Table  VIII. 

The  rule,  then,  to  find  the  deflection  in  inches  of  any  beam 
loaded  in  any  of  the  ways  shown  in  Table  VIII.  is  this : — Multiply 
together  the  cube  of  the  length  in  feet,  the  total  load  in  pounds,  the 
number  called  deflection  in  Table  VIII.,  and  the  number  called 
deflection  in  Table  VII.,  and  divide  the  product  by  the  breadth  of 
the  beam  in  inches,  and  by  the  cube  of  the  depth  in  inches. 

Example. — A  beam  20  feet  long,  10  inches  broad,  15  inches 
deep,  of  pitch  pine,  fixed  at  one  end  and  having  spread  all  over 
it  a  total  load  of  4,000  Ibs. — what  is  its  deflection  ?  Here  the 
number  in  Table  VIII.  is  6,  and  in  Table  VII.  it  is  '00035  ;  hence 
we  have  20  x  20  x  20  x  4,000  x  6  x  -00035  divided  by  10,  and 
again  divided  by  15  times  15  times  15,  which  gives  as  answer  1-99 
inch.  The  end  of  the  beam  would  be  deflected  this  distance. 

A  beam  is  said  to  be  stiff  if  its  deflection  is  small,  and  we 
say  that  the  stiffness  of  a  beam  supported  and  loaded  in 
the  various  ways  shown  in  Table  VIII.  is  for  the  various  cases 

—-,  -,  1,  1*6,  4,  8.  In  fact,  a  beam  of  a  certain  length  carrying 
lo  b 

a  certain  load  is  128  times  stiffer  when  it  is  fixed  at  the  ends 
and  loaded  uniformly  than  when  it  is  fixed  at  one  end  and 
loaded  at  the  other  end. 

It  is  well  to  remember  that  when  we  double  the  breadth  of 
a  beam  we  double  its  strength  and  also  its  stiffness,  but  if  we 
double  its  depth  we  get  four  times  the  strength  and  eight  times 
the  stiffness.  Beams  required  to  be  very  stiff  ought  to  be  very 
deep.  Care  must  be  taken,  however,  that  they  are  laterally 
supported,  else  they  will  buckle.  If  you  double  the  length  of 
a  beam  you  get  half  the  strength,  but  you  only  get  one-eighth 
of  the  stiffness. 

353.  The  student  must  see  that  the  heights  of  the  points 
on  the  force  polygon ;  for  example,  the  heights  of  the  points 
between  1  and  2,  2  and  3,  &c.  (Fig.  236),  above  N,  give  him  the 
ordinates  of  his  shearing  force  diagram  so  that  he  can  obtain 


432 


APPLIED    MECHANICS. 


this  diagram  by  horizontal  projection.  When  the  loads  are 
distributed,  at  points  like  H,  K  and  L  of  Fig.  237,  the  ordinates 
are  the  same  as  if  the  loads  were  detached,  and  they  may  be 
joined  by  a  curve. 

We  shall  now  consider  a  number  of  cases  where  arithmetic 
and  algebra  help  out  our  graphical  methods  of  working. 

354.  If  the  loads  on  a  structure  are  W1?  w2,  etc.,  the  stresses  and 
strains  everywhere  are  the  sum  of  those  that  would  be  produced  if  each 
load  acted  alone.  "We  find  this  true  in  all  cases  that  we  try,  unless, 
indeed,  in  certain  cases  where  instability  is  produced.  It  is  usually 

supposed  to  be  a  mere 
statement  of  the  mathe- 
matical law  of  superposi- 
tion of  small  effects.  We 
know,  at  all  events,  that 
the  bending  moment  at 
any  part  of  a  beam  due 
to  loads  Wj,  w?,  etc. ,  act- 
ing together,  is  the  sum 
of  the  bending  moments 
due  to  each  acting  singly. 
Exercise. — In  Fig.  239 
loads  B.  =  10  tons  and  s  = 


Fig.  239. 


15  tons  act  at  the  ends  c 
and  D  ;  supporting  forces 
p  and  Q  act  at  A  and  B. 
Draw  the  diagrams  of 


c  A  =  5  feet,  A  B  =  15  feet,  B  D  =  3  feet, 
bending  moment  and  shearing  force. 

Answer. — c  E  F  D  c  is  the  diagram  of  bending  moment,  where 
A  E  represents  50  foot-tons  and  B  F  represents  45  foot-tons.  The 
supporting  forces,  which  are  easily  found  graphically  and  analytic- 
ally, are  P  =  10'33  tons,  Q  =  14-67  tons.  Hence  the  diagram  of 
shearing  force  is  DGHIJKLCD,  where  D  o  =  -  15,  BI=-  0-33, 
AK  =  +  10,  all  in  tons. 

Exercise. — In   Fig.  240  a  weightless  beam  rests   on  supports 


c           A 

p             Fig. 

240.               Q 

»      D 

at  A  and  B;    torques  m^  and  m^  are  applied  at  c  and  D.     Find 
the  bending  moment  and  shearing  force  everywhere. 

Evidently  P  and  a  are  equal  and  opposite  ;  P  x  A  B  =  mi  -  mv 
—  -   —  TP  ,    „ 

If  P   is  an   upward  force, 


or  P  = 


Q  = 


A  B  A  B 

0,  must  be  a  force  holding  the  beam  down  at  B.      Thus,   let 
mj  =  30  ton  -feet,  wa  =  17  ton-feet;   let  AB  be    15   feet.      Then 


APPLIED   MECHANICS. 


433 


p  =  13/15  ton,  Q  =  -  13/15  ton.  s  is  zero  from  D  to  B,  -  13/15 
ton  from  B  to  A  ;  and  again  zero  from  A  to  c.  The  bending 

moment  from  D  to  B  is 
constant  and  equal  to  w2 ; 
from  B  to  A  it  increases 
to  my  and  remains  con- 
stant from  A  to  c. 

CGHJKDcis  the  dia- 
gram of  bending  moment, 
where  c  G  =  A  H  represents 
mly  and  J  B  =  K.  D  =  w2. 
D  B  E  F  A  c  D  is  the  dia- 

tgram  of    shearing   force 
where    B  E  =  A  F  =    - 
-6-  *«.  13/15. 

Exercise. — Let  the  beam  of  Fig.  240  have  moments  m1  and  mz 
applied  at  its  ends,  as  shown  in  Fig.  241 ;  also  let  it  have  a  load  of 
Wj  Ibs.  (say  5  tons)  at  T  (say  that  AT  is  6  feet).  This  of  itself 
would  produce  bending  moment,  as  shown  at  ALB  of  Fig.  242, 
where  TL  is  -  18  ton-feet, 
and  would  produce  support- 
ing forces  P  =  3  tons,  o,  =  2 
tons.  The  shearing  force 
diagram  is  BXVTUNAB, 
where  B  x  =  T  v  =  2  tons, 
and  AN=TTJ=  -  3  tons. 
Now  imagine  the  beam 
loaded  with  1  ton  per  foot 
on  c  A  and  0  '5  tons  per  foot  Fig.  242. 

on  B  D.     If  c  D'  =  -a?,   the 

bending  moment  at  D'  is  1  x  a;  x  \  x  or  £  #2,  and  at  A  it  is  |  (c  A)a 
or  12 -5  ton-feet.  The  bending  moment  curve  CE  is  parabolic. 
Similarly  the  bending  moment  curve  D  F  is  parabolic,  the  moment 

being  £  x  0*5  x  32  or  2*25 
ton-feet  at  B,  shown  at  B  F. 
The  bending  moment  due 
to  the  loads  on  the  ends 
would  be  c  E  F  D  over  the 
whole  beam,  and  the  sup- 
porting forces  would  be 
p  =  5-6833,  Q,  =  0-8167. 
The  shearing  force  dia- 
gram is  DIJKAECD  (Fig.  245),  where  BI  =  -  1£  tons,  BJ  =  AK 
=  -  0-6833  ton,  AE  =  5  tons.  " 

Now  imagine  the  beam  to  have  a  uniformly  distributed  load 
of  J  ton  per  inch  be- 
tween A  and  B.  The  sup- 
porting forces  required 
by  this  are  p  =  2-5  and 
Q  =  2-5  tons.  ^  The  bend- 
ing moment  diagram  is  a 
parabolic  curve  A  E  B,  where  Fig.  244. 


Fig.  243. 


434 


APPLIED    MECHANICS. 


D  E  =  -  9'375  ton-feet.     The  shearing  force  diagram  is  B  o  D  H  A  B 
(Fig.  244),  where  BO  =  2-5  tons,  AH  =  -  2'5  tons. 

Let  the  student  work  <    ...  - 

out  all  these  diagrams 
and  add  all  the  ordinates 
together. 

Now  let  him  obtain 
the  same  result  by  one 
graphical  construction. 
The  continuous  loading 


Fig.  245. 


he  will  break  up  into  a  number  of  detached  loads,  and  the  following 
example  will  show  him  how  to  work  the  problem.  Let  the  beam 
K  L  N  P  be  loaded  and  supported  as  shown  in  Fig.  246.  Use  Bow's 
lettering. 

355.  In  Fig.  246a  the  loads  are  given,  half -barbs  being  used  for 


Fig.  246. 

the  arrow-heads ;  the  missing  corner  of  the  force 
polygon  ABCDEFOHI  is  i.  Choose  a  pole 
4  0  and  join  with 
ABC,  etc.  We  can- 
not yet  draw  0  i.  Now 


Fig  246a. 


start  at  K,  or  any  point  in  the  line  oi 
load  AB  (Fig.  246),  and  draw  a  line 
through  the  space  A  parallel  to  0  A  (Fig. 
246a)  to  meet  (produced)  the  force  A  i 
in  Q.  Also  draw  K  x  through  the  space 
B  parallel  to  0  B  to  meet  the  load-line 
B  c  in  x.  Draw  x  w  through  c  parallel 
to  0  c,  and  so  on  to  R.  Now  join  R  and 
Q.  This  is  parallel  to  the  missing  line 
0  i  (Fig.  24  6a),  which  may  now  be  drawn 
and  i  found.  HI  and  IA  (Fig.  246a) 
are  the  amounts  of  the  supporting 
forces.  The  diagram  QAXWVUTSRQ 
is  the  diagram  of  bending  moment, 


APPLIED    MECHANICS. 


435 


vertical  distances  such  as  VY  representing  the  bending  moment 
at  each  corresponding  section.  The  scale  of  measurement  may 
be  computed  by  Art.  349.  We  have  positive  bending  moment 


Fig.  247. 

(tending  to  make  the  beam  convex  upwards)  between  K  and  z 
and  between  M'  and  P,  and  we  have  negative  bending  moment 
between  z'  and  M'.  z'  and  M'  are  places  of  no  bending  moment, 
where  the  beam  has  no  curvature ;  they  are  called  points  of 
inflexion.  The  shear  force  diagram  1,  2,  3,  4,  5  presents  no 
difficulty. 

It  is  usual,  when  we  finish  the  work,  to  draw  the  diagram  of 
bending  moment  as  in  Fig.  247,  the  ordinates  being  measured,  not 
from  a  broken  line  as  in  Fig.  246,  but  from  a  horizontal  line. 

Travelling  Loads. 

356.  I.  Suppose  the  load  w  (Fig.  248),  to  travel  over  the  beam 
from  A  to  B.  When  w  is  at  any  point  c,  between  A  and  D,  the 


Fig.  248. 

shearing  force  at  D  (say  SD  )  is  positive,  and  equal  to  the  reaction  Q, 
This  positive  shear  at  D  increases  with  Q  as  the  load  approaches  D, 

and  in  the  limit,  when  w  is  very  near  to  D,  it  has  the  value  -x. 

L 

At  the  instant  the  load   passes  r»,  the  shearing  force  at  D 

•vy  -m 

diminishes  by  the  amount  w,  and  becomes      x  -  w,  or   -  ~  (L  -  x), 

L  L 

thus  becoming  negative,  and  equal  to  the  reaction  at  P  ;  and  as  the 
load  moves  on  towards   B,   the  negative    shear  at  D  diminishes 


436  APPLIED   MECHANICS. 

numerically.  We  thus  see  that  the  greatest  positive  shearing  force 
at  any  place  occurs  when  the  load  is  just  to  the  left  of  the  place, 
and  the  greatest  negative  shear  when  the  load  is  just  to  the  right. 

Expressed  algebraically: — Max.  +  SD  =  — x;  max.  -  gD=-— (L-X). 

L  L 

These  are  the  equations  to  straight  lines,  and  the  corresponding  diagrams 
are  set  out  in  Fig.  248  ;  that  for  maximum  positive  shear  is  the  triangle 
ABE,  and  that  for  maximum  negative  shear  the  triangle  B  A  r. 

Next  consider  the   maximum  bending  moment  at  D  (say  max.  MD). 


Fig.  249, 

As  the  load  advances  from  A  towards  D,  MR  increases,  since  its  value  is 
Q  (L  -  x),  and  o,  increases.  When  the  load  passes  D,  MI(  diminishes,  since 
its  value  is  now  p  x,  and  P  diminishes.  Therefore  the  maximum  value 
of  MD  occurs  when  the  load  is  at  D,  and  its  value  is  given  by  the  equation 


=  P  x  or  Q  (L  -  x)  = 


x. 


This  is  the  equation  to  a  parabola  passing  through  A  and  n,  axis 
vertical,  the  ordinate  of  the  vertex  (see  Fig.  248)  being  w  L/4. 

II.  Two  travelling  loads  wl  and  w2  at  a  fixed  distance  I  apart ; 
draw  the  diagrams  of  maximum  positive  and  negative  shearing 
force  and  maximum  bending  moment. 

As  the  front  of  the  first  load  Wj  approaches  D,  +  SD  increases, 


APPLIED    MECHANICS.  437 

since  its  value  =  Q  ;  when  Wj  passes  D,  s  D  undergoes  a  sudden 
diminution  by  the  amount  Wi  ;  and  as  Wj  moves  from  D  towards  B 
the  value  of  +  s  D  increases,  on  account  both  of  the  advance  of  Wi 
towards  B  and  the  approach  towards  D  of  the  other  load  w2.  This 
increase  goes  on  until  w2  crosses  D,  when  there  is  a  second  sudden 
drop  in  the  value  of  +  s  D,  followed  by  a  gradual  increase  until  W2 
arrives  at  B.  We  see  that  the  maximum  positive  or  negative  s"£ 
occurs  when  the  front  or  back  respectively  of  Wi  or  w2  is  at  n. 

To  draw  the  diagram  :  Draw  the  diagrams  of  s  for  the  front 
and  back  of  each  of  the  loads  Wj  and  w2  in  its  passage  across  the 
beam  ;  the  boundary  line  is  the  diagram  required.  The  diagram 
for  front  of  Wi  will  be  a  straight  line  through  A,  the  equation  of 

^ 

which  is   +  s  =  Q  =  Wj  -,  until  W2   comes  on,  when  the   slope 

Ij 

suddenly  alters,  the  s  at  front  of  wl  then  being 

+   8  =   Q=W1£+Wa  ^-'  =    (WX   +   W2)  ^   -   W2  l-. 

The  slope  now  is  —  --  2,  and  the  diagram  of  s  in  front  of  Wj 

IJ 

continues  a  straight  line  in  this  direction.  Next,  for  s  at  back  of  Wi. 
This  is  equal  to  s  at  front  -  Wi  for  all  positions,  and  the  diagram 
is  a  broken  line  parallel  to  the  one  just  drawn  at  a  distance  below 
it,  equal  to  wt.  Next,  for  s  at  front  of  w2.  The  s  between  Wi  and  ws 
is  the  same  at  all  points  ;  that  is,  when  w2  is  just  coming  on,  the 
s  in  front  of  it  is  the  same  as  the  s  at  back  of  Wj.  Through  K 
draw  the  horizontal  line  k  m,  then  in  is  the  starting-point  for  the  s 

diagram  in  front  of  w2.    The  initial  slope  is  —  -  —  2  ;  this  extends 

L 

to  a  point  distant  horizontally  I  from  B,  when  Wi  passes  off  the 
beam,  and  the  slope  diminishes  to  —  -  .  The  diagram  for  the  rear  of 

L 

w2  is  parallel  to  the  line  just  drawn  and  at  a  distance  below  it, 
equal  to  w2.  As  a  test  of  accuracy,  it  is  to  be  noted  that  this  broken 
line  should  end  at  B. 

Consider  now  the  maximum  bending  moment.  Let  wt  +  w2 
at  t  be  the  resultant  of  ^  at  a  and  w2  at  b.  Let  ta  =  a  and 

tb  =  j8.     Then  ^  =  —  2,  that  position  being  shown  in  Fig.  249 
P      wi 

in  which  the  loads  Wi  and  w2  are  one  on  each  side  of  D.  Let  the 
distance  of  line  of  load  Wj  from  D  be  x.  The  maximum  MD  evi- 
dently occurs  either  when  Wi  or  w2  is  at  D,  or  for  some  intermediate 
position.  Consider  an  intermediate  position  as  shown.  MD  = 

d  (L  -  x)  -  w,  x  =  (wi  +  w2)  -  —  ~  —  (L  -  x)  -  w^  =  (wx  -j-  w2) 


-  x)  +  x  { 


Wl  +  w2)  ^-13  -  Wl  }  .     Itwillbeseen  that 


so  long  as  the  expression  in  the  last  bracket  is  positive,  MD  is  a 
maximum  when  x  is  greatest  —  that  is,  when  x  =  I,  or,  that  is,  when 


438 


APPLIED    MECHANICS. 


load  wa  is  at  D.   And  J  (w,  +  w2)  L-—  -  w1\  is  positive  so  long  as 


is  > 


t?  the  expression  in 


the  last  "bracket  is  negative,  and  the  value  of  MDis  therefore  a  maxi 
mum  when  x  is  zero  —  that  is,  when  load  Wj  is  at  D. 

To  draw  the  diagram,  take  point  T  in  A  B  such  that  —  =  —  • 

B  T  ^^i 

Let  A  T  be  called  the  field  of  w2,  and  B  T  the  field  of  W],  ;  then  MD  is 
a  maximum  for  any  point  D  in  the  field  A  T  when  the  load  w2,  which 
governs  that  field,  is  directly  over  point  D,  and  for  any  point  of  the 
field  B  T  when  wx  is  at  the  point,  proAided  in  each  case  the  load  can 
completely  traverse  its  field  without  the  other  load  leaving  the  beam, 
which  condition  requires  I  to  be  not  greater  than  the  smaller  of 

the  two  values  -  -  —  L,  or  -  ^—  L. 

Wi  +  W2  Wi  +  W2 

Curve  for  field  B  T.     Putting  x  =  o  in  the  expression  for  MD,  we 

•vtr       ,  I    "W"o 

have  maximum  MD  =  —  -  -  -  (x  -  o)  (L  -  x),  which  is  the  equa- 

tion to  a  parabola  passing  through  B  and  a  point  N  distant  a  from  A; 
and  by  symmetry  the  curve  for  field  AT  is  a  parabola  passing 
through  A  and  a  point  K  distant  /3  from  B  ;  the  two  parabolas  will 


be  found  to  intersect  at  i  directly  under  T.     The  distance  x  of  the 
mid  point  of  B  N  from  A  is  a  H  --  ^-^  =  —  —  -.     Putting  this  value 

W    4-  W 

in  the  equation  to  the  curve,  we  have  the  ordinate  H  Y  =  —  -  — 

[L  4-  a  L  +  ol       w,  +  w, 

—2  --  a        L  -   -g—     = 


4L 


. 

(L  -  a)2  ;  the  correspond- 


ing  value  for  G  x,  or  depth  of  other  parabola  A  K,  is  o  x 

(L  -  £)2.    When  the  distance  I  is  greater  than  the  shorter  of 


+   W 


APPLIED    MECHANICS. 


439 


two  fields,  there  is  then  a  third  parabola  through  A  and  B  corre- 
sponding to  the  greater  of  the  two  loads  taken  alone  "by  method 
shown  in  Case  I. 

III.  A  load  as  of  a  travelling  train,  w  Ib.  per  unit  length,  comes 
upon  a  girder  A  R  from  the  left,  covering  it  from  end  to  end,  and 
then  leaving  it.  Show  that  the  greatest  positive  shearing  force  at 
a  section  D  occurs  when  the  front  of  the  train  reaches  n,  and  that 
the  greatest  negative  shearing  force  occurs  when  the  rear  of  the 
train  leaves  D.  Also  find  the  maximum  bending  moment  at  D. 

We  have  seen  that  any  load  produces  positive  s  if  it  is  to  the  left 
of  D,  and  negative  s  if  it  is  to  the  right  of  D.  Hence,  to  produce  the 
greatest  positive  s  at  D,  there  ought  to  be  no  load  to  the  right  of  D, 
and  for  greatest  negative  s  at  D  there  ought  to  be  no  load  to  the  left  of 
D,  so  the  proposition  is  proved.  When  the  train  covers  A  D,  s  at  D  =  Q, 

at*  -v2 

or  ^-  ....  (1),  being  shown  by  DO  in  Fig.  250.    When  the 

w 
train  covers  DB,  s  at  D  is  -  p,  or  -  ^—  (L  -  x)2  .  .  .  .  (2),  being 

2t  L 

shown  at  D  H  in  Fig.  250.  B  E  and  p  A  are  numerically  equal  to  half 
the  load  when  it  covers  the  span.  The  curves  AE  and  BF  are 
parabolic,  the  equations  of  which  are  given  in  (1)  and  (2)  respect- 
ively. The  student  ought  now  to  add  to  the  ordinates  of  Fig.  250. 
the  shearing  force  due  to  a  uniformly  distributed  constant  load, 


Fig.  251. 


and  write  out  in  words  actually  what  s  is  as  a  train  rolls  on,  covers, 
and  rolls  off  the  bridge. 

Next  consider  the  maximum  M  at  D.  Any  load  anywhere  on  a 
girder  increases  the  bending  moment  anywhere.  Hence  where 
load  due  to  a  travelling  train  comes  upon  a  girder,  covers  the  girder, 
and  leaves  it,  the  bending  moment  at  any  place  is  never  greater 


440  APPLIED   MECHANICS. 

than  what  it  is  when  the  whole  girder  is  covered.  Therefore,  the 
maximum  bending  moment  will  occur  when  the  girder  is  fully 
loaded.  A  curve  to  this  is  known  to  be  a  parabola  through  A  and 

B,  the  depth  of  vertex  being  -5-. 

o 

IV.  Travelling  load  of  uniform  intensity  w  Ib.  per  unit  length, 
of  length  I  less  than  the  span  L.  The  maximum  positive  s  occurs 
when  the  front  of  the  load  is  at  D,  and  maximum  negative  when  the 
rear  of  the  load  is  at  D  ;  so  that  while  the  load  is  only  partially  on 
the  beam  the  diagram  of  maximum  +  s  will  be  a  parabola  similar 
to  that  of  Case  III.,  the  equation  being  for  a  distance  I  from  A 

maximum  +  s  =  -^—  .  When  the  load  comes  wholly  on  the  beam,  as 
in  Fig.251,  the  diagram  alters ;  maximum  +  S  =  Q  =  —  (x-^Jor 

L    \          // 

The  equation  to  a  straight  line  slope  — ,  intersecting 

L 

A  B  at  a  distance  ^  from  A,  and  therefore  tangential  to  the  parabola. 

The  diagram  of  positive  s  is  most  readily  set  out  by  first  drawing  the 
line  E  c,  as  shown  in  figure,  then  drawing  the  parabola  to  touch  it 
at  E,  and  the  line  A  B  at  A.  A  similar  curve  set  out  downwards 
from  B  will  be  the  diagram  of  maximum  negative  s. 

Maximum  bending  moment  at  D.  It  is  easily  seen  that  the 
maximum  MD  occurs  either  when  the  front  a  or  the  rear  b  of  the 
load  is  at  D,  or  for  some  intermediate  position,  as  in  Fig.  251.  To 
find  value  of  M  D  for  an  intermediate  position  let  the  condition  of 
the  co-ordinates  be  as  in  Fig.  251. 

MD  =  Q(L-X)-^-=^- 

For  a  maximum,  -j-2  =  0  (x  being  constant), .-.  maximum  M  D  occurs 

.       wl  ,  N  .        L  -  x          x  B  D       a  D 

when  —  (L  -  x)  -  wx  =  0,  or  —    —  =  , ;  i.e.  when  —  =  -, — , 

L   v  x  I  -  x'  AD       bo' 

or  the  maximum  M  D  occurs  for  such  a  position  of  the  load  that  D 
divides  it  and  the  beam  into  segments,  the  ratio  of  which  are 
respectively  equal. 

Putting  this  value  of  x,  namely  x  =  I.  ^-^,  into  the  expres- 

L 

sion  for  MD,  we  have  maximum  MD 


=  llI-2ljX(l-X) 

equation  to  a  parabola  passing  through  A  and  B,  the  depth  of  the 
vertex  being 


441 


CHAPTER    XX. 

MORE    DIFFICULT    CASES    OF    BENDING    OF    BEAMS 

357.  WE  do  not  usually  trouble  ourselves  as  to  whether  we  call  the 
bending  moment  which  makes  a  beam  convex  upwards  positive  or 
negative,  representing  it  by  upward-drawn  or  by  downward-drawn 
ordinates.  There  is  the  same  sort  of  choice  in  regard  to  shearing 
force.  But  for  the  sake  of  having  plus  signs  in  the  following 
expressions  (1)  and  (2),  we  had  better  adhere  to  the  following 
definitions.  A  section  which  is  being  sheared  is  supposed  to  be  at 
the  positive  distance  x  to  the  right  of  the  zero  point.  Loads  are 
positive  forces  ;  supporting  forces  are  negative.  Positive  shearing 
force  s  means  that  the  material  to  the  left  of  a  section  acts  with 
do wn ward  force  on  the  material  to  the  right  of  the  section. 
Positive  bending  moment  M  causes  a  beam  to  be  convex  upwards ; 
positive  y  at  a  place  is  a  downward  displacement.  When  this  is 
the  case  we  know  that 


k 


In  the  same  way  we  can  show  that 

To  prove  (3)  and  (4),  consider  the  equilibrium  of  the  portion  oi 
beam  between  the  sections  A  B  and  CD.  At  A  B  there  is  the  bending 
moment  M  and  shearing  force  s,  and 
at  CD  there  are  M  +  SM  and  s  +  5s, 
and  0  0'  is  5  x.  The  forces  acting  on, 
this  portion  of  beam  are  shown  in 
Fig.  252.  The  load  being  w  per 
unit  length,  the  resultant  load  here 
is  w  .  fix.  Hence,  considering  the 
vertical  forces,  S  s  —  w  .  Sa;  or  d  s/dx 
—  w  .  .  .  .  (4).  Taking  moments 
about  0',  M  +  s .  Sa;  +  |  w  (8x)'2  =  M  + 
5  M  or  5  M/5*  =s  -{-  £  w .  8x ;  and  in 
the  limit,  as  dx  gets  smaller  and 

smaller,  —  =  s  .  .  .  .  (3). 

358.  We  saw  (Art.  349)  that  if  we 
have  a  diagram  of  w  we  can  find  easily, 
graphically,  the  diagram  of  M.  We 
now  see  that  if  the  value  of  M  at 
every  section  be  divided  by  the  value 


Fig.  252. 


of  E  i  there,  and  if  we  treat  this  diagram,  showing  M/E  i  every- 
where, exactly  as  we  treated  the  w  diagram,  we  obtain  y. 


442  APPLIED   MECHANICS. 

This  graphical  method  of  working  is  quick  and  accurate.  If  we 
only  possessed  an  accurate  mechanical  integrator,  such  that  when 
given  a  curve  showing  x  and  v,  v  being  a  function  of  x,  we  could 
at  once  draw  another  curve  whose  ordinate  for  a  particular  value 
of  Xi  represented  the  area  of  the  v  curve  up  to  that  place  from  some 
datum  value  of  a?,  we  could  easily  solve  more  difficult  problems.  I 
have  often  done  this  by  counting  squares  on  squared  paper ;  also  I 
have  worked  to  obtain  a  number  of  points  in  the  new  curve  by 
using  a  planimeter  a  number  of  times.  We  see  now  that  if  we 
know  w,  the  integral  of  w  shows  s,  the  integral  of  s  shows  M,  the 

M 

integral  of  —  shows  i,  and  the  integral  of  i  shows  y,  the  shape  of 

E  I 

the  beam.  When  we  integrate,  however,  we  must  settle  the  start- 
ing value  of  the  new  ordinate,  and  this  is  what  usually  gives 
trouble.  Thus  the  starting  value  of  s  (when  we  integrate  w)  is 
not  zero,  but  depends  upon  the  supporting  forces. 

The  student  must  see  very  clearly  that  the  change  in  t,  in  going 
from  one  value  of  x  to  another,  is  equal  to  the  area  of  the  M/-E  i 
curve  between  those  places. 

359.  I  have  found  the  arithmetical  method  of  Art.  214  very  satis- 
factory. Its  accuracy  depends,  of  course,  on  the  number  of  ordinates 
taken.  A  student  ought  to  test  for  himself  the  accuracy  of  the 
method  on  some  such  exercise  as  the  following,  of  which  he  knows 
the  answer. 

A  beam  of  rectangular  section  1|  inch  broad,  2  inches  deep,  and 
of  length  15  inches,  is  fixed  at  one  end  and  is  loaded  uniformly  with 
10  Ib.  per  inch.*  Its  E  is  25  x  108,  find  M  everywhere  and  y.  Here 
i  =  2-25  x  43  ^  12,  or  i  =  12.  a;  is  distance  in  inches  measured 
from  the  free  end.  Imagine  the  free  end  to  be  on  our  left  and  the 
rest  of  the  beam  on  our  right.  The  table  gives  the  whole  work, 
and  needs  almost  no  explanation. 

The  student  who  thinks  will  have  no  difficulty  in  working  out 
a  problem  of  this  kind.  To  find  such  a  number  as  M  for  x  =  9,  for 
example,  we  add  the  average  value  of  130  and  140  to  the  previous 
M,  and  so  get  855. 

We  know  that  i  is   0   when  x  =  15,    and   so   we  subtract 

3,752  x  10~  from  all  the  found  values  of  i  -\-  c.  In  the  same 
way  to  get  the  last  column  we  add  39,801  to  every  number  of  the 
previous  column. 

Let  us  compare  our  results  with  the  true  answers,  which  the 
student  can  easily  work  out  as  in  Art.  339  :  s  =  50  +  10  x,  M  =  50 


Thus  when  x  =  15,  M  =  1,875,  as  in  the  table.     When  x  =  0,  i  or 
dyldx  =  —  3,750  x  10"~  ,  whereas  the  table  gives  —3,752  x  10~~8. 


•  See  Appendix. 


APPLIED    MECHANICS. 


443 


8 


Again,  when  x  =  0,  y  —  39,844  x  10    ~>  whereas  the  table  gives 

39,801  x  108,  which  is  the  same  for  all  practical  purposes.  After 
having  worked  this  example  a  student  must  feel  confidence  in 
using  this  method  of  integration  which  gives  us  answers  so  readily. 


X 

w 

8 

M 

M 

is7 
xio' 

i  +  c 
x  10 

i 
xlO8 

y  +  c' 

x  10 

V 
x  10" 

0 

50 

0 

0 

-3,752 

0 

39,801 

10 

1 

GO 

55 

18 

9 

-3,743 

-  3,747 

36,054 

10 

2 

70 

120 

40 

38 

-3,714 

-7,476 

32,325 

10 

3 

80 

195 

65 

91 

-3,661 

-11,163 

28,638 

10 

4 

90 

280 

93 

170 

-3,582 

-  14,785 

25,016 

10 

5 

100 

375 

125 

279 

-3,473 

-18,312 

21,489 

10 

6 

110 

480 

160 

421 

-3,331 

-21,714 

18,087 

10 

7 

120 

595 

198 

600 

-3,152 

-24,956 

14,845 

10 

8 

130 

720 

240 

819 

-2,933 

-27,998 

11,803 

10 

9 

140 

855 

285 

1,082 

-2,670 

-30,800 

9,001 

10 

10 

150 

1,000 

333 

1,391 

-2,361 

-33,315 

6,486 

10 

11 

160 

1,155 

385 

1,750 

-  2,002 

-35,497 

3,304 

10 

12 

170 

1,320 

440 

2,162 

-1,590 

-37,293 

2,508 

10 

13 

180 

1,495 

498 

2,631 

-1,121 

-38,648 

1,153 

10 

14 

190 

1,680 

560 

3,160 

-592 

'-39,505 

296 

10 

15 

200 

1,875 

625 

3,753 

0 

-39,801 

0 

360.  Beams  Fixed  at  the  Ends. — If  the  loading  on  a  symmetrical 
beanrts  symmetrical  so  that  we  find,  graphically  or  any  other  way, 
the  diagram  of  bending  moment  m  as  if  it  were  supported  at  the 
ends,  we  know  that  equal  and  opposite  torques  are  needed  to  fix 
the  ends.  Thus,  if  A  B  is,  say,  a  beam  of  uniform  section  and  has 
a  loading  of  any  kind  whatsoever  which  will  produce  (the  beam 
being  only  supported  at  the  ends)  a  diagram  of  bending  moment  m 


444 


APPLIED    MECHANICS. 


such  as  is  shown  in  A  c  D  E  B  (Fig.  253),  and  if  equal  and  opposite 
torques  are  applied  to  fix  the  ends  such  as  alone  would  produce  the 
diagram  of  bending  moment  shown  to  scale  in  B  G  F  A,  then  the 

algebraic  sum  of  the  two  (for 
the  A  c  D  E  B  is  negative  and 
A  B  G  F  A  positive)  is  shown. 
in  A  F  c'  D'  E'  G  B  A,  being 
positive  from  A  to  c'  and 
from  E'  to  B,  so  that  in  these 
parts  the  beam  is  convex 
upwards;  and  being  nega- 
tive from  c'  to  E',  where 
the  beam  is  concave  upwards. 

Fig.  253.  We   want,    then,    to    know 

exactly  how  nmch    G  B  or 

AF  must  be  to  fix  the  ends  of  the  beam.  Now,  the  difference 
of  slope  of  beam  between  A  and  B  is  nothing  if  the  ends  are  both 
fixed,  and  therefore  the  total  area  of  the  bending  moment  diagram 
must  be  zero  (since  EI  is  constant,  we  say  M  instead  of  M/EI). 
Hence,  the  area  of  the  portion  c'  D'  E'  being  negative,  must  be 
numerically  equal  to  the  sum  of  the  areas  A  c'  F  -f  B  E'  °-  IQ  ^ac^> 
the  average  ordinate  of  the 
M  curve  must  be  zero,  and 
therefore  we  raise  the  tn 
curve  A  c  D  E  B  A  by  its 
average  height  to  get  the 
M  curve. 

Example    1. — Beam    of 

length  I,  with  load    w  in  Fig-  254. 

the  middle,    fixed    at    the 

ends.  The  diagram  of  m  is  A  D  B  A  (Fig.  254),  where  D  D"  re- 
presents 4  wZ ;  raise  it  therefore  by  the  amount  |  w£,  and  we  find 
A  F  c'  D'  E'  G  B  A.  Evidently  A  F  =  -  D"  D'  =  GB  =  |  w/.  The 
beam  is  convex  at  the  ends,  concave  in  the  middle,  equally  ready 
to  break  at  ends  and  middle,  and  the  points  of  inflexion  are  half- 
way between  ends  and  middle  (see  Art.  342) 

Example  2. — Beam  A  B  of 
length  I,  with  total  load  w 
spread  uniformly.  A  D  B 
shows  the  m  curve,  a  parabola, 
the  diagram  of  bending  mo- 
ment if  the  beam  were  merely 


supported  at  the  ends,   D  D" 
being  |wj.     The  average  or- 
t  dinate  of  A  D  B  A  is  f  D  D". 

Fig.  255.  Raising  the  diagram  by  this 

amount,    we    find    the    true 

diagram  AFC'  D'E'  G  B  A  of  M  for  the  beam  fixed  at  the  ends.  There 
is  ^  wJ  at  each  end  and  -_L  w;  at  the  middle  (see  Art.  342).  The 
points  of  inflexion  are  at  c'  and  E',  no  longer  exactly  half-way 
between  ends  and  middle. 

361.  Students  will  do  well  to  work  at  least  one  complicated 


APPLIED    MECHANICS. 


445 


example  of  a  uniform  beam  fixed  at  the  ends  with  symmetrical  loading. 
If  the  beam  varies  in  section,  but  is  symmetrical — that  is,  if  at  two 
points  equally  distant  from  the  two  ends  the  sections  are  the  same, 
and  if  the  loading  is  symmetrical,  first  obtain  the  m  curve  graphi- 
cally and  measure  m  at  a  number  of  equi-distant  points.  Thus,  for 
some  beam  of  20  feet  long,  let  us  suppose  that  the  values  of  m  as 
measured  are  given  in  the  following  table.  We  need  not  use  E,  as 
we  will  suppose  it  to  be  constant. 


Distance 

—  m 

Values  of 

from  end 
in  feet. 

in 
ton-feet. 

in  inches  to  tne 
4th  power. 

—  7H 

T 

T 

M. 

1 

15 

500 

•03 

•002 

17-85 

3 

25 

300 

•0833 

•00333 

7-85 

5 

35 

250 

•140 

•004 

-   2-15 

7 

40 

320 

•125 

•003125 

-  7-15 

9 

44 

360 

•122 

•002778 

-11-15 

11 

44 

360 

•122 

•002778 

-11-15 

13 

40 

320 

•125 

•003125 

-  7-15 

15 

35 

250 

•140 

•004 

-  2-15 

17 

25 

300 

•0833 

•00333 

+  7-85 

19 

15 

500 

•03 

•002 

+  17-85 

Total  

1-0006 

•03046 

Average 

0-10006 

•003046 

We  see  that  the  area  of  the  —  curve  is  to  be  zero,  and  if  ml  is 
the  unknown  bending  moment  applied  at  each  end  to  fix  it, 

M  =  m  +  ml ;  so  that  the  average  value  of  l  must  be  zero, 

or  the  average  value  of must  be  equal  to  MI  multiplied  by  the 

average  value  of  — .    Hence  mi  is  equal  to  the  average  value  of 

-  divided  by  the  average  value  of-. 

In  the  above  case  this  is  0-10006  -f-  0  003046,  so  that 
*»!  =  32-85  ton-feet;  and  M  =  32-85  +  m  (algebraical  sum)  is 
the  true  bending  moment  everywhere. 

362.  The  shearing  force  diagram  in  the  symmetrical  cases  is  the 
same  whether  a  beam  is  fixed  or  onlv  supported  at  the  ends ;  and 

as  -^-  is  not  altered  by  the  fixing,  the  shearing  force  and  the  deflec- 
tion due  to  shear  are  everywhere  the  same  in  the  beam.    (See  Art.  369.) 


446  APPLIED    MECHANICS. 

The  solution  just  given  is  applicable  to  a  Learn  of  which  the  I 
of  every  cross-section  is  settled  beforehand  in  any  arbitrary  manner, 
so  long  as  i  and  the  loading  are  symmetrical  on  the  two  sides  of  the 
middle.  Let  us  give  to  i  such  a  value  that  the  beam  shall  be  of 

uniform  strength  everywhere ;  that  is,  that  -  z  =/ (2),  where 

2  is  the  greatest  distance  of  any  point  in  the  section  from  the 
neutral  axis  on  the  compression  or  tension  side,  and  /  is  the 
constant  maximum  stress  in  compression  or  tension  to  which  the 
material  is  subjected  in  every  section.  Taking  z  =  %d,  where  d 

is  the  depth  of  the  beam,  (2)  becomes  *  4  =  +  2/ (3),  the 

-f  sign  being  taken  over  parts  of  the  beam  where  M  is  positive, 
the  —  sign  when  M  is  negative. 

As  the  area  of  the  —  curve  from  end  to  end  of  the  beam  is 
i 

M 

to  be  zero,  and  —  —  2f/d,  we  see  that  the  area  of  a  curve  show- 
ing everywhere  the  value  of  +  Ifd  ought  to  be  zero,  the  positive 
sign  being  taken  from  the  ends  of  the  beam  to  the  points  of 
inflexion,  and  the  negative  sign  being  taken  between  the  two 
points  of  inflexion.  We  see,  then,  that  to  satisfy  (4)  we  have 
only  to  solve  the  following  problem. 

In    the    figure,    EATUCGEis    a    diagram  whose   ordi- 

nates  represent  the  values  of  -  or  the  reciprocal  of  the  depth 

of  the  beam  which  may  be  arbitrarily  fixed,  care  being  taken, 

however,  that  d  is  the  same 

A c      at  points  which  are  at  the 

same  distance  from  the 
centre.  E  F  o  E  is  a  diagram 
of  the  values  of  what  the 
bending  moment  m  would 
be  if  the  beam  were  merely 
supported  at  its  ends.  We 
are  required  to  find  a  point 
p  such  that  the  area  of 
EPTA  =  area  of  POO'T, 
where  o  is  in  the  middle  of 

the  beam.  When  found,  this  point  P  is  a  point  of  inflexion,  and 
p  a  is  what  we  have  called  m^  That  is,  m  -  p  R  is  the  real 
negative  bending  moment  M  at  every  place,  or  the  diagram  E  F  G 
must  be  raised  vertically  till  R  is  at  p  to  obtain  the  diagram  of  M. 
Knowing  M  and  d,  it  is  easy  to  find  i  through  (3). 

It  is  evident  that  if  such  a  beam  of  uniform  strength  is  also  of 
uniform  depth,  the  points  of  inflexion  are  half-way  between  the 
middle  and  the  fixed  ends. 

363.  In  the  most  general  way  of  loading,  the  bending  moments 
required  at  the  ends  to  fix  them  are  different  from  one  another. 


APPLIED    MECHANICS. 


447 


Thus  in  Fig.  257  let  A  r  c  G  B  A  be  what  the  bending  moment  m 
would  be  if  the  beam  were  merely  supported  at  its  ends  ;  let  fixing 
moments  mi  =  A  H  and  w2  =  E  B  be  applied  at  the  ends,  producing 
of  themselves  (see  Art.  354)  a  bending  moment  diagram  shown  by 
A  H  E  B  A,  or,  if  A  D  is  a;,  then  R  D,  the  bending  moment  produced  by 


the  end  couples  is  m1  + 


#,  if  Hs  the  length  A  B.    Let  us 


call  this  wij  +  bx.     Hence  M  or  R  D  -  D  s  is 

M  =  w.  +  bx  -  m  .  .  .  .  (1), 


or        E  _ ?  =  -J 


Fig.  257. 

Now,  as  I  is  known,   let  AJiJ2J3J4BA  be  drawn  to  represent 
the  value  of  —  everywhere.     Let  it  be  integrated  ;  that  is,  let  such 

an  ordinate  as  D  K1  (call  it  Y)  represent  the  area  of  A  Jl  J2  D,  and  so 
obtain  A  KX  K2  K3  B  A,  and  let  Y!  be  the  value  of  this  when  x  =  A  B. 

Also  integrate  -,  and  say  that  the  ordinate  of  the  resulting  curve 


Fig.  258. 

A  LI  L2  L3  B  A  is  x,  and  let  xx  be  the  value  of  this  when  x  =  A  B. 
Integrate  also  the  curve  whose  ordinate  is  —  everywhere,  calling 

P 


448  APPLIED    MECHANICS. 

the  answer  /t,  and  /ij  its  value  for  x  =  A  B  ;   then  (2)  becomes, 

since  -~  is  the  same  at  both  ends, 
cLx 

0  =  mj  YJ  +  b  Xj  -  ^, (3). 

Again  integrating  Y,  x,  and  p  (but  it  is  only  necessary  to  get  the 
areas  over  the  whole  length  at  once),  calling  the  answers  YP  Xj, 
and  MJ,  we  see  that,  since  at  the  two  ends  y  is  0, 

0    =    MjYj    +    i*!    -    M,...       (4). 

The  two  unknowns,  m^  and  b,  can  now  be  found  from  (3)  and  (4).* 
We  give  in  Art.  365  an  example  completely  worked  out.  The 
column  headed  m  represents  the  bending  moment  in  ton-feet  due 
to  a  given  set  of  loads,  if  the  beam  were  merely  supported  at  its 
ends.  These  values  may  be  found,  of  course,  by  the  graphical 
method  of  Art.  349.  The  values  of  i  are  supposed  to  be  given  us, 
and  they  are  in  inches  to  the  fourth  power. 

384.  We  shall  now  consider  a  beam  fixed  at  one  end,  B,  and 
merely  supported  at  the  other,  A,  which  is  on  the  same  level  as  B.  If, 
as  before,  m  is  the  bending  moment  at  any  place,  D,  which  would 
exist  if  the  beam  were  supported  at  each  end,  and  if  wa  is  the 
fixing  couple,  the  true  bending  moment  is 

Take,  first,  a  simple  case,  a  uniform  beam  uniformly  loaded  with 
w  Ib.  per  inch.  It  is  easy  to  prove,  as  in  Art.  339,  that  m  = 
-  ^  wlx  +  \  wx2,  and 


Hence 


*  We  have  used  the  symbols  p.,  x,  Y,  A*b  *i»  TI,  M,  x,  Y,  MI,  x,,  Yh  fearing 
that  students  are  still  a  little  unfamiliar  with  the  symbols  of  the  calculus. 
Perhaps  it  would  have  been  better  to  put  the  investigation  in  its  proper 
form,  and  asked  the  student  to  make  himself  familiar  with  the  usual  symbol, 
instead  of  dragging  in  fresh  symbols. 

After  (3)  above,  write  as  follows  :— 


Again  integrating  between  limits, 


The  integrations  indicated  in  (4)  and  (5)  being  performed,  the  unknowns  m\ 
and  6  can  be  calculated  and  used  in  (1).  The  student  must  settle  for  himself 
which  is  the  better  course  to  take—  to  use  the  formidable-looking  but  really 
easily  understood  symbols  of  this  note,  or  to  introduce  the  letters  whose 
meaning  one  is  always  forgetting. 


APPLIED    MECHANICS.  449 

and  .c 


We  need  not  add  a 
constant  here,  because  y 
is  0  when  a;  is  0.  In 
3)  and  (4)  we  insert  the 
conditions  that 


/  =  o  when  *  =  I  )    ,_. 
^  }    (»)>  Fig.  269. 

y  =  o  when  x  =  I  J 

and  we  find 

«....  (6)' 


From  these  we  find  that  c  =  -^-g  wl3,  m%  =  |  w;/2,  so  that  the  true 
bending  moment  everywhere  is  given  by  (2),  and  the  shape  of  the 
beam  by  (4).  The  slope  of  the  beam  at  the  supported  place  is 
wls/4S  E  i. 

If  the  loading  is  of  any  kind  whatsoever,  and  if  the  section 
varies  in  any  way,  it  is  necessary  to  be  able  to  integrate  a  function 
when  given  as  the  ordinate  of  a  curve.  We  have,  as  before,  in  (1), 


Let  the  integral  of  -  be  called  a,  and  the  integral  of  -  be  called  /*, 

and  let  curves  be  drawn  showing  the  values  of  x  and  p.     Let  their 
values  when  x  —  I  be  Xj  and  /ij.     Then 


We  can  now  integrate  (9)  again,  and  obtain  Ey  ;  but  if  we  do 
not  actually  wish  to  find  the  shape  of  the  beam,  we  need  only  use 
the  two  ideas  —  first,  that  y  =  0  when  x  =  0,  and  this  shows  that  no 
new  constant  needs  to  be  added  ;  secondly,  that  y  =  0  when  x  =  /; 
and  hence,  if  the  areas  of  the  whole  x  and  /*  curves  from  x  —  0  to 
z  =  J  are  Xj  and  MM  we  have 

0  =     2x1-!-M1  +  ^.    ..(10). 


450  APPLIED    MECHANICS. 

Also,  using  in  (9)  the  statement  that  -~  is  zero  when  a?  =  I,  we 
have 

O  =  ^XI  +  /AI  +  ,....  (11). 

From  (10)  and  (11)  we  can  find  c  and  -3.     Using  the  value  of  m2 

so  found,  we  find  the  "bending  moment  everywhere  given  in  (1).* 

If  in  the  above  exercise  we  imagine  the  supported  end  at  A  to 
be  raised  a  distance  a  above  its  original  position,  or  that  B  has 
settled  downwards  by  this  amount,  find  (supposing  the  beam,  fixed 
at  one  end,  B,  to  be  unloaded)  what  upward  force  at  A  (call  it  p) 
will  cause  it  to  rise  through  the  distance  a.  We  have  only  to 
assume  that  there  is  an  additional  supporting  force  of  this  amount, 
and  that  the  bending  moment  due  to  it  acts  as  well  as  the  other  — 
in  fact,  instead  of  (1),  we  have  as  the  bending  moment 


365.  It  is  very  important  that  the  student  should  work  out  care- 
fully such  an  exercise  as  the  following.  A  beam  is  given  with  loads, 
and  we  know  m  and  i  at  every  place.  The  integrations  to  be 
performed  are  much  the  same  whether  it  is  a  case  of  a  beam  fixed 
at  the  ends  or  fixed  at  one  end  and  merely  supported  at  the  other, 
and  therefore  we  give  both.  The  two  results  are  stated  imme- 
diately after  the  table. 

Without  using  the  letters  /i,  x,  /tb  Xi,  etc.,  the  above  investigation  is 


Integrating  (9)  between  the  limits  0  and  I,  and  recollecting  that  y  is  zero  at 
both  limits, 

0  =  E 

Also  using  in  (9)  the  statement  that  -     is  zero  when  x  =  I,  we  have 


The  integrations  in  (10)  and  (11)  being  performed,  the  unknowns  *—  and  c 
can  be  calculated.  The  true  bending  moment  everywhere  is  what  we  started 
Witt,  Hi  4- 


APPLIED    MECHANICS 


451 


X 

r/t 

i 
+ 

1 

+ 

£ 

+ 

X 

T 

+ 

/; 

+ 

m 

M  = 

JV= 

M 

Beam 
Fixed  at 
Eiids. 

M 

Beam  Fixed 
at  one  End, 
Supported  at 
the  Other. 

•o 

r5 

8 

500 

•00200 

•00100 

•01600 

+  22-30 

-  7-13 

•00200 

•00100 

•01600 

1-5 

15 

450 

•00222 

•00333 

03330 

+  14-73 

-  12-38 

•00422 

•00433 

•04930 

2-5 

21 

400 

•00250 

•00625 

•05250 

+  8-16 

-  16-64 

•00672 

•01058 

•10180 

3-5 

25 

350 

•00286 

•01001 

•07150 

+  3-59 

-  18-90 

•00958 

•02059 

•1733 

4-5 

28 

320 

•00313 

•01404 

•08740 

+  -03 

-  20-15 

•01271 

•03467 

•2607 

5-5 

32 

300 

•00333 

•01826 

•1066 

-  4-54 

-  22-41 

•01604 

•05289 

•3673 

6-5 

34 

300 

•00333 

•02159 

•1132 

-  7-11 

-  22-66 

•01937 

•07448 

•4805 

7-5 

30 

320 

•00313 

•02340 

•1123 

-  9-67 

-  22-92 

•02250 

•09788 

•5928 

8-5 

37 

350 

•00286 

•02431 

•1058 

-  11-24 

-  22-18 

•02536 

•1222 

•6986 

9-5 

35 

400 

•00250 

•02375 

•0875 

-  9-81 

-  18-43 

•02786 

•1459 

•7861 

10-5 

32 

400 

•00250 

•02625 

•0800 

-  7-38 

-  13-69 

•03036 

•1722 

•8661 

11-5 

31 

400 

•00250 

•02875 

•0775 

-  6-95 

-  10-94 

•03286 

•2009 

•9436 

12-5 

30 

380 

•00263 

•03288 

•0789 

-  6-51 

-  8-20 

•03549 

•2338 

1-0225 

13-5 

28 

360 

•00278 

•03753 

•0774 

-  5-08 

-  6-46 

•03827 

•2713 

1-0999 

14-5 

20 

330 

•00303 

•04393 

•0788 

-  3-65 

-  -71 

•04130 

•3153 

1-7187 

15-5 

24 

300 

•00333 

•05161 

•0799 

-  2-22 

+  3-03 

•04463 

•3669 

1-2586 

16-5 

18 

330 

•00303 

•05000 

•0545 

+  3-22 

+  10-78 

•04766 

•4169 

1-3131 

17-5 

12 

380 

•00263 

•04602 

•0316 

+  8-65 

+  18-52 

•05029 

•4629 

1-3447 

18-5 

5 

400 

•00250 

•04625 

•0125 

+  15-08 

+  27-26 

•05279 

•5092 

1-3572 

195 

•1 

500 

•00200 

•03900 

•0080 

+  15-52 

+  30.01 

•05479 

•5482 

1-3652 

YI  = 

•5748 

Xl   = 

4-062 

Mi  = 

15-276 

452  APPLIED    MECHANICS. 

Beam  Fixed  at  the  Ends.  —  /*i  +  ml  YJ  +  p  Xj  =  0,  where 
the  bending  moment  where  x  =  0,  w2  where  g  =  i,  p  =  m2  ~ 


or  -  1-365  +  -0548  ml  +  -548  p  =  0  ;  also  -  MJ  +  ma  y2  +  p  xi  =  0, 

Hence,     Wj 
P  =   -  -5673,  w2  =  19-23.     Adding  p  £  to  m,  we  find  the  true 


or     -  15-276  +  -5748  ml  +  4-062  p  =  0.       Hence,     Wj  =  30-58, 


bending  moment  M  given  in  the  tenth  column  of  the  table.  It  will 
be  found  that  if  every  value  of  M  be  divided  by  the  corresponding 
value  of  i,  the  algebraic  sum  for  the  whole  beam  is  -  -0013 
instead  of  0—  a  very  close  approximation  to  accuracy.  The 
student  may  easily  proceed  to  find  the  shape  of  the  beam. 

Beam  Fixed  at  One  End.  —  Where  x  —  I  and  the  bending 
moment  is  w2,  supported  merely  at  the  other,  where  x  —  0, 

-  fr  +  PXX  +  e  =  0, 
where  c  is  E^  at  x  =  0,  or  -  1-365  +  -5482  p  +  e  =  0; 

-  MI  +  P  .  X!  +  el  =  0, 
or     -  15-276  +  4-062  p  +  20  c  -  0. 

Hence  p  =  ™2  =  1-744,  or  mz  =  34-88,  and  c  -  -4091.    Adding 

p  x  to  m  everywhere,  we  find  the  true  bending  moment  M  given  in 
the  eleventh  column  of  the  table.  The  student  may  easily  proceed 
to  find  the  shape  of  the  beam. 

366.  As  we  have  proved  (Art.  357),  since 

dy  -  i     di  _   v.     du  _        ds  _ 
Jx  ~  *»  Tx  ~  iTi'    Tx  ~  S'   to  -  "> 

we  have  a  succession  of  curves  which  may  be  obtained  Irom 
knowing  the  shape  of  the  beam  y  by  differentiation,  or  which  may 
be  obtained  frcm  knowing  w,  the  loading  of  the  beam,  by  integra- 
tion. Knowing  w,  there  is  an  easy  graphical  rule  for  finding  M  ; 

At 

knowing  —  ,  we  have  the   same  graphical  rule  for  finding  y. 

Some  rules  that  are  obviously  true  in  the  w  to  M  construction  and 
f 


need  DO  mathematical  proof  may  at  once  be  used  without  mathe- 


matical proof,  in  applying  the  analogous  rule  from  —  to  y.  Thus 
the  area  of  the  —  curve  between  the  ordinates  ^  and  #2  is  the 

increase  of  t  from  x^  to  #2,  and  tangents  to  the  curve  showing  the 
shape  of  the  beam  at  xl  and  #2  meet  at  a  point  which  is  vertically 
in  a  line  with  the  centre  of  gravity  of  the  portion  of  area  of 

M 

the  curve  in  question.     The  whole  area  of  the  —    curve  in  a 

EI 

span  H  j  is  equal  to  the  increase  in  -j-  from  one  end  of  the  span  to 
the  other,  and  the  tangents  to  the  beam  at  ite  ends  H  j  meet  in  a 


APPLIED   MECHANICS.  453 

point  P,  which  is  in  the  same  vertical  as  the  centre  of  gravity  of  tho 
whole  —  curve.  These  two  rules  may  be  taken  as  the  starting- 
point  for  a  treatment  of  the  most  difficult  problems  in  beams 
by  graphical  methods. 

If  the  vertical  from  this  centre  of  gravity  is  at  the  horizontal 
distance  HG  from  H  and  GJ  from  j,  then  P  is  higher  than  H 
by  the  amount  H  G  x  i  H,  the  symbol  i  H  being  used  to  mean 
the  slope  at  H;  j  is  higher  than  P  by  the  amount  GJ  x  i  at  j. 
Hence  j  is  higher  than  H  by  the  amount  HQ.»H  +  oj.ij 
a  relation  which  may  be  useful  when  conditions  as  to  the  relative 
heights  of  the  supports  are  given,  as  in  continuous  beam  problems. 

367.  Theorem  of  Three  Moments. — For  some  time  railway  engi- 
neers, instead  of  using  separate  girders  for  the  spans  of  a  bridge, 
fastened  together  contiguous  ends  to  prevent  their  tilting  up,  and 
so  made  use  of  what  are  called  continuous  girders.  It  is  easy  to 
show  that  if  we  can  be  absolutely  certain  of  the  positions  of 
the  points  of  support,  continuous  girders  are  much  cheaper  than 
separate  girders.  Unfortunately  a  comparatively  small  settlement 
of  one  of  the  supports  alters  completely  the  condition  of  things. 
In  many  other  parts  of  applied  mechanics  we  have  the  same 
difficulty  in  deciding  between  cheapness  with  some  uncertainty 
and  a  greater  expense  with  certainty.  Thus  there  is  much  greater 
uncertainty  as  to  the  nature  of  the  forces  acting  at  riveted  joints 
than  at  hinged  joints,  and  therefore  a  structure  with  hinged  joints 
is  preferred  to  the  other,  although,  if  we  could  be  absolutely 
Certain  of  our  conditions,  an  equally  strong  riveted  structure 
might  be  made  which  would  be  much  cheaper. 

Students  interested  in  the  theory  of  continuous  girders  will  do 
well  to  read  a  paper  published  in  the  "  Proceedings  of  the  Royal 
Society,"  cxcix.,  1879,  where  they  will  find  a  graphical  method  of 
solving  the  mosb  general  problems.  "We  shall  take  here,  as  a  good 
example  of  the  use  of  the  calculus,  a  uniform  girder  resting  on 
supports  at  the  same  level,  with  a  uniform  load  distribution 
on  each  span.  Let  ABC  be  the  centre  line  of  two  spans,  the 
girder  originally  straight,  supported  at  A,  B,  and  c.  The  distance 
from  A  to  B  is  li  and  from  B  to  c  is  £2,  and  there  are  any  kinds  of 
loading  in  the  two  spans.  Let  A,  B,  and  c  be  the  bending  moments 
at  A,  B,  and  c  respectively,  counted  positive  if  the  beam  is  convex 
upwards. 

At  the  section  at  P  at  the  distance  x  from  A  let  m  be  what  the 
bending  moment  would  have  been  if  the  girder  on  each  span  were 
quite  separate  from  the  rest.  We  have  already  seen  that  by 
introducing  couples  mz  and  ml  at  A  and  B  (tending  to  make  the 
beam  convex  upwards  at  A  and  B)  we  made  the  bending  moment  at 
p  really  become  what  is  given  in  Art.  363.  Our  w2  =  A,  m\  =  B, 
and  hence  the  bending  moment  at  p  is 


454  APPLIED    MECHANICS. 

where  m  would  be  the  bending  moment  if  the  beam  were  merely 
supported  at  the  ends ;  and  the  supporting  force  at  A  is  lessened  by 
the  amount 


Assume  E  i  constant  and  integrate  with  regard  to  *,  and  we  have 

• 

m  .  dx  +  A  x  +  A  x2  — -, —   +  <?,  =  E  i  .  -/•  .  .  .  .  (3). 

'•  /  *  f?f  *     ' 


Using  the  sign   I    \mdx.dxto  mean  the  integration  of  the 

curve  representing  I  m  .  dx,  we  have 
I   1  m  .  dx  .  dx  +  £  A*2  +  J  x*  ?-IL^  +  ClX  +  e  =  E  i  .  y  ____  (4). 

As  y  is  0  when  x  =  0  and  it  is  evident  that  I  I  m  .  dx  .  dx  =  C 
when  x  =  0,  e  is  0.  Again,  y  =  0  when  x  =  ^.  Using  the 
symbol  /*i  to  indicate  the  sum  I  \rn.dx.dx  over  the  whole  span, 

/*i  +  IA^  +  K2  (B  -  A)  +  ^  =  0  .  .  .  .  (5). 
From  (3)  let  us  calculate  the  value  of  E  i  ~  at  the  point  B,  and 
let  us  use  the  letter  a^  to  mean  the  area  of  the  m  curve  over  the 


Ji 
m  .  dx: 
0 


m  .  dx:  so  that  E  i  ^  at  B  is 
0  dx 

Oj   +   Af,    +    Ml(B    -    A)    +  Cl  ----  (6). 

But  at  any  point  Q  of  the  second  span,  if  we  had  let  B  Q  =  x,  wo 
should  have  had  the  same  equations  as  (1),  (3),  and  (4),  using  the 
letters  B  for  A  and  c  for  B  and  the  constant  c2.  Hence,  making 

this  change  in  (3),  and  finding  E  i  -~  at  the  point  B  where  x  —  0. 

we  have  (6)  equal  to  cs  or 

cz  -  c~=  a,  +  A  Jj  +  K  (B  -  A)  ----  (7), 
and  instead  of  (5)  we  have 

F2  +  |B^  +  J/22  (C  -  B)  +  cJ2  =  0  .  .  .  .  (8). 
Subtracting  (5)  from  (8),  after  dividing  by  ^  and  Ja>  we  have 

(C-B)  ....  (0) 


APPLIED    MECHANICS.  455 

The  equality  of  (7)  and  (9)  is 

A*!  +  2B  ft  +  Q  +  c  I,  =  6         -«i-....  (10), 


an  equation  connecting  A,  B,  and  c,  the  bending  moments  at  three 
consecutive  supports.  If  we  have  any  number  of  supports,  and  at 
the  end  ones  we  liave  the  bending  moments  0  because  the  girder  is 
merely  supported  there,  or  if  we  have  two  conditions  given  whioh 
will  enable  us  to  find  them  in  case  the  girder  is  fixed  or  partly 
fixed,  note  that  by  writing  down  (10)  for  every  three  consecutive 
supports  we  have  a  sufficient  number  of  equations  to  determine  all 
the  bending  moments  at  the  supports. 

Example.  —Let  the  loads  be  wl  and  w2  per  unit  length  over  two 
consecutive  spans  of  lengths  ^  and  /2.    Then 

m  =  -  |  wlx  +  J  wx2,    I  m  .  dx  —  -  \  wlx*  +  &  wx9, 
and     a,  =  -  *^  If,    \  \  m  .  dx  .  dx  =  -  T 


and  m  - 

Hence    ^  +  ^  -  ^  becomes  -  &  w.2l/  -        /,»  +  ^*Vi8» 

or  -  J*  W  +  «^i3), 

and  hence  the  theorem  becomes  in  this  case 

Af,    +    2  B  (/!    +    It)    +   C  /2   =  i  (W2?23   +   W^jS)  ----  (10). 

If  the  spans  are  similar  and  similarly  loaded,  then 
A  +  4B  +  c  =  |w»^....  (11). 
Case     1.  —  A  uniform    and  uniformly  loaded  beam    rests    on 


three  equidistant  supports.     Here  A  =  c  =  0  and  B  =  + 

m  =   -  \  w  (Ix  -  a:2),  and  hence  the  bending  moment  at  a  point 


p  distant  x  from  A  is  -  |  w  (Ix  -  #2)  +  0  -  -  |  wV.  The  sup- 
porting force  at  A  is  lessened  from  what  it  would  be  if  the  part  of 
the  beam  A  B  were  distinct  by  the  amount  shown  in  (2)  —  -  — 

or  \  wl.  It  would  have  been  \  wl,  so  now  it  is  really  f  wl  at  each 
of  the  end  supports,  and  as  the  total  load  is  2  wl,  there  remains 
^wl  for  the  middle  support.  (See  also  Example  on  next  page.) 

Case  2.  —  A  uniform  and  uniformly  loaded  beam  rests  on  four 
equidistant  supports,  and  the  bending  moments  at  these  supports 
are  A,  B,  c,  D.  Now,  A  =  D  =  0,  and  from  symmetry  B  =  c. 


Thus  (11)  gives  us  0  +  5B  =  |«^^  or  B  =  C  =  ^wP.  If 
the  span  A  B  had  been  distinct,  the  first  support  would  have  had 
the  load  \  wl  ;  it  now  has  \  wl  -  ^  wl  or  ^  wl.  The  supporting 
force  at  D  is  also  ^  wl.  The  other  two  supports  divide  between 
them  the  remainder  of  the  total  load,  which  is  altogether  3  wlt  and 

P* 


456 


APPLIED    MECHANICS. 


so  each  receives  ii  wl.     The  supporting  forces  are  then   ^  wl, 

iJfrf,  ii«;/,andT%«;Z. 

Exercise. — If  a  beam  ABC  has  any  kind  of  loading,  and  varies  in  section 
in  any  way,  and  is  supported  at  three  places,  A  and  c,  on  the  same  level,  B, 
on  a  level  b  inches  below  A  or  c,  first  find  the  diagram  of  bending 


Fig.  260. 

moment  and  the  deflection  y  everywhere  of  the  beara,  assuming  the 
support  B  not  to  exist.  Find,  in  particular,  y1}  the  deflection  at  the  point 
B.  Now  consider  a  new  problem — the  same  beam  supported  at  two  places, 
and  with  only  an  upward  force  at  B.  Find  what  the  force  B  must  be  to 
cause  a  deflection  upwards,  yt  -  b,  at  B,  and  what  the  upward  deflection  e 
everywhere  is.  This  force  B  is  evidently  the  supporting  force  at  B  in  the 
real  problem,  and  the  deflection  in  the  real  problem  everywhere  is  y  -  z. 

Example.— Uniform  loading  w  per  inch  on  A  B  and  on  B  c,  each 
of  length  I;  beam  of  uniform  section,  the  supports  all  on  same 
level.  (1)  If  prop  B  is  absent,  the  deflection  at  B  would  be  (Art. 

Call  this  b.    Each  of  the  supporting  forces  is 


tv  1.     (2)  Beam  of  length  21,  supported  at  the  ends.     If  an  upward 
force  B  produces  an  upward  deflection,  we  know  that  5  =  Jg-  —  (2  Z)8. 

EI 

and  B  =  |  wl — that  is, 


384 


EI 


when  a  uniform  and  uniformly  loaded  beam  is  on  three  equidistant 
supports,  the  supporting  forces  are  $wl,  ^wl,  and  $wl. 

Mr.  George  Wilson  (Proc.  Royal  Soc.,  Nov.,  1897)  describes  a 
method  of  soh  ing  the  most  general  problems  in  continuous  beams 
which  is  simpler  than  any  other.  Let  there  be  supports  at  ABODE. 
(1)  Imagine  no  supports  except  at  A  and  E,  and  find  the  deflections  at 
B  c  and  D.  Now  assume  only  an  upward  load  of  any  amount  at  B,  and 
find  upward  deflections  at  B  c  and  D.  Do  the  same  for  c  and  D. 
These  equations  enable  us  to  calculate  required  upward  loads  at  B  c 
and  D,  which  will  just  bring  these  points  to  their  proper  levels. 


457 
CHAPTER    XXI. 

BENDING  AND   CRUSHING. 

368.  Stress  over  a  Section. — When  any  portion  of  a  column 
or  beam  or  arch  on  one  side  of  a  section,  B  c,  is  acted  upon  by 
loads  and  supporting  forces,  we  can  generally  find  one  force, 
representing  the  resultant  of  the  stresses  at  the  section,  which 
will  balance  them  all.  If,  instead  of  a  force,  we  merely  get  a 
couple,  then  the  section  is  exposed  solely  to  bending  moment, 
and  we  know  now  how  to  find  the  effect  of  this.  If  the  force 
is  parallel  to  the  section,  then  we  know  that  the  section  is 
either  exposed  to  mere  shearing  strain  or  shearing  and  bend- 
ing, as  in  a  horizontal  beam  with  vertical  loads ;  but  if  the 
force  is  inclined  to  the  section,  there  will  usually  be  shearing 
and  bending,  and  besides  this  a  uniform  distribution  of  com- 
pression  or  extension  all  over  the  section.  In  practice  we 
generally  find  that  compression  and  bending  alone  have  to  be 
considered.  Thus,  if  BC  (Fig.  261)  is  the 
edge  view  of  the  section  of  a  structure, 
o  being  a  line  at  right  angles  to  the  paper 
through  its  centre  of  gravity,  and  if  P  is 
the  resultant  of  all  the  forces  supposed  in 
the  plane  of  the  paper  which  act  on  the 
structure  to  the  right  of  the  section,  let 
p  and  s  be  the  resolved  parts  of  P  normal 
to  and  tangential  to  the  section ;  then  s  is 
balanced  by  an  equal  and  opposite  shear-  * 
ing  force  which  must  be  exerted  by  the  Pigt  26i. 

material  to  the  left.  P  is  a  compressive 
load  which  is  spread  uniformly  over  the  section,  producing 
a  compressive  stress  P/A  if  A  is  its  area ;  but  besides  this  we 
have  in  the  section  the  varying  compressive  stress  on  the  B 
side  of  o  and  the  tensile  stress  on  the  c  side  of  o,  which  the 
bending  moment  P  .  o  D  produces.  In  fact,  the  compressive 
stress  at  any  place  which  is  at  the  distance  y  from  o  on  the 

.     P          P  .  OD 

compression  side  is  -  +  y  ....  (1). 

A  I 

The  student  ought  to  draw  a  slice  between  two  parallel 
cross-sections  like  B  c  near  one  another,  and  draw  the  change 


458  APPLIED    MECHANICS. 

of  shape,  first  making  it  uniformly  thinner  because  of  P/A,  and 
then  making  it  wedge-shaped.  We  have,  in  fact,  the  wedge- 
shape  c  e'f  d'  of  Fig.  207,  where  H'  j'  =  H  J,  and  in  addition 
we  have  to  imagine  e'f  moved  parallel  to  itself  in  the  direc- 
tion c  d'  ;  but  students  must  draw  this  for  themselves.  They 
will  see  that  the  result  may  be  —  compression  everywhere,  but 
much  more  at  B  than  at  c  ;  or  compression  everywhere  and 
just  no  stress  at  c;  or  compression  at  B,  tension  at  c,  and 
a  neutral  line  somewhere  between.  Most  people  have  the 
habit  of  calling  the  line  through  o  the  neutral  line  of  the 
section,  although  it  is  the  neutral  line  only  when  the  com- 
ponent P  is  zero. 

The  proof  of  the  above  statement  is  this  :  Assume  that  a  plane 
section  remains  plane,  it  follows,  as  it  did  in  Art.  319,  that  there  is 
a  neutral  line,  say  at  o',  at  right  angles  to  the  paper.  Let  any 
point  H  be  at  the  distance  z  from  o'  ;  the  compressive  stress  there 
is  cz  say,  where  c  is  some  constant.  The  force  on  a  small  area  a 
there,  is  cza.  Then  P  =  c  2  za  .  .  .  (2)  and  P  .  o  D  =  c  2  za  .  y  .  .  .  (3), 
because,  if  o  H  =  y,  za  .y  is  the  moment  of  za  about  o.  Now 
z  =  y  +  o  o',  so  that  equation  (2)  is 


=  e.oo'.A....  (4), 

because  2  ya  —  o  ;  and  P  .  o  D  =  c  2  y^a  +  c^oo'.y,a  =  ci....  (5) 
if  i  is  the  moment  of  inertia  of  the  area  about  o.  We  see  that 

cy  from  (5)  is  —  -  —  y  and  c  .  o  o'  from  (4)  is  -  ;  but  the  sum  of 

these  is  cz,  the  compressive  stress,  and  so  we  have  proved  (1). 

If  the  section  is  rectangular,  the  dimension  at  right  angles  to 
the  paper  being  1,  and  BC  being  d,  then  i  =  1  x  d*/12  and 
A  =  1  x  d.  The  compressive  stress  is  least  at  c  where  y  =  -  |  d, 

or  by  (1),  the  compressive  stress  at  c  is  -r  -  —  -^  —  ....  (6). 

As  o  D  gets  greater,  the  compressive  stress  at  c  becomes  less  until 
there  is  a  value  of  o  D  which  just  causes  c  to  have  no  stress  ; 
a  greater  value  of  o  D  than  this  would  create  tensile  stress  at  c. 
We  usually  take  it  that  in  a  masonry  joint  there  ought  to  be 
no  tensile  stress,  and  hence  in  a  masonry  joint  the  limiting  value 
of  o  D  is  given  by  putting  (6)  equal  to  zero  ;  that  is,  6  .  o  D  =  d. 

Hence  D  must  fall  within  the  middle  third  of  the  masonry 
joint,  B  c,  if  there  is  to  be  stability.  This  is  the  fundamental 
rule  in  the  design  of  arches  and  buttresses.  Another  condition 
of  stability  for  a  masonry  joint  is  that  s  shall  not  exceed  the 
f  rictional  resistance,  or  the  angle  J  D  L  must  be  less  than  the 
angle  of  repose  (see  Art.  96)  for  the  materials. 

The  above  rule  is  very  generally  useful  in  machine  design, 


APPLIED    MECHANICS. 


459 


but  we  need  not  give  many  examples.    Fig.  262  is  a  crane-hook. 
The  section  at  B  c  is  not  usually  elliptic ;  rather  like  an  ellipse, 
with  the  end  at  B  blunter  than  that 
at  c.    If  o  is  a  line  at  right  angles  to 
the  paper  through  the  centre  of  grav- 
ity of  the  section,  and  w  is  the  load, 
the  stress  at  any  place  is  that  due  to 
a  bending  moment  w  .  o  D,  together 
with  a  tensile  stress  due  to  w  being 
spread  uniformly  over  the  section. 

When  a  weight  w  hangs  from  a 
bracket  as  in  Fig.  263  the  strength 
at  any  section,  such  as  o,  is  merely 
calculated  from  the  bending  moment 
W .  D  B,  because  when  the  distance 
D  B  is  con- 
siderable the 
stresses  due 
to  the  bend- 
ing are  usu- 
ally much 
greater  than 
those  due  to 

w,  divided  by  the  area  of  cross-section. 
369.  Shear  Stress  in  Beams.— Let  the 
distance  measured  from  any  section  of  a 
beam,  say  at  o  (Fig.  264),  to  the  section 
at  A  be  x,  and  let  on  =  x  +  8x.  Let  the 
bending  moment  at  c'  A  c  be  M,  and  at 
D'  B  D  be  M  +  SM.  Let  A  c  be  the  com- 

pressive  side  of  c  c'.  Let  o  AB  (Fig.  264)  and  A  A  (Fig.  265)  repre- 
sent the  neutral  surface.  We  want  to  know  the  tangential  or  shear 
stress  fs  at  E  on  the  plane  c  A  c'.  Now  it  is  known  that  this  is  the  same 

F 


0- 


w 


Fig.  263. 


-B 

F 


Fig.  264. 


.  266. 


as  the  tangential  stress  in  the  direction  E  F  on  the  plane  E  r,  which 
is  at  right  angles  to  the  paper,  and  parallel  to  the  neutral  surface 
at  A  B.  Consider  the  equilibrium  of  the  piece  of  beam  E  c  D  p,  shown 


460  APPLIED    MECHANICS. 

in  Fig.  265  as  E  c  E,  and  also  shown  magnified  in  Fig.  266.  We 
have  indicated  only  the  forces  which  are  parallel  to  the  neutral 
surface,  or  at  right  angles  to  the  sections.  The  tctal  pushing  forces 
on  DF  are  greater  than  the  total  pushing  forces  on  CE,  the  tangential 
forces  on  E  F  making  up  for  the  difference.  "We  have  only  to  state 
this  mathematically,  and  we  have  solved  our  problem. 

At  a  place  like  H  in  the  plane  c  A  c',  at  a  distance  y  from  the 

M 

neutral  surface,  the  compressive  stress  is  known  to  be  %>  =  —  y  >  and 

if  b  is  the  breadth  of  the.  section  there,  shown  as  H  H  (Fig.  265),  the 
total  pushing  force  on  the  area  E  c  E  is 


fAC  Mf 

=          b^y.dy,  orp= 
JAB    r  'J 


AC 

by  .  dy  .  .  .  .  (I). 
AB 


Observe  that  if  b  varies,  we  must  know  it  as  a  function  of  y  before 
we  can  integrate  in  (1).  Suppose  we  call  this  total  pushing  force 
on  E  c  by  the  name  P,  then  the  total  pushing  force  on  D  F  will  be 

p  +  Sx  .  -j-.  The  tangential  force  on  E  p  is  /f  x  area  of  E  F,  01 
f9  .  Sx  .  E  E,  and  hence 


Example.  —  Beam  of  uniform  rectangular  section,  of  constant 
breadth  b  and  constant  depth  d.     Then 


_ 
'' 


and  hence 


so  that  f$  is  known  as  soon  as  M  is  known.     As  -^  is  the  shearing 

force  over  the  whole  section,  we  may  regard  (3)  as  telling  us  what 
fraction  of  the  total  s  there  is  on  every  square  inch. 

As  to  M,  let  us  choose  a  case  —  say  the  case  of  a  beam  supported 
at  the  ends,  and  loaded  uniformly  with  w  Ib.  per  unit  length  of 
the  beam.  We  saw  that  in  this  case,  x  being  distance  from  the 


middle,  -  M  =  $wP  -  %wx2.     Hence        =  wxt  so  that  (3)  is 

^  =  ^r(^-AE3)^....(4). 

If  we  like,  we  may  now  use  the  letter  y  for  the  distance  A  E  ;  and 


APPLIED   MECHANICS.  461 

we  see  that  at  any  point  of  this  beam  x  inches,  measured 
horizontally  from  the  middle,  and  y  inches  from  the  neutral  axis, 
the  shear  stress  is 

/.-jptt*-*1)-.. ..(«)• 

Observe  that  where  y  =  0  the  shear  stress  is  greater  than 
at  any  other  point  of  the  section — that  is,  at  points  in  the 
neutral  axis.  The  shear  stress  is  zero  at  c.  Again,  the  end 
sections  of  the  beam  have  greatest  shear.  A  student  has  much 
food  for  thought  in  this  result  (5).  It  is  interesting  to  find  the 
directions  and  amounts  of  the  principal  stresses  at  every  point  of 
the  beam — that  is,  the  interfaces  at  right  angles  to  one  another  at 
any  point  across  which  there  is  only  compression  or  only  tension 
without  tangential  stress. 

We  have  been  considering  a  rectangular  section.  The  student 
ought  to  work  exercises  on  other  sections  as  soon  as  he  is  able  to 
integrate  by  with  regard  to  y  in  (1)  where  b  is  any  function  of  y. 

fAC 

He  will  notice  that  1       by  .  dy  is  equal  to  the  area  of  E  H  c  H  E 
JAE 

(Fig.  265)  multiplied  by  the  distance  of  its  centre  of  gravity  from 
A  A.  Taking  a  flanged  section,  the  student  will  find  that/s  is  small 
in  the  flanges,  and  gets  greater  in  the  web.  Even  in  a  rectangular 
section  fs  became  rapidly  smaller  further  out  from  the  neutral  axis ; 
but  now  to  obtain  it  we  must  divide  by  the  breadth  of  the  section, 
and  this  breadth  is  comparatively  so  great  in  the  flanges  that  there 
is  practically  no  shearing,  the  shear  being  confined  to  the  web  ; 
whereas  in  the  web  itself  /  does  not  vary  very  much.  The  student 
already  knows  that  it  is  our  usual  custom  to  calculate  the  areas  of 
the  flanges,  or  top  and  bottom  booms  of  a  girder,  as  if  they  merely 
resisted  compressive  and  tensile  forces,  and  the  web  or  the  diagonal 
bracing  as  if  it  merely  resisted  shearing.  He  will  note  that  the 

shear  in  a  section  is  great  only  where  -=- ,  or  rather  —  (-),  is 

great.     But,  inasmuch  as  in  Art.  357  we  saw  that  —  =  s,  the 

total  shearing  force  at  the  section,  there  is.  nothing  very  extra- 
ordinary in  finding  that  the  actual  shear  stress  anywhere  in  the 
section  depends  upon  its  value.  (See  Appendix.) 

Deflection  of  Beams  due  to  Shear.— If  a  bending  moment 
M  acts  at  a  section  of  a  beam,  the  part  of  length  $x  gets  the 
strain-energy  ^  M2  SX/E  i,  because  M  5#/E  i  is  the  angular  change, 
and  therefore  the  whole  strain-energy  in  a  beam  due  to  bending 
moment  is 


dx (6). 

If  /  is  a  shear  stress,  the  shear  strain-energy  per  unit  volume 


462  APPLIED   MECHANICS. 

is  /2/2N  ....  (7),  and  by  adding  we  can  therefore  find  its  total 
amount  for  the  whole  beam. 

By  equating  the  strain-energy  to  the  loads  multiplied  by  half 
the  displacements  produced  by  them  we  obtain  interesting  rela- 
tions. Thus  in  the  case  of  a  beam  of  length  I,  of  rectangular 
section,  fixed  at  one  end  and  loaded  at  the  other  with  a  load  w ; 
at  the  distance  x  from  the  end,  M  —  w#,  and  the  energy  due 
to  bending  is 

1      /^™.2~2 

....  (8). 


1    fZw^2 

21*      ~T'dx  = 
•J  n 


The  above  expression  (5)  gives  for  the  shearing  stress 


The  shear  strain-energy  in  the  elementary  volume  b.  8z.  Sy  is 
b.  Sx.  5y./2/2N.  Integrating  this  with  regard  to  y  from-^  to 
+  \d,  we  find  the  energy  in  the  slice  between  two  sections  to  be 
3  w2£.  5#/5  N  b  d,  so  that  the  shear  strain-energy  in  the  beam  is 
3  w2//5  N  b  d  ____  (10). 

If  now  the  load  w  produces  the  deflection  z  at  the  end  of  the 
beam,  the  work  done  is  £w*  .  .  .  .  (11).  Equating  (11)  to  the 
sum  of  (8)  and  (10)  we  find 

W  I*       6     w  I 


Note  that  the  first  part  of  this  due  to  bending  is  the  deflection 
as  calculated  in  Art.  339,  Example  1.  We  believe  that  the  other 
part  due  to  shearing  has  never  before  been  calculated. 

370.  Springs  which  Bend.  —  We  consider  the  bending  in  springs 
of  regular  shape,  such  as  spiral  springs,  later,  in  Art.  521.  But  it 

seems  natural  to 
consider  certain 
irregularly  shaped 
springs  here.  Let 
Fig.  267  show  the 
centre  line  of  a 
spring  fixed  at  A, 
loaded  at  B  with 
a  small  load  w 
in  the  direction 
Fig.  267.  shown.  To  find 

the     amount      of 

yielding  at  B,  the  load  and  the  deflection  are  supposed  to  be  very 
small.  Consider  the  piece  of  spring  bounded  by  cross-sections  at 
p  and  Q.  Let  p  Q  =  8s,  the  length  of  the  spring  between  B  and 
*  being  called  *. 


APPLIED   MECHANICS.  463 

The  bending  moment  at  P  is  w  .  P  R  or  w .  x,  if  x  is  the  length 
of  the  perpendicular  from  p  upon  the  direction  of  w.  Let  B  R  be 
called  y.  Consider  first  that  part  of  the  motion  of  B  which  is  due 
to  the  change  of  shape  of  QP  alone;  that  is,  imagine  AQ,  to  be 
perfectly  rigid  and  p  B  a  rigid  pointer.  The  section  at  Q  being 
fixed,  the  section  at  p  gets  an  angular  change  equal  to  8s  x  the 

change  of  curvature  there,  or  8s  —  or    S  '  W—  ....  (1),  where 

B  is  Young's  modulus  and  i  is  the  moment  of  inertia  of  the  cross- 
section.  The  motion  of  B  due  to  this  is  just  the  same  as  if  p  B 
were  a  straight  pointer ;  in  fact,  the  pointer  p  B  gets  this  angular 
motion,  and  the  motion  of  B  is  this  angle  multiplied  by  the  straight 

distance  p  B,  or          W3? .  p  B  .  .  ,  .  (2).     Now  how  much  of  B'S 

EI  pK 

motion  is  in  the  direction  of  w  ?    It  is  its  whole  motion   x  — 

PB 

or  x  — ,  and  hence  B'S  motion  in  the  direction  of  w  is  — .  . .  (3). 

PB  E  i 

Similarly  B'S  motion  at  right  angles  to  the  direction  of  w  is 

EI  *.; 

In  the  most  general  cases  it  is  easy  to  work  out  the  integrals  of 
(3)  and  (4)  numerically.  We  usually  divide  the  whole  length  of 
the  spring  from  B  to  A  into  a  large  number  of  equal  parts  so  as  to 
have  all  the  values  of  8s  the  same,  and  then  we  may  say  (*  being 

the  whole  length  of  the  spring)  that  we  have  to  multiply  —  ~ 
upon  the  average  values  of  —  and  —  for  each  part.  In  a  well- 
made  spring,  if  b  is  the  breadth  of  a  strip  at  right  angles  to  the 
paper  and  t  its  thickness,  so  that  i  =  ^  bfl,  we  usually  have  the 

spring  equally  ready  to  break  everywhere,  or  —^-  =  /,  a  constant. 
When  this  is  the  case  (3)  and  (4)  become  St_l  _ .  _  an(j 
~~ —  .  -.  And  if  the  strip  is  constant  in  thickness,  varying  in 

EC  _ 

breadth  in  proportion  to  x,  then  ~*~ —  .  x  is  (3)  and  -^ —  .  y 

E  t  E  t 

is  (4).  If  aTand  y  are  the  x  and  y  of  the  centre  of  gravity  of  the 
curve  (see  Art.  110),  -^—  is  the  total  yielding  parallel  to  w,  and 

this  is  what  we  generally  desire  to  know.  '  •    is  the  total 

yielding  at  right  angles  to  w. 

371.  Struts. — As  we  have  already  said,  when  a  very  short 
column  is  loaded  (as  it  usually  is)  in  such  a  way  that  the 
material  is  prevented  from  swelling  laterally,  it  will  withstand 
exceedingly  great  loads  without  fracture.  It  is  only  when  the 


464  APPLIED    MECHANICS. 

length  of  a  square  or  round  prism  is  three  or  four  times  its 
diameter  that  the  material  gets  a  chance  of  showing  how  it 
behaves  under  compressive  stress ;  up  to  a  length  of  about  ten 
times  its  diameter  its  breaking  load  is  now  nearly  the  same,  and 
it  remains  the  same  for  any  length  if  slight  lateral  restraints 
against  bending  are  provided.  In  a  tie-rod,  want  of  uniformity 
of  stress  causes  yielding,  and  this  is  the  same  in  a  strut ;  but 
in  a  tie  the  effect  of  yielding  is  to  remedy  the  defect,  the  tie 
gets  straighter,  the  stress  better  distributed,  whereas  in  a  strut 
local  failure  causes  bending,  instability,  and  a  worse  distribu- 
tion of  stress.  In  Art.  243  we  have  discussed  the  behaviour 
of  material  under  uniform  axial  compressive  stress  merely,  and 
also  in  columns  which  are  too  short  for  mere  axial  compressive 
stress.  We  now  know  that  if  B  c,  Fig.  261,  is  a  cross  section  of 
a  strut,  and  if  the  resultant  load  on  the  part  of  the  strut  on 
one  side  of  B  c  is  F  or  p,  not  only  have  we  the  usually  assumed 
stress  P/A  over  the  section,  but  also  the  bending 
moment  stress.  What  this  may  amount  to  de- 
pends upon  o  D ;  and  this  depends  on  two  things 
— first,  the  want  of  accuracy  in  applying  the  load 
at  the  ends  of  the  strut,  a  matter  which  cannot  be 
taken  into  account  in  calculation  unless  we  know 
how  much  error  there  is;  secondly,  the  bending 
of  the  strut. 

372.  Bending  of  Struts. — Consider  a  strut  per- 
O         fectly  prismatic,  of  homogeneous  material,  its  own 
weight  neglected,  the  resultant  force  F  at  each  end 
passing  through  the  centre  of  each  end.     Let  A  c  B, 
Fig.  268,  show  the  centre  line  of  the  bent  strut.  Let 
p  Q  =  y  be  the  deflection  at  p  where  o  Q  =  x.  Let  o  A 
=  o  B  =  I }  2/is  supposed  everywhere  small  in  com- 
\j  parison  with  the  length  21  of  the  strut.     Let  E  be 

AB         Young's  modulus  for  the  material,  and  I  the  least 
FJ  moment  of  inertia  of  the  cross-section  everywhere, 

•   r  2Cg         about  a  line  through  the  centre  of  gravity  of  the 
section.    Then  the  result  of  the  following  theory  is 
that  the  load  F,  which  will  produce  bending,  is  E  i  w2/4J3. 

p  y  is  the  "bending  moment  at  p  and  —  is  the  curvature  there, 

E  I 

Then  as  in  Art.  339  the  curvature  being  -  — |  we  have 

iy  -  _  fa.        %n 

B  I  "        dx*     '   '  '  '  ll* 


APPLIED    MECHANICS.  465 

Notice  that  when  we  choose  to  call  3-5-  the  curvature  of  a  curve, 

ax* 

if  tho  expression  to  which  we  put  it  equal  is  essentially  positive,  we 
must  give  such  a  sign  to  — ^  as  will  make  it  also  positive.  Now  if 
the  slope  of  the  curve  of  Fig.  268  be  studied  as  we  studied  the 
curve  of  Art.  338,  we  shall  find  that  j-f  is  negative  from  x  =  0  to 

X  =  o  A,  and  as  y  is  positive  so  that  —  is  positive,  we  must  use 
-  -~2  on  the  right-hand  side. 

If  the  student  tries  he  will  find  that 

y  =  a  cos.  x  \  /  —    .         .  (2) 


satisfies  (1)  whatever  value  a  may  have.  When  x  =  0  we  see  that 
y  =  a,  so  that  the  meaning  of  a  is  known  to  us  ;  it  is  the  deflection 
of  the  strut  in  the  middle.  The  student  is  instructed  to  follow 
carefully  the  next  step  in  our  argument.  When  x  ~  I,  y  =  0. 
Hence 


How  can   this  be  true?      Either  a  =  0,   or  the   cosine   is  0. 
Hence,  if  bending  occurs,  so  that  a  has  some  value,  the  cosine  must  be 

0.     Now  if  the  cosine  of  an  angle  is  0,  the  angle  must  be  -  or  — 

2         2 

or  — ,  etc.     If  we  confine  our  attention  to  •=,  the  condition  that 
bending  occurs  is 

TT  El  IT2. 


is  the  load  which  will  produce  bending.  This  is  called  Euler's  law 
of  strength.  It  is  easy  to  see  why  we  confine  our  attention  to  -^ 

as  it  gives  the  least  value  of  w-  The  meaning  of  the  other  cases  is 
that  y  is  assumed  to  be  0  one  or  more  times  between  x  =  0  and 
x  =  I,  so  that  the  strut  has  points  of  inflexion. 

It  will  be  found  that  the  complete  solution  of  any  such  equation 

as  (1),  which  may  be  written  ^  +  n2y  =  0,  is  y=  a  cos.  nx  +  b  sin.  nx 

where  a  and  b  are  arbitrary  constants,  a  and  b  are  chosen  to  suit 
the  particular  problem  which  is  being  solved.  In  the  present  case 
it  is  evident  that  as  y  =  0  when  x  =  I  and  also  when  x  —  -  \ 
0  =  a  cos.  nl  +  b  sin.  tdt  0  =  a  cos.  nl  -  b  sin.  nl,  so  that  b  is  0. 


466  APPLIED   MECHANICS. 

Now  let  us  consider  a  strut  fixed  at  the  ends.  The  resultant 
force  F  at  an  end  no  longer  acts  through  the  centre  of  gravity  of 
the  end  section.  The  solution  is  as  before 

y  =  a  cos.  nx  +  b  sin.  nx  .  .  .  .  (5). 
Differentiating  we  have 

dy 

—  =  -  na  sin.  nx  +  nb  cos.  nx  .  .  .  .  (6). 

dv 

Now  j-  =  0  for  x  =  0,  x  =  I,  x  =  -  I,  and  hence  0  =  nb,  so  that 

b  =  0  or  y  =  a  cos.  nx  .  .  .  .  (7). 
Also  0  =  -  na  sin.  nl  =  na  sin.  nl (8). 

Now  (3)  tells  us  that  a  is  the  deflection  at  c,  where  x  is  0,  and  if  a 
has  any  value,  (4)  tells  us  that  sin.  nl  i&  0 — that  is,  nl  —  v,  or 

(9). 


Hence  if  a  strut  is  fixed  at  both  ends  the  load  which  it  will 
stand  before  bending  is  the  same  as  for  a  strut  of  half  the  length 
hinged  at  the  ends.  In  fact,  if  G  c  H  is  a  strut  fixed  at  the  ends, 
its  strength  is  like  that  of  a  strut  K  L  hinged  at 
K  and  L,  or  again  like  that  of  a  strut  G  K  L  fixed  at  a 
and  hinged  at  L.  The  strength  to  resist  bending  of 
a  strut  of  length  2  I  hinged  at  the  ends  may  be  cal- 
culated from  (4) ;  this  is  the  strength  of  a  strut  of 
length  3  I,  hinged  at  one  end  and  fixed  at  the 
other,  or  of  a  strut  of  length  4 1  fixed  at  botli  ends. 
It  is  only  necessary,  then,  to  remember  the  rules  for 
struts  hinged  at  the  ends. 

The  load  given  by  (4)  will  produce  either  very 
little  or 'very  much  bending  equally  well,  and  we 
may  take  it  that  F  given  by  (4)  is  the  load  which 
will  break  a  strut  if  it  breaks  by  bending.     If  /  is 
the  compressive  stress  which  will  produce  rupture 
and  A  is  the  area  of  cross  section,  the  load  /A  will 
break  the  strut  by  direct  crushing,  and  we  must  take 
the  smaller  of   the  two  answers.      In  fact,  we  see 
Fig.  26&        that  fA  is  to  be  taken  for  short  struts  or  for  struts 
which  are  artificially  protected  from  bending,  and 
(4)  is  to  be  taken  for  long  struts.     Now,  even  when  great  care  is 
taken  we  find  that  struts  are  neither  quite  straight  nor  homo- 
geneous, nor  is  it  easy  to  load  them  in  the  specified  manner. 
Consequently,  when  loaded  they  deflect  with  even  small  loads, 


APPLIED    MECHANICS.  467 

and  they  break  with  loads  less  than  either  /"A  or  that  given  by 
(4).  Curiously  enough,  however,  when  struts  of  the  same 
section  but  of  different  lengths  are  tested,  their  breaking  loads 
follow,  with  a  rough  approximation  to  accuracy,  some  rule  as 
to  length.  Let  us  assume  that  as  P  =/A-  for  short  struts  and 
what  is  given  in  ^4)  for  long  struts,  then  the  formula 

P  -        /A      ....  (10) 


EITT* 

may  be  taken  to  be  true  for  struts  of  all  lengths  because  it  is 
true  both  for  short  and  for  long  ones.  For  if  I  is  great  we 
may  neglect  1  in  the  denominator  and  our  (10)  is  really  (4)  ; 
again,  when  I  is  small,  we  may  regard  the  denominator  as  only 
1,  and  so  we  have  w  =/A.  We  get  in  this  way  an  empirical 
formula  which  is  found  to  be  fairly  right  for  all  struts.  To  put 
it  in  its  usual  form,  let  I  =  A  &2,  k  being  the  least  radius  of 
gyration  of  the  section,  then 

F=   A     ....(ii), 


where  a  is  4//E  ir2. 

If  F  does  not  act  truly  at  the  centre  of  each  end,  but  at  the 
distance  h  from  it,  our  end  condition  is  that  y  =  h  when  x  =  I. 
This  will  be  found  to  explain  why  struts  not  perfectly  truly 
loaded  break  with  a  load  less  than  what  is  given  in  (4). 
Students  who  wish  to  pursue  the  subject  are  referred  to 
pages  464  and  513  of  the  Engineer  for  1886,  where  initial 
want  of  straightness  of  struts  is  also  taken  account  of. 

When  we  consider,  therefore,  how  the  rule  (11)  has 
been  arrived  at,  it  is  evident  that  it  needs  to  be  tested 
by  practical  experiment,  the  constants  «  for  various  materials 
and  kz  for  various  kinds  of  section  being  also  determined  by 
experiment.  This  has  been  done,  and  on  the  whole  we 
feel  fairly  well  satisfied  when  the  rule  is  put  in  the  following 
form  :  —  For  a  strut  whose  ends  are  hinged,  or  a  column  whose 
ends  are  not  fixed,  the  breaking  load  in  pounds  is  equal  to  the 
breaking  stress  per  square  inch  given  in  Table  IX.  multiplied 
by  the  area  of  cross  section  in  square  inches,  and  divided  by 
1  +  WB  where  n  is  given  in  Table  XI.,  and  B  is  given  in 
Table  X 


468 


APPLIED    MECHANICS. 


Exercise  for  Advanced  Students. — Test  in  the  six  cases  of 
Table  XL  to  what  extent  this  rule  agrees  with  (11). 

TABLE  IX. 


Breaking  Stress,  in  pounds 
per  square  inch. 

Cast  Iron                 .  .         ...         ...         ... 

80,000 

36,000 

Timber 

7,200 

TABLE  X. 

Value  of  B  for  struts  of  the  sections  shown  in  Table  XI.      The  first  column 
gives  the  length  of  the  strut  divided  by  its  least  lateral  dimension. 


Length  divided  by 
Lateral  Dimension  d. 

B  for 
Cast  Iron. 

B  for 
Wrought  Iron. 

B  for  Strong  Dry 
Timber. 

10 

0748 

0-132 

1-0 

15 

1-fiP 

0-300 

3-6 

20 

3-00 

0-532 

6-4 

25 

4-64 

0-832 

10-0 

30 

6-76 

1-200 

14-4 

35 

9-20 

1-632 

19-6 

40 

12-00 

2-132 

25-6 

45 

18-72 

3-332 

40-0 

EXERCISES. 

1.  Find  the  breaking  load  for  a  cylindrical  strut  of  wrought  iron 
3  inches  diameter  and  10  feet  long,  supposing  it  to  be  not  fixed  at  the 
ends.  Am.,  30-1  tons. 

2.  A  hollow  cast-iron  pillar  10  feet  long,  and  fixed  at  the  ends,  has  an 
external  diameter  of  6  inches;  what  should  be  the  thickness  of  the  metal 
to  carry  a  load  of  30  tons,  allowing  a  factor  of  safety  of  8  ?     Ans.,  -42  in. 

3.  "What  is  the  safe  load  for  an  angle  iron,  least  breadth  3  inches  and 
7 1  feet  long,  acting  as  a  strut,  firmly  fixed  at  both  ends  ?    Factor  of 
safety,  6.  Ans.,  3-51  tons. 

4.  The  diagonal  brace  of  a  Warren  girder  is  10  feet  long,  and  is 
composed  of  two  tee-bars   6*  x  3"  x  %",   placed  back    to    back    and 
riveted  together.     Find  the  maximum  compressive  working  load  which 
may  be  applied  to  it  when  the  ends  are  firmly  riveted  to  the  boom.     Use 
a  factor  of  safety  of  4.  Ans.,  17'5  tons. 

5.  Section  4,  Table  VI.,  has   flanges    4*02   inches   broad,    -56  inch 
thick,  depth  over  all  8  inches,  web  '42  inch  thick.     The  greatest  moment 
of  inertia  about  a  line  through  its  centre  of  gravity  is  74 '2,  according  to 
the  rule  of  the  table.     Its  least  moment  of  inertia  is  about  a  'ine  at  right 
.ngles  to  the  tot,  and  is   *L> L2?i3  x  2       «"88  x    -42 


12 


12 


APPLIED    MECHANICS. 


469 


The  area  of  the  section  is  7*39  square  inch.  The  ultimate  stress  being 
taken  as  45,000  Ibs.  per  square  inch,  and  the  Young's  modulus  as 
30  x  106  Ibs.  per  square  inch,  the  following  are  the  breaking  loads, 
according  to  Euler's  theory,  if  it  is  the  section  of  a  strut  fixed  at  the  ends 
of  the  following  lengths.  For  these  loads  to  be  withstood,  it  is  necessary 
to  carefully  adjust  the  application  of  load  at  the  ends  so  as  to  have 
absolutely  no  bending  until  the  breaking  loads  are  reached,  w  =  A/  or 
4  E  i 7T2/L2,  if  L  is  whole  length  of  strut  fixed  at  the  ends ;  the  lesser 
answer  of  the  two  to  be  taken. 


L  in  inches  ... 

96 

120 

144 

180 

216 

240 

288 

Breaking  load  in  tons 

148-5 

148-5 

148-5 

99-5 

69-1 

56 

38-9 

TABLE   XI. 

Values  of  n  for  struts  and  pillars  of  the  following  sfctions  : — 


Square  of  side  d,  or  rectangle  with  smallest  side  d    ... 


Hollow  rectangle,  or  square  with  thin  sides    ...         ... 


Circle,  diameter  d  

Thin  ring,  external  diameter  a  ... 


Angle  iron,  smallest  side  d 
Cruciform,  smallest  breadth  d 


2-00 


If  we  want  the  breaking  load  for  a  strut  whose  ends  are  not. 
hinged,  it  is  necessary  to  find  in  what  way  it  tends  to  bend,  and 
to  use  the  above  rule  regarding  the  strut  as  hinged  at  two 
points  of  contrary  flexure.  Thus  in  Fig.  269  the  strut  or  column 
E  is  as  strong  as  a  strut  hinged  or  rounded  at  both  ends,  whose 
length  is  only  K  L.  The  rule  becomes — For  a  strut  fixed  at  both 


470  APPLIED    MECHANICS. 

ends,  calculate  by  the  above  rule,  but  take  n  one  fourth  of  what 
I  have  given  in  Table  XI.  For  a  strut  one  end  of  which  is 
fixed  and  the  other  is  only  hinged  ;  calculate  the  breaking  load 
as  if  both  its  ends  were  hinged  ;  then  calculate  as  if  both  its 
ends  were  fixed,  and  take  the  mean  value  of  the  two  answers. 

373.  Struts  with  Lateral  Loads.  —  If  the  lateral  loads  are  such 
that  by  themselves,  and  the  necessary  lateral  supporting  forces,  they 
produce  a  bending  moment  which  we  shall  call  </>  (#),  then  (1)  Art. 

•372  becomes  py  +  <f>  (x)  —  -  EI^.     When  we  know  the  lateral 

loads  we  know  c/>  (x),  and  it  is  quite  easy  to  integrate.  Thus  let  a 
strut  be  uniformly  loaded  laterally,  as  by  centrifugal  force  or  its 
own  weight,  and  then  <f>  (x)  =  \w  (I2-  a?2)  if  w  is  the  lateral  load 
per  unit  length.  We  find  it  slightly  more  convenient  to  take 

4>  (x)  =  J  w  I  cos.  —  x,  where  w  is  the  total  lateral  load.  This  ia 
not  a  very  different  law.  Hence 


We  find  here  that 

cos.      *....  (2). 


Observe  that  when  F  =  0  this  gives  the  shape  of  the  beam.     The 
deflection  in  the  middle  is 


and  the  greatest  bending  moment  /x  is  /j.  =  F  yl  +  £  w  I,  or 


If  -w  =  0,  and  if  p  has  any  value  whatever,  the  denominator  of  (4) 
must  be  0.  Putting  it  equal  to  0,  we  have  Euler's  law  for  the 
strength  of  struts,  which  are  so  long  that  they  bend  before  break- 
ing. If  Euler's  value  of  F  be  called  u,  or  u  =  E  1  1^/4  Z2,  (4)  becomes 


If  zc  is  the  greatest  distance  of  a  point  in  the  section  from  the 
neutral  line  on  the  compressive  side,  or  if  i  -?-  zc  =  z,  the  least 
strength  modulus  of  the  section,  and  A  is  the  area  of  cross-  section, 
and  if  /  is  the  maximum  compressive  stress  to  which  any  part  of 
the  strut  is  subjected,  J*?—/  Using  this  expression,  if  ft 
stands  for  -  (that  is,  Euler's  breaking  load  per  square  inch  of 


APPLIED    MECHANICS.  471 

section),  and  if  w  stands  for  -  (the  true  breaking  load  per  inch  of 


This  formula  is  not  difficult  to  remember.     From  it  w  may  be 

found. 

EXERCISES. 

1.  Every  point  in  an  iron  or  steel  coupling  rod  of  length  2b  inches 
describes  a  radius  of  r  inches.  Its  section  is  rectangular,  d  inches  in  the 
plane  of  the  motion,  and  b  at  right  angles  to  this.  We  may  take 
w  =  Ibdrn2  -j-  62,940  in  pounds  where  n  =  number  of  revolutions  per 
minute.  Take  it  as  a  strut  hinged  at  both  ends  for  both  directions  in 
which  it  may  break.  (1)  For  bending  in  the  direction  in  which  there  is 

,.  .       ,  «  .    db^      _.  ,    ,         ,  E  db^-n2      „ 

no  centrifugal  force  where  i  is  -^  .     Euler's  rule  gives  -jg-«-  •     ^ow  we 

shall  take  this  as  the  endlong  load,  which  will  cause  the  strut  to  break  in 
the  other  way  of  bending  also,  so  as  to  have  it  equally  ready  to  break 
both  ways.  (2)  Bending  in  the  direction  in  which  bending  is  helped  by 
centrifugal  force.  Our  w  is  the  above  quantity  divided  by  bd,  or,  taking 

E  =  3  x  107,  w  =  6-17  x  jz  x  106.      Taking  the  proof  stress  /  for  the 

steel  used  as  20,000  Ibs.  per  square  inch  (remember  to  keep/  low,  because 
of  reversals  of  stress),  and  recollecting  the  fact  that  i  in  this  other 

direction  is  —  ,  we  have  (6)  becoming 

8-4  x  108  (i  _  308  ^  (l  _  ^  =  n*Pr  +  d  ..  ..  (7). 

Thus,  for  example,  if  b  =  1,  I  =  30,  r  =  12,  the  following  depths  d  inches 
are  right  for  the  following  speeds.  It  is  well  to  assume  d^  and  calculate  n 
from  (7).  d 


d\l\  1-5 
n  \  0  I  205 


277 


2-5 

327 


368 


437 


545 


2.  A  round  bar  of  steel,  1  inch  in  diameter,  8  feet  long,  or  I  —  48 
inches.  Show  that  an  endlong  load  only  sufficient  of  itself  to  produce  a 
stress  of  1,910  Ibs.  per  square  inch,  and  a  bending  moment  which  by  itself 
would  only  produce  a  stress  of  816  Ibs.  per  square  inch,  if  both  act  together 
produce  a  stress  of  23,190  Ibs.  per  square  inch. 

Students  will  find  that  this  subject  will  well  repay  further 
study.  The  effect  of  a  small  lateral  load  on  a  strut  is  sometimes 
very  striking.  Again,  in  tie-bars  it  is  very  important  to  consider 
the  effect  of  a  lateral  load.  The  subject  is  treated  more  fully  in  a 
paper  in  the  Philosophical  Magazine,  March,  1892.  (See  Appendix.) 

374.  Beams  without  Compression. — In  Art.  368  we  have 
seen  that  a  tensile  load  applied  to  extend  a  beam  may  not  only 
diminish  the  greatest  compressive  stress,  but  also  the  tensile 
stress.  Again,  there  are  many  cases  of  beams  or  infinitely 


472  APPLIED   MECHANICS. 

flat  arches  in  which  there  is  no  tensile  stress  anywhere.  Ju 
such  a  case,  of  course,  the  earth  takes  the  necessary  tensile 
load.  When  the  pneumatic  wheel  tyre  was  invented,  Pro- 
fessor Fitzgerald  pointed  out  to  me  that  columns  to  support 
loads,  and  military  bridges  easy  to  pack  and  unpack,  might  be 
made  of  inflated  tubes,  the  solid  material  being  everywhere  in 
tension.  I  consider  this  a  notion  of  great  importance.  In  a 
thin  straight  tube  of  circular  section  if  the  greatest  bending 
moment  is  M  and  R  is  the  radius,  t  the  small  thickness  of  the 
material,  the  compressive  stress  anywhere  due  to  bending  is 

AT 

•  — _-  y  where  y  is  the  distance  from  the  diameter  which  is  the 
irRst 

neutral  line  of  the  section  on  the  compressive  side.  The 
greatest  compressive  stress  is  M/TT  R2  t.  Now  imagine  the  tube 
to  be  subjected  to  internal  fluid  pressure  P  above  that  of  the 
atmosphere ;  there  is  a  tensile  endlong  stress  p7rR2-r27rRZ  or 
p  R/2  t,  and  hence  the  greatest  compressive  stress  is  M/TT  R2  t 
-  P  R/2  t.  This  is  just  o  when  p  =  2  M/TT  R3.  The  greatest 
tensile  endlong  stress  is  then,  of  course,  P  R/2 ;  but  this  is  equal 
to  the  lateral  tensile  stress  which  the  mere  internal  pressure 
produces.  When,  therefore,  the  internal  pressure  is  just 
sufficient  to  remove  all  compressive  stress  in  the  material,  the 
tensile  stress,  where  it  is  greatest,  is  the  same  in  all  directions 
and  is  2  M/TT  R2 1.  We  see,  therefore,  that  great  loads  may  be 
carried  by  inflated  tubes  of  thin  material  if  they  are  only  large 
enough  in  diameter,  or  by  a  bundle  of  small  tubes.  Shearing 
forces  might  be  taken  up  by  side  frameworks  like  lazy-tongs. 
I  make  no  attempt  here  at  exact  theory.  What  I  give  is 
sufficiently  correct  to  show  the  general  value  of  the  suggestion. 
One  may  go  far  in  speculation  on  this  idea — rigidity  gained  by 
using  thin  material  and  subjecting  it  to  internal  fluid  pressure, 
so  that  there  shall  be  no  compressive  stress.  The  great  ships 
of  the  future  may  owe  their  stiffness  and  strength  to  the 
general  use  of  fluid  pressure  in  those  parts  of  them  where 
cargo  is  stored,  and  the  same  pressure  which  gives  strength 
may  serve  to  keep  out  the  sea  in  case  of  a  leak.  It  is  the 
means  by  which  the  leaves  of  plants  are  made  rigid. 
Similarly,  large  flat  areas  might  be  made  of  considerable  size 
by  fastening  together  two  plane  sheets  by  means  of  many 
connecting  ties  so  that  they  may  not  balloon  out,  and  then 
inflating  them  like  an  air  cushion.  Aeroplanes  of  sufficient 
size  to  support  a  man  by  Lilienthal's  method  can  be  made  with 


APPLIED   MECHANICS.  473 

comparatively  small  internal  fluid  pressures  and  are  not  liable 
to  make  splinters  when  they  fall  to  the  ground,  these  splinters 
being  a  cause  of  considerable  risk  with  aeroplanes  made  with 
sticks  as  stiffeners.  Kites  much  larger  than  those  suggested 
for  military  purposes  might  be  made,  in  which  the  whole  kite 
might  be  like  an  air  cushion,  or  thin  tubes  with  compressed  air 
might  take  the  place  of  the  present  bamboo  framework.  The 
inflation  might  be  maintained  automatically. 

375.  Inflated  Columns. — Again,  a  thin  tube  of  radius  R  and 
thickness  t  has  to  act  as  a  column  carrying  a  load  w,  and  this 
is  the  load  which  is  carried  when  there  is  no  axial  tensile 
stress.  The  pressure  of  the  fluid  inside  being  p,  we  have 
TT  R3  P=W  .  .  . .  (1).  Also  the  lateral  tensile  stress  produced  in 

w 

the  material  is  p  n/t  or  -,  so  that  great  loads  may  be  sup- 
ported by  inflated  tubes  of  thin  material  if  they  are  large 
enough  in  diameter.  Thus,  for  example,  I  find  that  a  tower 
of  thin  steel  1,000  feet  high  would  have  in  it  a  lateral 
tensile  stress  of  only  3  tons  to  the  square  inch,  due  to  its 
own  weight  and  the  necessary  fluid  pressure.  Being  all  in 
tension  there  is  no  danger  of  instability  such  as  exists  in 
ordinary  pillars.  If  large  in  diameter,  the  hemispherical  top 
cap  becomes  of  importance  as  a  load.  Any  moderate 
diameter  like  20  feet  would  bear  many  tons  on  the  top  in 
addition  to  the  weight  of  the  structure  itself.  Thus,  1,000 
feet  high  and  20  feet  in  diameter  and  '01  foot  thick  would  itself 
weigh  about  125  tons.  Its  hemispherical  cap  would  weigh  6 '3 
tons,  and  it  would  support  325  tons  on  its  top.  The  internal 
pressure  would  be  23  Ibs.  per  square  inch  and  the  tensile  stress 
10  tons  per  square  inch.  There  would  be  no  compressive  stress. 

Neglecting  lateral  stiffness,  whether  we  assume  the  adiabatic 
or  constant  temperature  law  for  increasing  pressure  down- 
wards in  the  fluid,  we  are  led  to  rules  as  to  the  relationship  of 
radius  of  column  to  thickness  of  metal  if  the  column  is  to  have 
the  same  stress  in  its  material  at  all  levels.  In  fact,  Fitzgerald's 
idea  gives  rise  to  a  number  of  easy  and  interesting  problems 
which  I  have  worked  out,  but  for  which  I  have  no  space. 

In  all  probability  it  would  be  found  cheaper  to  use 
long  stays  from  the  top,  and  possibly  from  several  places  at 
different  heights  to  resist  lateral  motion  due  to  wind  pressure, 
etc.,  than  to  stiffen  as  with  lazy  tongs  the  sides  of  the  tube 
itself.  It  is  certain  that  the  thing  is  practical. 


474  APPLIED   MECHANICS. 

376.  Tapering  Column  of  Circular  Section. — If  #is  the  distance 
downwards  to  any  section  where  t  is  the  thickness  and  r  the 
radius,  p  then  being  the  internal  pressure  due  to  fluid  and  q  the 
external  pressure  due  to  the  atmosphere ;  if  fl  is  the  compressive 
stress  in  the  material  and  wl  the  weight  per  unit  volume  of  the 
metal  and  w  of  the  inside  fluid, 

Ix .  2  -nrt .  w1  +  vr2  .  5x  .  w  +  2  irr.Sr.q  =  Sx  .  ~  (pitr*  +  2  irrtf1). 

It  follows  from  this,  if  w  =  ^/7>  as  wl  *s  constant  and  q  is  a 
known  function  of  x,  that 


dx  ^  2  I  dx  ^  •   -J  .      ., 

if  a  •==.  1  +  2/1//,  where  /  is  the  greatest  stress  in  the  material ; 

where  s  •=.  — ^—- ,  taking  the  same  fluid  inside  and  outside.     The 
result  is,  if  q  is  taken  to  be  small, 

^  x  +  (1  +  *)  s  log.  (1  -  te)  +  2  log.  r  =  constant. 
af  a 

If  q  is  not  small,  the  integration  seems  troublesome. 

377.  The  limiting  length  of  a  vertical  prism  supporting  its  own 
weight. — h  being  measured  downwards  to  any  place  from  the 
centre  of  gravity  of  the  uppermost  section,  let  y  be  the  deflection. 

Let  the  load  per  unit  length  be  w,  and  let  w  =  I  w  .  dh,  the  total 


=  I  w  .  dh, 


load  above  any  section.    Considering  bending  moment  M  and  M  +  8  M 
at  the  sections  at  h  and  h  4-  Sh,  we  see  that  w.  8y  +  £  w.Sh.Sy  =  5  M, 


^.. 

dy       dh      dy  i 

In  the  most  general  case,  where  n  and  i  vary,  and  where  there 
is  a  load  p  on  the  top, 


Let  i  and  w  be  constant,  and  let  P  be  0, 


an  equation  whose  solution  is  useful  in  other  problems.  If  H  is 
the  total  length  of  the  column  whose  end  is  fixed  in  the  ground, 
lot  x  =  h/n  and  (2)  becomes 

=—  -- 


El 

The  solution  of  (3)  in  series  may  be  indicated  by 


It  we  put  ~  =  0  when  x  —  0  at  top,  we  find  B  =  0,  and  the 


APPLIED    MECHANICS.  475 


other  part  of  (4)  is 


,  f 

*l        27374  +  2.3.5.6.7     2  .  3  .  5  .  6.  8.  9.  10  + 

In  this,  if  we  put  •/  =  0  when  #  =  1,  we  find 
cix 

0  =  1  -  JOL  +        m*     -  -  m*         -  +  etc., 

2.32.3.5.6       2.3.5.6.8.9^ 

and  by  trial  we  find  that  this  is  satisfied  by  m  =  7'85.*  Hence 
if  bending  occurs,  m  HS/E  i  =  7  '85  ;  so  that  this  gives  us  the 
limiting  height  H  of  a  uniform  column.  Thus  we  find  K  in 
feet  =  8  x  106/L2  for  thin  steel  tubes,  and  H  =  4  x  rO«/La  for 
solid  rods  of  steel,  where  L  is  the  ratio  of  length  to  diameter. 
Thus,  for  example,  the  maximum  height  of  a  tvfbe  8  feet  in 
diameter  is  800  feet. 

Let  A  k2  =  i  :  a  =  diameter  in  inches.     Greenhill  give* 


p  is  ~.     First  put  p  =  ro*  then  put  x3  =  r2 ;  we  get 

:  -£  - 

This  is 


Bessel's  equation,  where  A2  =        ^2,  w2  =  J.     Solution  of  (3)  in 
Consequently 


_^  =  0  when  a;  =  0,  makes  A  =  0  ; 


Put  ^  =  0  when  x  =  h,  the  lowest  point,     j  __  ,  (£A  ^  =  0.     If 

<?  is  the  least  root  of  j    ,  (0)  =  0,  then  c  =  kh  %.     Greenhill  gives 
the  expansion  of  j  _  j  (c),  and  finds  root  by  trial. 

378.  Stability  of  Shafts.  —  Rankine  considered  the  effect  of  cen- 
trifugal force  on  shafting  in  1869.  Professor  Greenhill  has  investi- 
gated the  stability  of  a  shaft  between  bearings  (see  his  Paper  to  the 
Institution  of  Mechanical  Engineers,  1883).  He  took  into  account 

*  Prof.  Maurice  Fitzgerald  published  curves  showing  the  values*  of  y  and 
^  when  x  =  1  for  various  values  of  m  (Proo.  Phys.  Soc.  of  London,  Oct., 

1892).  Prof.  Greenhill  had  previously  given  the  solution,  not  only  for 
uniform  solid  cylinders,  but  also  for  cones  and  paraboloids  of  revolution 
(Proc.  Camb.  Phil.  Soc.,  1881). 


476  APPLIED    MECHANICS. 

an  end  thrust,  F,  a  twisting  moment,  T,  and  centrifugal  force.  He- 
found  in  practical  examples  of  propeller  shafts  that  the  twisting 
moment  was  not  nearly  so  important  as  the  thrust  in  determining 
the  maximum  length  of  shaft  for  stability  ;  in  fact,  that  the  shaft 
might  be  designed,  merely  like  a  strut,  by  Euler's  'formula.  I 
consider,  therefore,  that  such  a  shaft  ought  to  be  studied  under 
the  rules  of  Art.  372. 

379.  Whirling  Loaded  Shaft. — If  a  uniform  shaft,  originally 
straight,  is  w  Ib.  per  unit  length,  and  has  an  angular  velocity  of 
a  radians  per  second ;  if  B  is  its  flexural  rigidity,  or  E  i ;  if  the 
deflection  from  straightness  is  y  at  a  point  distant  x  from  the 

centre,  then  the  load  due  to  centrifugal  force  is  —  o2y  per  unit 

length.  If  we  also  take  into  account  the  effect  of  gravitjr,  there  is 
one  position  in  a  revolution  when  the  centrifugal  force  and  the 
weight  produce  their  greatest  effects.  We  may  approximately 
take  it  that  the  path  of  every  point  is  a  circle,  and  that  the  weight 
is  a  radial  force  always  acting  in  the  same  direction  as  the  centri- 
fugal force.  If  we  take  into  account  also  an  -ondlong  thrust,  F,  we 
are  led  to  the  equation 

dty       F  cPy       w  o2          w 

and  to  the  solution 

y  =  A!  sin.  /3a?  +  A2  cos.  j8  x  +  A3  e  "fx  +  A^  »~~^x  —  gjc?  ....  (2) 
where 


T2 


=  _  JL  +          /  r2         wa2 


*2!  +  -( 

(3). 


y  is  to  have  the  same  values  for  equal  positive  and  negative  values 
of  x,  so  that  A]  =  0,  A3  =  A4.  Putting  y  •=.  0  where  x  =  I  and  con- 
fining our  attention  to  shafts  whose  bearings  do  not  constrain 
them  in  direction  at  the  ends,  so  that  cPyldx*  =:  0  where  x  =  I  we 
find  AJ  =  0,  Aa  =  7^/a2  (72  +  02)  cos.  01,  A3  =  \4  =  /8^/2aa  (y2  +  02) 
cosh  yl. 

The  bending  moment  anywhere  is  Eid2*//^2,  and  hence  the 
bending  moment  in  the  middle,  where  it  is  evidently  greatest,  is 
M0=iEi(-)82A2+2A3-x2),  or 


/a2 


The  greatest  stress  at  the  middle  is 


where  z  is  the  strength  modulus  of  the  section,  and  a  is  the  area  of 
the  section  ;  or  if  u  is  outside  radiui)  of  a  circular  shaft  we  have 


APPLIED    MECHANICS.  477 

Exercise.—  Show  that  in  the  case  of  a  propeller  shaft  13-4  inches 
diameter,  of  length  98  feet  between  its  bearings,  with  endlong  thrust 
50,000  Ibs.,  if  o  =  2  v,  or  60  revolutions  per  minute,  tho  important  terms 

in   the    above   calculation   are    F2/4B2=30  x  10~14,   and  wa?lffu  = 

89  x  10  ~      .     So  that  the  centrifugal  force  is  300  times  as  important  as 
the  endlong  thrust. 

If  we  altogether  neglect  twisting  moment  and  endlong  thrust 
and  write  »4  for  w  <£\g  B,  we  have  to  solve 


Our  old  /3  =  y  =  «,  and  the  greatest  stress  is 

/        1  IN 

/=EM>R(  —  —  --  -      -     /2Bn2  ____  (8). 
\cosh  nl       cos.  nlj  ' 

This  is  infinite  if  cos.  nl  is  0,  that  is  if  nl  =  ^  ;  that  is,  if 

II  -,('-2$*.       -(9). 
\wa2/ 

This  limiting  length  may  be  obtained  very  simply  by  leaving  out 
the  constant  term  in  (1),  as  Rankine  does. 

Mr.  Dunkerley  (Phil.  Trans,  for  1894)  discusses  the  stability 
under  centrifugal  force  of  shafts  loaded  with  pulleys,  and  he  has 
illustrated  his  results  experimentally.  (See  Appendix.) 

EXERCISES. 

1.  If  the  critical  length  of  a  shaft  13-4  inches  in  diameter,  subjected 
to  endlong  thrust  alone,  is  equal  to  the  critical  length  when  subjected  to 
centrifugal  force  alone,  show  that  F  =  68,500  o.     Show  that  if  the  length 
is  98  feet,  the  critical  F  is  327,600,  and  the  critical  o  is  nearly  4*1  radians 
per  second,  or  46  revolutions  per  minute. 

2.  Professor  Greenhill's  result  is  that  if  F  is  the  endlong  force,  and  T 
the  twisting  moment,  the  limiting  length  of  shaft  being  2  I,  we  have 

^  —  JL          ^ 

4T2  ~~  E!  +  4~W2< 

The  propeller  shaft  of  the  Cunard  steamer  Servla  is  of  wrought  iron, 
22^  inches  diameter.  The  pitch  of  screw  is  35^  fact.  At  53  revolutions 
per  minute  the  indicated  horse-power  was  10,350.  Assuming  that  all 
this  power  is  really  utilised—  an  assumption  which  is,  of  course,  quite 
wrong—  prove  that  F  =  181,530  Ibs.,  and  that  T  =  12'3  x  106  pound- 
in^hea  Take  E  =  29  x  106,  and  show  that  the  above  formula  becomes 

—  =  4-98  x  10-7  +  2-8  x  10~10, 

so  that  the  twisting  moment  term  is  quite  inconsiderable  compared  with 
the  thrust  term.  Show  that  2  I  =  4,454  inches.  Consider  this  shaft  of 
22|  inches  diameter,  4,454  inches  long  between  bearings,  subjected  to  no 
thrust  or  twisting  moment,  and  not  even  to  its  own  weight  ;  prove  that  il 
cannot  be  rotated  at  a  higher  speed  than  a=  '14,  or  1^  revolution  per 
minute,  without  fracture  by  centrifugal  force. 


478 


CHAPTER  XXII. 

METAL    ARCHES. 

380.  THE  student  must  examine  drawings  and  actual  specimen* 
of  "bridges  to  see  how  the  weight  of  a  roadway  is  brought  to  bear 
upon  the  arched  ribs  whose  function  is  to  carry  weight,  trans- 
mitting it  with  certain  horizontal  forces  to  the  abutments.  Our 
problem  is,  given  the  distribution  of  vertical  loads  on  an  arch-ring 
of  which  the  shape  of  the  centre  line  and  the  shape  of  cross-section 
everywhere  are  known,  to  find  the  stress  everywhere.  If  K  z  o  is 
the  centre  line  of  the  rib  shown  in  Fig.  270,  then  any  cross-section 
B  c  is  supposed  by  our  theory  to  remain  plane.  The  resultant  of 


Fig.  270. 


all  the  loads  acting  to  the  right  of  B  c,  together  with  the  resultant 
force  R  at  the  abutment,  is  a  force  whose  direction  is  shown  at  D 
by  the  direction  of  the  line  of  resistance  there,  and  the  force 
polygon  shows  its  amount.  We  saw  that  in  Fig.  261  this  force, 
which  we  there  called  F,  produces  a  bending  moment  M  at  the 
section  whose  amount  is  p  .  o  D.  But  evidently  this  is  the  same  as 
F  .  oj  (Fig.  271)  if  o  j  is  perpendicular  to  F.  A  much  more 
important  thing  for  our  present  purpose  is  to  know  that  it  is  also 
equal  to  the  horizontal  component  H  of  the  force  F  multiplied  by 
the  vertical  distance  o  L.  This  the  student  will  easily  prove  for 
himself. 

Now  if  all  the  loads  are  vertical,  we  know  from  Art.  349  that 
(1),  the  resultant  force  F  at  every  joint  has  the  same  horizontal 
component  H  as  at  any  other.  This  is  represented  on  the  force 
polygon  (Fig.  236)  by  OH.  It  is,  of  course,  also  the  horizontal 


APPLIED    MECHANICS. 


479 


Fig.  271. 


thrust  on  each  of  the  abutments,  and  is  generally  called  the 
horizontal  thrust  of  the  arch.  Of  course  the  force  polygon  shows 
all  the  forces  F  at  all  the  sections.  (2)  If  two  vertical  lines  he  drawn 
through  K  and  G,  as  in  Fig.  270,  and  if  any  line  of  resistance  he  drawn 
whose  ends  K'  and  G'  are  in  these  lines,  and 
if  K'  and  G'  are  joined,  the  vertical  height 
anywhere,D  T,of  this  link  polygon  is  inversely 
proportional  to  the  o  H  of  the  force  polygon 
by  means  of  which  it  is  drawn ;  so  that  if 
two  such  figures  as  K'Z'G'  are  drawn,  the 
vertical  ordinates  D  T  are  all  in  the  same 
proportion.  We  see,  therefore,  that  since 
the  bending  moment  at  B  c  is  the  horizontal 
thrust  multiplied  by  o  L,  the  vertical  distance 
from  the  centre  line  anywhere  to  the  line 
of  resistance  there  represents  to  some  scale 
the  bending  moment  at  the  section  there. 
At  v  and  u  the  lines  cross.  These  are  points 
of  no  bonding.  From  K  to  u,  and  from  G  to  v  the  bending 
moment  tends  to  make  the  arch  more  convex  upwards.  From 
v  to  u  the  bending  moment  tends  to  make  the  arch  less  convex 
upwards.  If  we  can 
only  find  the  true  line 
of  resistance  K'  Z'G', 
we  know  by  Art.  129 
the  stress  in  every 
section.  The  true  po- 
sition of  K'  z'  o'  is 
determined  by  these 
conditions : — 

1.  In    an     arch 

hinged  at  the  ends,  if  K  and  o  are  the  centres  of  the  hinges, 
we  know  that  the  line  of  resistance  must  pass  through  them. 
We  only  need  one  other  condition,  and  that  is  given  by  the 
statement :  the  yielding  everywhere  of  the  arch  must  be  such  that 
the  distance  K  a  remains  constant. 

2.  In  an  arch  fixed  at  the  ends,  if  the  fixing  is  perfect,  we  have 
the  above  condition  that  the  distance  x  G  remains  constant,  and 
also  the  condition  that  the  inclinations  of  the  centre  line  at  K  and 
G  remain  constant.     Just  as  in  the  case  of  beams  fixed  at  the  ends, 
it  is  exceedingly  difficult  to  fix  the  arch  at*  the  ends  so  perfectly 
that  we  can  rely  upon  this  condition  being  fulfilled ;  and  hence,  on 
account  of  our  uncertainty,  we  prefer  to  use  arches  hinged  at  the 
ends. 

3.  The  simplest  case  of  all  is  that  in  which  there  are  three 
hinges  in  the  arch,  one  at  each  abutment  K  and  G,  and  another  at 
the  crown.     Even  when  the  loads  are  not  vertical  it  is  easy  to  find 
the  line  of  resistance,  because  two  corners  and  a  point  in  it  are 
given,  and  we  have  merely  the  exercise  of  Art.  106. 

Any  system  of  oblique  loads  is  given,  acting  upon  an  arch 
whose  end  hinges  are  K  and  G  and  whose  crown  hinge  is  L. 
Find  the  resultant  of  all  the  loads  from  K  to  L,  and  let  it  be  A  B 


Fig.  272. 


480  APPLIED   MECHANICS. 

(Fig.  273).  Find  the  resultant  of  all  the  loads  Irom  L  to  o,  and  let 
it  be  B  c.  Draw  A  B  and  B  c  in  the  force  polygon  (Fig.  274). 
Choose  any  pole  o.  Join  o  A,  OB,  and  o  c.  Draw  K  p,  p  o.,  and  a  R 
parallel  to  o  A,  o  B,  and  o  c.  The  resultant  of  A  B  and  B  c  passes 
through  s.  Now  join  a  s  and  draw  the  new  diagram,  c  o'  parallel 


fcPfg.  274. 


Fig.  273. 


to  o  s,  and  join  o'  B  ;  so  that  K  p,  p  Q',  of  Q  correspond  to  o'  A,  o'  B, 
and  o'c.  We  have  then  a  link  polygon.  But  it  does  not  pass 
through  L.  Produce  P  Q'  to  meet  K  G  in  T.  Join  T  L  and  let  it  meet 
the  given  forces  in  p"  and  Q".  Join  K  P"  and  Q,"  G,  and  K  p"  Q"  G  is 
the  link  polygon  or  line  of  resistance  required.  I  set  the  problem 
as  an  exercise  for  students,  and  one  of  them,  Mr.  Stansfield,  gave 
me  this  solution. 

If  the  loads  are  vertical,  and  K,  G,  and  z  are  the  given  hinges,  Fig. 
272,  draw  any  line  of  resistance  K  z'  G',  its  corners  z'  and  G'  being  in 
the  verticals  from  z  and  G.  K  z"  G'  is  a  straight  line.  Draw  a  vertical 
a'  z  x  z"  through  z.  Construct  a  figure  on  the  base  K  G  such  that 
it  B'  is  any  point  in  it,  B'  c  is  in  the  same  proportion  to  B  D  that 
z  x  is  to  z'  z".  This  will  evidently  be  the  true  line  of  resistance, 


Fig.  275. 


and  the  vertical  ordinates  between  it  and  the  centre  lines  of  the 
portions  of  the  arch  will  be  the  bending  moments.  One  example 
of  each  of  these  must  be  worked  out  by  each  student. 

381.  Arch  Hioged  at  the  Ends. — Imagine  the  small  slice  of 


APPLIED    MECHANICS. 


481 


the  arch  between  two  sections  B  o  c  and  B'  o'  c',  at  the 
small  distance  5*  asunder,  to  yield,  and  study  the  effect  of 
the  yielding  of  this  portion  alone,  exactly  as  we  did  with 
our  spring  in  Art.  370.  That  is,  we  imagine  the  part  o' G 
(Fig.  275)  to  he  absolutely  rigid  and  fixed ;  the  part  o  K  to  be 
rigid,  but  to  move  about  o  as  centre,  as  if  OK  were  merely 
a  pointer,  the  motion  K  K"  being  the  angular  motion  of  B  c 
relatively  to  B'  c',  multiplied  by  the  straight  distance  o  K.  The 

M 

angular  change  from  o  to  o',  if  o  o7  is  5*,  is  —  5s,  if  M  is  the 
bending  moment,  i  the  moment  of  inertia  of  B  c  about  o  through 


its  centre  of  gravity,  and  B  is  Young's  modulus;   so  that  KK" 
js  _^L  gs  .  o  K.      The  horizontal    component   of    this    is  K  K'"   or 

K  K"  cos.  K"  K  K'".     But  this  angle  is  the  same  as  K  o  N,  whose 
cosine  is  o  N/O  K.     Hence  the  horizontal  motion  of  K  due  to  the 


yielding  of  the  little  slice  is 


M .  5«  .  O  K      ON         M.ON.5S 


.  .   .  .  (1). 


El  OK  El 

And  as  K  does  not  yield  at  all,  we  make  this  sum  zero.  We  beg 
to  point  out  that  we  cannot  in  the  same  way  state  the  vertical 
displacement  of  K.  If  o  N  is  called  y,  and  if  K  N  is  called  #,  then 

(1)  is  2  —  y  .  5*  =  0 ;  and  we  might  in  the  same  way  think  that, 
as  —  x  .  8*  is  the  horizontal  displacement  of  K  due  to  the  yielding 

K  I 

of  the  slice ;  therefore  the  sum  of  all  these  terms  ought  also  to  be  0. 
If  the  end  G  is  fixed,  this  is  true,  and  it  enables  us  to  calculate  the 

fixing  moment.     But  in  truth  2  —  x  .  8«  is  not  0  if  o  is  hinged ; 

E  I 

it  is  equal  to  trio  angular  movement  at  G  multiplied  by  K  G.  Note 
that  for  the  horizontal  displacements  the  angular  yielding  at  a 
produces  no  effect. 


482  APPLIED   MECHANICS. 

Since  M  is  represented  to  scale  by  the  distances  OL,  (1)  becomes 
3  °L'  °N'  S  =0 (2).     In  Fig.  276  K  z  G  is  the  centre  line 

of  the  arch  of  which  K  and  G  are  the  hinges.  K.  z'  G  is  the  Tine  of 
resistance  which  we  want  to  find.  If  a  student  wishes  to  keep  (2) 
in  his  memory,  let  him  imagine  that  the  line  K  z  G  has  positive  and 
negative  density,  represented  by  M/E  i  per  unit  length.  Then  (1) 
or  (?)  tells  us  that  if  the  whole  positive  part  of  the  linear  mass  is 
mpt  and  if  yp  is  the  distance  of  its  centre  of  gravity  from  x  G,  and 
if  the  whole  negative  part  of  the  linear  mass  is  m^  and  if  yn  is  the 
distance  of  its  centre  of  gravity  from  K  G,  then  mp  yp  —  mn  yn. 
To  find  K  z'  G,  let  K  z"  G  be  any  line  of  resistance  through  K  G 
drawn  at  random.  Then  L  N  =  k  .  P  N,  where  Jc  is  an  unknown 
constant;  and  if  we  know  k,  knowing  P  N,  we  can  find  L  N. 
Divide  the  given  centre  line  K  o  z  G  into  any  number  of  equal 
parts,  and  draw  an  ordinate  N  o  p  at  the  middle  of  each  of  them. 
All  values  of  5s  are  now 

v..     ON.OL       .            /ON        _ON(LN-ON) 
equal,  and  (2)  is  2 =  0  ....  (3),  or  2 — '  =  0  ; 

so  that,  as  LN  is  k  .  PN,  we  have  k$ON '**  =  2™2 (4), 

and  k  is  evidently  2 -r  2 '- ....  (5)       This  is  easily 

calculated.  Thus,  for  example,  take  K.  z  G,  an  arc  of  a  circle,  span 
200  feet,  rise  30  feet.  For  a  given  system  of  unsymmetrical  loads, 
K  P  z"  G  was  drawn.  Dividing  K  z  G  into  sixteen  equal  parts  and 
raising  perpendiculars  at  the  middles  of  the  parts,  we  found  the 
following  values.  The  curves  as  drawn  were  measured  in  inches, 
no  attention  being  paid  to  scale.  The  values  of  i  are  in  inches  to 
the  fourth  power.  We  see  that  k  =  -03014  ^  -07447  or  0-405. 
Multiplying  each  value  of  p  N  by  this,  we  find  the  values  of  L  N. 
The  horizontal  thrust  corresponding  to  this  true  line  of  resistance 
is  greater  than  the  one  for  KZ"G  in  the  ratio  1  :  0'405,  and  its 
force  polygon  may  be  drawn.  The  student  needs  no  further  hints 
for  the  completion  of  the  calculation.  The  loads  taken  were  a3 
follows : — 


APPLIED    MECHANICS. 


483 


O  10  «5  O  05  r-<  O 
O  O  O  O  O  O  O 

o  o  p  p  p  p  p 


G)  ^ 


SI' 


oo—  iT*eoc<iaoiMcoocot^c<ia»—  toco 

T^Oi-^iOSC^OOSi—  I 
t~~t^.OOOOOiOiOO 
O  O  O  O  O  O  O  >-H 


O 

oooooo 
oooooo 


CO  >O  •—  I  -—  I 


i—  '  •—  I  tO  CO 


OOOOO 

oooooooooooooo 

00000000000'- 


O-^i-H«5<M^-iOOOOi-i<MCOr-iT^O  aj 

O>Of-H«5(MOO^OOT}iOOC^COr-|kOO  jg 


^Hl^OOOOOO«OOCCOCOt^i-l 


484  APPLIED    MECHANICS. 

382.  Arch  Ring  Fixed  at  the  Ends. — In  the  case  of  a  symme- 
trical loading,  the  end  fixing  moments  being  called  m^  we -simply 
have  this  added  on  or  subtracted  everywhere  to  the  bending  moment 
considered  in  the  last  case ;  or,  rather,  the  bending  moment  at  a 
place  o  is  o  L  -  MJ,  instead  of  being  merely  o  L.  (2)  becomes 

^  (PL  -  mi)  ON.  5s  _  Q  ^ 

(3)  becomes 

ao-("-iO'-*"  =  0....(»). 

and  (4)  becomes 

is^Z2'_2»JL2_Ml2«'J'  =  0....(4). 

But  we  have  another  condition  to  satisfy.     The  angular  change  in 
the  length  5s  is       5s,  and  the  integral  of  thii 

from  x  to  Y  is  0.     Hence  2 =  0,  or 


the  length  5s  is  M  5s,  and  the  integral  of  this  along  the  centre  line 


When  we  find  the  summations  in  (4)  and  (5),  it  is  easy  to  calculate 
the  two  unknowns  Wj  and  k.     When  found,  ml  will  be  to  the  same 


Fig.  277. 


scale  as  that  to  which  the  distances  o  L  represent  bending  moment, 
and  the  curve  of  Fig.  277,  x  z'  y,  obtained  when  we  know  k  must  be 
displaced  downwards  everywhere  by  the  constant  displacement  m1} 
till  its  appearance  is  that  shown  in  x'  M  Y'.  A  thoughtful  student 

will  see  that  the  5  —  K  N  .  5s  summation  is  also  0  ....  (6)  in  this 

case,  but  that  (5)  and  (2)  cannot  both  be  true  unless  (6)  is  also  true, 
because  the  loading  is  symmetrical. 

In  the  case  of  an  unsymmetrical  loading,  the  bending  moment 
everywhere  is  less  than  o  L  by  an  amount  which  is  wj  at  the  end  x, 
and  is  w2  at  the  end  Y,  and  which  changes.  A  little  consideration 
will  show  that  we  must  lower  the  curve  K  z'  G  of  Fig.  276  by 
amounts  shown  for  each  vertical  line  by  the  ordinates  of  such  « 


APPLIED    MECHANICS.  485 

diagram  as  K  K'  G'  a  of  Fig.  271.     If  a;  is  the  horizontal  distance  of 
the  o  of  any  section  from  K,  and  I  is  the  whole  span,  then  we  must 

subtract  the  amount  m,  +  x  -  2   . — -  from  the  OL  of  Fig.  276. 

I 

Call  this  mi  +  p  x,  then  (4)  becomes 
(5)  becomes 

i          i      mi   7  ~~       x  —          •  v  /• 

We  also  now  find  another  condition  by  stating  that  the  sum  of  the 
vertical  displacements  of  K  is  0  when  G  is  held  fixed — that  is, 

2  —  K  N  .  5s  =  0  .  .  .  .  (6),  and  this  gives  us 


ij^      -  S_       _  MlS    -  P2     =  o  . . . .  (6); 

and  the  summations  indicated  in  (4),  (5),  and  (6)  being  effected,  we 
have  no  difficulty  in  solving  for  Jc,  mv  and  P,  and  so  finding  the 
true  line  cf  resistance. 

The  table  shows  these  summations.  It  is  for  the  same  arch 
ring,  with  same  loads,  as  in  Art.  381,  and  therefore  some  of  the 
columns  are  the  same. 

The  student  will  note  that  if  y1  is  the  ordinate  of  the  centre 
line  of  the  unloaded  arch  at  any  place,  and  if  y  is  the  ordinate  of 
the  loaded  arch,  and  if  yl  -  y,  the  downward  motion  of  the  point, 
be  called  z,  then,  unless  in  some  case  there  is  very  great  slope 

indeed,   we  may  take   — |  -^  {  1  +  (~j^)  j  ^  as  the  change  of 

curvati're.  The  proof  of  this  is  easy.  Equating  this  to  M/E  i,  we 
see  that  we  have  only  to  imagine  i  at  every  place  divided  by 

j  1  +(-j~ )   j  s  there,  calling  it  by  the  new  name  i1,  and  we  take 
-r-|  =  M/E  i1.     Having  a  diagram  showing  everywhere  the  value  of 

M/E  i,  it  is  easy  to  obtain  the  diagram  of  z  by  graphical  statics,  just 
as  in  Fig.  217  we  obtained  the  shape  of  a  beam  from  a  diagram 

Of  M/E  I. 


486 


CHAPTER  XXIII. 

MEASUREMENT  or  A  BLOW 

383.  IN  Ait.  46  we  considered  what  occurs  when  a  chisel  is 
cutting  metal.  We  shall  now  consider  the  same  subject  from  a 
point  of  view  which  at  first  seems  different.  Consider  how  it 
is  that  a  blow  of  a  hand  hammer  will  indent  a  steel  surface, 
whilst  a  steady  force  applied  to  the  same  hammer-head  would 
require  to  be  very  great  to  produce  any  indentation.  The 
pressure  between  the  hammer  and  the  steel  is  very  great, 
and  it  must  be  all  the  greater  because  the  time  of  contact 
is  very  short.  Indeed,  if  a  hammer  weighs  2  Ibs.,  so  that 
its  mass  is  2  -^  32-2  or  -0621,  and  if,  just  before  touching 
the  steel,  its  velocity  was  10  feet  per  second,  then  we 
know  that  its  momentum  was  10  x '0621,  or  '621.  Now,  if 
one-ten-thousandth  of  a  second  elapses  from  the  time  of 
actual  contact  until  the  hammer's  motion  there  is  destroyed 
— that  is,  until  the  elasticity  of  the  steel  is  just  about  to  send 
the  hammer  back  again  a  little — the  momentum  -621  is  destroyed 
in  -0001  second ;  hence  the  average  force  (notice  that  it  is  a 
time  average)  acting  between  hammer  and  steel  during  this 
short  time  must  have  been  *621  -f-  *0001,  or  6,210  Ibs.  It  is 
certain,  however,  that  this  average  force  is  less  than  what  the 
force  actually  was  for  some  very  small  portion  of  the  time. 
We  observe,  then,  that  we  cannot  tell  the  average  force  of 
an  impact  unless  we  know  two  things — the  momentum  and 
the  time  in  which  it  was  destroyed.  Now  the  duration  of 
an  impact  depends  greatly  upon  the  nature  of  the  objects 
which  strike  one  another,  and  we  see  that  the  average  force 
of  a  blow  is  less  as  the  time  is  greater.  Sometimes,  instead  of 
a  great  force  acting  for  a  very  short  time,  what  we  require  is 
a  smaller  force  acting  for  a  longer  time.  For  instance,  when 
cutting  wood  we  obtain  this  result  by  using  a  wooden  mallet 
and  a  chisel  with  a  long  wooden  handle,  because  the  force 
required  to  make  the  chisel  enter  the  wood  is  not  very  great, 
and  we  wish  this  force  to  act  for  some  time,  so  that  much 
wood  may  be  cut  at  one  blow.  In  chipping,  we  have  the  time 
short,  because  considerable  force  is  required  to  cause  the  chisel 
to  enter  metal.  The  duration  of  an  impact  depends  on  the 


APPLIED    MECHANICS.  487 

shapes  of  the  bodies  and  their  masses,  and  on  the  elasticity  of 
their  materials. 

384.  Why  is  it  that  in  driving  a  nail  into  wood  our  blows 
seem  to  be  of  no  effect  unless  the  wood  is  thick  and  rigid,  or  un- 
less it  is  backed  up  by  a  piece  of  metal  or  stone  ?     It  is  because 
the  wood  yields  quite  readily,  and  so  prevents  the  hammer 
losing   its   momentum    rapidly.       There   are   few  subjects  in 
which  people  are  so  apt  to  have  erroneous  ideas  as  in  this  one 
of  impact.     Thus  a  man  will  speak  of  the  force  produced  by  a 
weight  falling  through  a  height  without  having  any  idea  of 
the  time  during  which  the  motion  of  that  weight  is  being 
stopped — in  fact,  without  considering  what  time  the  weight 
is  allowed  for  delivering  up  its  momentum.    Now,  a  little 
consideration  will  show  that  the  mean  force  of  the  blow  will 
be  quite  different  according  as  the  weight  falls  on  a  long  and 
yielding  bar  or  on  a  short  and  more  rigid  one.     If  we  could 
imagine  bodies  to  be  formed  of  perfectly  unyielding  materials, 
then  the  slightest  jar  of  one  against  the  other  would  produce 
an  infinitely  great  pressure  between  them ;   and  in  the  blow 
produced  by  a  falling  body  there  may  be  every  gradation  from 
exceedingly  great  pressures  to  very  small  ones,  depending  on 
the  yielding  power  of  the  body  that  is  struck.     Everybody  is 
acquainted  with  the  sensation  pvoduced  by  suddenly  placing 
one's  foot  on  a  level  floor  when  one  was  preparing  for  a  step 
downwards.       The    downward    momentum    of    the    body   is 
suddenly  destroyed,  and  there  are  great  pressures-  in  all  the 
bones  of  the  body.     Carriages  are  hung  on  springs  for  the 
purpose  of  preventing  their  losing  or  gaining  momentum  with 
too  great  rapidity  when  the  carriage  wheels  pass  over  obstacles. 
When  we  are  sitting  on  a  hard  seat  in  a  third-class  railway 
compartment,   and  the  carnage  gets  a  slight  jerk  upwards, 
momentum  is  given  much  too  rapidly  to  our  bodies  for  perfect 
comfort,  and  to  sit  on  cushioned  seats  is  preferable.    A  cannon- 
ball  is  safely,  because  comparatively  slowly,  stopped  by  sand- 
bags or  bales  of  cotton. 

385.  Example. — A  pile  driver  of  300  Ibs.  falls  through  a' 
height  of  20  feet,  and  is  stopped  during  O'l  second.     What 
average  force  does  it  exert  upon  the  pile  <\     A  body  which  has 
fallen  freely  through  a  height  of  20  feet  has  acquired  a  velocity 
equal  to  the  square  root  of  64*4  x  20,  or  35-89  feet  per  second. 
Its  momentum   is   35-89  x  300  -r  32-2,    or  3344,    and   this 
divided  by  0-1   gives   3,344   Ibs.,   which,   together   with   the 

Q* 


488  APPLIED    MECHANICS. 

weight  300  Ibs.,  is  3,644  Ibs.,  the  answer.  From  the  instant 
when  the  motion  of  the  driver  ceases  to  diminish,  the  force 
exerted  by  it  is  its  own  weight.  We  have  considered  the 
time  average  of  the  force.  The  average  force  of  friction  m 
pounds  between  the  pile  and  the  ground,  multiplied  by  the 
distance  in  feet  through  which  the  pile  descends  during  the 
stroke,  is  equal  to  300  x  20,  or  6,000  foot-pounds,  if  we  neglect 
the  loss  due  to  vibrations  of  the  body  and  the  energy  carried 
off  to  the  ground  to  be  wasted  in  earth  vibration,  and  if  we 
also  ignore  the  fact  that  the  weight  really  descends  a  little 
farther  than  20  feet.  The  neglected  energy  may  be  the 
whole  of  the  energy  if,  for  example,  the  blow  does  not  make 
the  pile  move  further  into  the  ground.  We  can  only  be 
certain  about  a  time  average  force,  and  even  in  this  case  we 
must  assume  that  there  is  an  instant  at  which  one  of  the 
bodies  has  the  same  average  velocity  in  all  its  parts. 

385a.  Example. — A  column  of  water  in  a  pipe  6  inches 
diameter,  30  feet  long,  was  moving  behind  a  piston  at  15  feet 
per  second.  The  piston's  motion  is  stopped  in  0-1  second. 
What  is  the  time  average  of  the  pressure  due  to  the  stoppage  1 

Ans.  The  area  of  the  circular  section  of  the  column  of 
water  being  28*274  square  inches,  the  quantity  of  water  in 
motion  is  30  x  28-274  -•-  144  or  5-89  cubic  feet,  or  5-89  x  62-3 
or  366-9  Ibs. ;  and  on  the  assumption  that  all  the  water  has 
exactly  the  same  motion,  its  momentum  is  its  mass  366*9 -4-32*2 
multiplied  by  15,  or  1-70  units.  The  time  average  of  the 
force  required  to  stop  the  motion  in  O'l  second  is  therefore 
(62-3  x  30  x  28-274  x  15)  -f  (144  x  32-2  x  0-1)  Ibs.,  which 
is  equivalent  to  a  pressure  of  60*5  Ibs.  per  square  inch  over 
the  area  of  the  piston. 

386.  Suppose  a  body  A  to  strike  another  B,  and  that  we 
can  neglect  the  actions  of  outside  bodies  upon  them  both.  If 
A  loses  momentum,  B  must  gain  the  same  amount  because 
their  mutual  pressures  are  equal  and  opposite  during  the  time 
of  impact.  It  is  our  knowledge  of  this  fact  that  enables  us  to 
calculate  the  motions  of  bodies  after  they  strike  one  another. 
Again,  for  the  same  reason,  if  from  any  internal  cause  the 
parts  of  a  body  separate  from  one  another,  either  violently  or 
gently,  the  total  momentum  remains  as  it  was ;  it  is  only  the 
relative  momentum  which  alters.  Hence,  when  a  shell  bursts 
in  the  air,  some  parts  move  in  the  same  direction  more  rapidly 
than  before,  but  others  less  rapidly  ;  one  part  may  double  its 


APPLIED   MECHANICS.  489 

velocity  and  another  may  drop  nearly  vertically,  its  forward 
motion  being  stopped,  but,  on  the  whole,  the  total  forward 
momentum  is  what  it  was  originally.* 

387.  Examples. — If  a  cannon  were  perfectly  free  to  move 
backward  when  the  shot  leaves  it,  the  backward  momentum  of 
the  cannon  would  be  exactly  equal  to  the  forward  momentum 
of  the  shot.  Thus,  if  a  shot  of  20  Ibs.  leaves  a  cannon  whose 
weight  is  2,240  Ibs.  with  a  velocity  of  1,000  feet  per  second, 
the  velocity  of  the  cannon  backward  would  be  1,000  x  20 
-r-  2,240,  or  about  9  feet  per  second,  neglecting  the  fact  that 
the  gases  leave  the  gun  also  with  a  certain  momentum.  When 
a  ship  fires  her  broadside,  each  gun  runs  back,  communicating, 
as  it  is  stopped,  its  momentum  to  the  ship,  which  heels  over 
in  consequence.  A  gun  firing  the  above  shot  of  20  Ibs. 
directly  astern  from  a  ship  whose  total  weight  is  600  tons 
gives  to  the  ship  (neglecting  the  momentum  of  water  moving 
with  the  ship)  so  much  momentum  that  its  speed  is  increased 
(neglecting  her  friction  with  the  water)  9  -f-  600,  or  '015  foot 
per  second.  We  see,  then,  that  a  ship  might  propel  herself 
by  means  of  her  guns.  The  steamship  Waterwitch  had  power- 
ful steam  pumps,  wherewith  she  brought  a  great  quantity  of 
water  in  nearly  vertically,  and  sent  it  out  backwards  on  the 
two  sides  below  water  level.  The  momentum  given  to  the 
water  backwards  was  equal  to  the  momentum  given  in  the 
other  direction  to  the  ship.  It  is  on  this  principle  that  Hero's 
steam-engine  and  Barker's  mill  work,  the  momentum  given  to 
jets  of  fluid  passing  out  of  certain  pipes  being  equal  to  the 
momentum  given  in  an  opposite  direction  to  the  vessel  from 
which  the  fluid  passed.  In  all  such  cases  the  propelling  force 
in  pounds  is  numerically  equal  to  the  momentum  of  the  fluid 
which  passes  out  in  a  second.  Thus,  if  from  a  vessel  moving 
with  a  velocity  of  14  feet  per  second  water  comes  through 
orifices  of  4  square  feet  in  area  with  a  velocity  (relative  to  the 
orifices)  of  20  feet  per  second,  then  the  quantity  passing  out 
in  one  second  is  4  x  20,  or  80  cubic  feet — that  is,  80  x  62 -3,  or 
4,984  Ibs.  Now,  recollecting  that  this  water  was  first  brought 
in  and  is  now  sent  out,  what  is  the  velocity  which  we  have 
really  impressed  upon  it  in  the  process  ]  At  the  beginning  it 

*  Of  course  the  kinetic  energies  of  the  parts  of  the  shell  added  together 
are  greater  than  they  were  before  the  shell  burst ;  we  are  now  merely 
speaking  of  the  momentum.  The  total  momentum  of  two  equal  bodies  going 
in  opposite  directions  with  the  same  velocity  is  nothing,  whereas  their  total 
kinetic  energy  is  double  that  of  one  of  them. 


490  APPLIED    MECHANICS. 

was  motionless  with  respect  to  the  sea ;  it  now  has  a  velocity 
of  20 — 14,  or  6  feet  per  second  with  respect  to  the  sea,  so  that 
the  momentum  given  to  it  is  its  mass,  4, 9  84 -s- 3  2 -2  multiplied 
by  6,  or  9 28 '7  :  hence,  as  this  momentum  is  given  every 
second,  928-7  Ibs.  is  the  propelling  force  exerted  on  the  ship. 
In  one  second  the  ship  moves  through  14  feet,  so  that  the 
useful  mechanical  work  done  is  14  x  928'7,  or  13,002  foot- 
pounds. We  have  given  to  4,984  Ibs.  of  water  a  velocity  of 
6  feet  per  second,  the  kinetic  energy  of  this  water  is  wasted, 
and  this  kinetic  energy  is  J  of  4,984-^32-2x6x6,  or  2,786 
foot-pounds.  In  fact,  we  have  altogether  spent  15,788  foot- 
pounds, and  13,002  of  this  have  been  usefully  employed,  so 
that  the  efficiency  of  the  method  is  13,002 -i- 15,788,  or  -824, 
or  82*4  per  cent.  As  a  matter  of  fact,  however,  the  friction 
in  pumps  and  pipes  usually  causes  a  third  of  the  actual  horse- 
power given  out  by  the  engine  to  be  wasted,  so  that  the 
true  efficiency  of  this  method  of  propulsion  is  two-thirds  of 
the  above,  or  0'55,  or  55  per  cent.,  neglecting  the  friction 
of  the  engine  itself.  You  will  remember  a  fact  which  has 
come  in  casually  here :  if  the  water  leaves  any  turbine, 
water-wheel,  or  any  propeller  of  a  vessel  with  a  velocity 
relative  to  the  still  water  into  which  it  passes,  or  if  it  has 
any  other  form  of  energy,  this  energy  has  been  wasted. 

388.  By  calculation  you  will  find  that,  when  two  free  and 
inelastic  bodies  strike,  the  momentum  communicated  from  one 
to  the  other  is  their  relative  velocity  multiplied  by  the  product 
of  their  masses  and  divided  by  the  sum  of  their  masses,  and 
this  quotient  divided  by  the  time  of  the  impact  gives  the  mean 
pressure.     This  pressure  acts  equally  on  both,  of  course,  but  it 
may  not  hurt  both  equally.     If  the  bodies  are  surrounded  by 
water,  like   ships,  they  can   no   longer  be  regarded   as  free 
bodies,  and  it  is  not  easy  to  say  in  a  few  words  how  much 
mass  we  must  add  to  the  bodies  to  represent  the  mass  of  the 
water,  which  has  also  to  undergo  change  of  motion.     In  the 
case  of  a  ship,  the  mass  of  water  to  be  moved  broadside  on  is 
much  greater  than  when  the  ship  is  struck  stem  on. 

389.  A  body  falling  into  a  liquid  sets  it  in  motion,  and 
this  motion  appears  at  distant  places  more  and  more  nearly 
instantaneously  as  the  liquid  becomes  more  and  more  incom- 
pressible.    The  nature  of  this  motion  is  known  to  us  if  we 
know  the  velocity  of  yielding  at  the  place  of  contact,  and 
from  this   the  total    momentum  given   to  the   liquid.      This 


APPLIED   MECHANICS.  491 

represents  a  very  considerable  pressure  applied  at  the  place 
of  contact,  and  this  pressure  becomes  greater  as  the  velocity  of 
the  body,  before  it  touches  the  liquid,  increases.  Hence  a 
cannon-ball  fired  at  sea  rebounds  from  the  water  as  from  a 
rigid  body.  Hence  also  a  man  diving  unskilfully,  as  he  falls 
prone  on  the  water,  gets  a  very  unpleasant  shock,  whereas  a 
skilful  diver  enters  in  such  a  way  as  to  make  the  momentum 
of  the  moving  water  as  small  as  possible,  and  to  make  the 
creation  of  this  momentum  gradual. 

390.  If  a  body  of  inertia  or  mass  M  with  velocity  v  overtakes 
a  body  of  mass  M1  with  velocity  v1,  the  motion  being  in  the 
same  direction,  there  is  an  instant  when  they  move  at  the 
same  speed,  and  then  they  usually  separate.  The  equal  and 
opposite  forces  equalise  their  velocities  until  they  are  both 
moving  with  the  velocity  v,  such  that 

(M  +  M1)  v  =  M  v  +  M1  v1  .  .  .  .  (1), 


M  +  M1 

There  has  been  a  communication  of  the  momentum,  M  (v  —  v)t 
or  M1  (v  —  v1),  and  this  is  the  amount  of  the  impact  or  the 
time  integral  of  the  force  from  either  on  the  other.  We 
usually  imagine  the  contact  surfaces  to  be  normal  to  the 
direction  of  motion.*  During  the  impact  the  total  kinetic 
energy, 

E!  =  J  M  v2  +  |  MI  vi2  ____  (3) 


becomes  lessened  to 

E  =  \  (M  +  Ml)  &  .  .  .  .  (4), 

the  amount  Ea  —  E  being  stored  as  strain  energy.  In  truth, 
much  of  it  travels  off  and  vibrations  take  place  (see  Art.  486), 
but  let  us  speak  of  EJ  —  E  as  the  stored  energy.  Assume 
that  the  energy  of  the  bodies  when  free  after  collision 

*  If  not  normal  we  may  speculate  on  the  connection  between  the  ideas  of 
force  as  rate  of  communication  of  momentum,  the  direction  of  the  force 
being  the  same  as  that  of  the  momentum  communicated,  and  yet  the 
direction  of  communication  being  different  from  both.  Students  who  have 
leisure  will  find  three  quite  neglected  letters  in  Nature  for  1878,  by  Prof. 
K.  H.  Smith,  which  will  give  them  novel  ideas  on  the  subiect—  ideas  likely 
to  be  of  use  to  engineers. 


492  APPLIED   MECHANICS. 

A  n 

E2  =  |  (M  u    -f  M1  u1  )  is  less  than  EJ  by  an  amount  which 
is  a  fraction  of  the  stored  energy,  or 

EI  -  E2  =  k  (E!  -  E)  ----  (5), 

where  k  is  a  constant  depending  on  the  nature  of  the  materials 
and  the  shapes  of  the  bodies.     We  know  also  that 

MU  + 


From  (2),  (5),  and  (6)  we  can  calculate  u,  u1  and  v  in  terms 
of  the  initial  velocities,  and  we  find  that 

ui  -  u  =  (v  -  vi)  yT^fc. 

Or,  "  the  relative  velocity  of  separation  is  </  1  —  k  times  the 
previous  relative  velocity  of  approach."  This  was  Newton's 
assumption.  It  is  usual  to  denote  the  ratio  </  1  —  k  by  the 
letter  e,  and  to  call  e  the  "  elasticity,"  but  this  is  unscientific. 
391.  Newton  found  by  experiment  that  e  =  -f  for  com- 
pressed wool,  iron  nearly  the  same,  -^|  for  glass.  Hence  for 
these  substances  our  k  has  the  values  -^  and  -J.  That  is,  in 
the  compressed  wool  or  iron  T7^  of  the  stored  energy  is  wasted  ; 
in  glass  only  -J  of  the  stored  energy  is  wasted.  It  is  very 
difficult  for  us  to  imagine  how  this  difference  between  iron 
and  glass  should  occur  ;  and  there  are  no  recent  experiments. 
Note  that  k  =  1  or  e  =  0  indicates  the  case  of  maximum  loss 
of  energy  ;  in  fact,  there  is  maximum  loss  of  energy  when  the 
two  bodies  continue  to  move  together  as  the  bullet  and  bob  of 
a  ballistic  pendulum  do.  In  this  case  the  kinetic  energy 
before  collision  is  J  M  v3,  and  after  collision  it  is 


Suppose  v1  =  0,  or  we  have  M  impinging  on  M1  at  rest,  the 
energy  remaining  is  J  M  v2  /(  1  +  —  J.      Hence,  the  greater  M1 

is  in  comparison  with  M  the  greater  the  loss.     In  fact, 

lost  energy         _  MJ 
remaining  energy       M 

So  that  the  larger  the  stationary  body  before  collision,  the 
greater  the  loss. 

392.  Example.  —  Thirty  gallons  of  water  per  second  enter 
a  wheel  in  a  direction  A  B  from  a  horizontal  pipe  4  inches  in 


APPLIED    MECHANICS.  493 

diameter,  the  shortest  distance  from  the  axis  of  pipe  to  vertical 
axis  of  wheel  being  1  '3  feet.  The  water  leaves  the  wheel  in  a 
'horizontal  direction,  c  D,  with  a  velocity  of  3  feet  per  second, 
the  shortest  distance  between  c  D  and  the  axis  of  wheel  being 
0-8  feet.  W.hat  is  the  turning  moment?  And  what  is  the 
power  if  the  wheel  makes  150  revolutions  per  minute? 

OQA 

Ans.   The  30  gallons,  or  300  Ibs.,  or  ^7o  cubic  feet,  have 

„  300  144 

an  initial  velocity  of  7*7^0  •  42  x  -7854  ^er  seconc*>  an(* 

mass  o^o-     The  product  is  the  momentum  per  second,  and  is 

force.  Multiply  by  1-3,  and  we  have  the  moment  of  the  force 
due  to  the  entering  water,  or  668 '2  pound-feet.  The  same 
mass  multiplied  by  3  and  by  0'8  gives  22-36  pound-feet;  and 
subtracting  this  from  the  former,  we  have  the  resultant 
moment  (or  moment  of  momentum  per  second),  6 45 -8  pound- 
feet.  Multiplying  this  by  150  x  2  TT  radians  per  minute,  and 
dividing  by  33,000,  we  find  18-4,  the  horse  power. 

393.  Example. — A  ball  of  4  Ibs.,  moving  to  the  north  on  a 
smooth,  level  table  with  a  velocity  of  6  feet  per  second,  strikes 
another  ball,  and  after  the  collision  is  found  to  be  moving  at  5 
feet  per  second  to  the  east.  What  was  the  amount  of  the 
impulse?  If  the  collision  lasted  0'002  second,  what  was  the 
average  force  of  the  blow  ? 

Ans.  Subtracting  the  second  velocity  (in  the  way  vectors 
are  subtracted)  from  the  first,  we  find  that  the  sudden  gain  of 
velocity  was  ^/  61  feet  per  second  in  a  direction  making 
tan."1!^  east  of  south.  The  impulse  given  to  the  ball  was 

4       

therefore  in  this  direction,  and  of  the  amount  -o^  y/61  pound- 
seconds.  The  force,  therefore,  was  in  this  direction,  and  of 

4        

the  average  amount  ^^  J  61  -f-  0'002  lb.,  or  485  Ibs. 

EXERCISES. 

1.  Compare  the  amounts  of  momentum  in  a  pillow  of  20  Ibs.  which 
has  fallen  from  a  height  of  1  foot,  and  an  ounce  bullet  moving  at  200  feet 
per  second.  Ans.,  12*8  to  1. 

2.  A  ball  of  56  Ibs.  is  projected  with  a  velocity  of  1,000  feet  per 
second  from  a  gun  weighing  8  tons.     What  is  the  maximum  velocity  of 
recoil  of  the  gun?  Ans.,  3-1  feet  per  second 


494  APPLIED    MECHANICS. 

3.  There  are  two  bodies  whose  masses  are  in  the  ratio  of  2  to  3,  and 
their  velocities  in  the  ratio  of  21  to  16.     What  is  the  ratio  of  their 
momenta  ?     If  their  momenta  are  due  to  constant  forces  acting  on  the 
bodies'  respectives  for  times  which  are  in  the  ratio  of  3  to  4,  what  is  the 
ratio  of  these  forces  ?  Ans.,  7  to  8 ;  7  to  6.  <% 

4.  A  body  weighing  10  Ibs.  impinges  on  a  fixed  plane  with  a  velocity 
of  20  feet  per  second.     If  the  coefficient  of  restitution  e  is  0-5,  find  the 
velocity  of  rebound,  and  how  many  foot-pounds  of  energy  are  wasted  in 
the  collision.  Ans^  10  feet  per  second ;  93-17.>  T  £ 

5.  A  ball  of  6  ounces  strikes  a  bat  with  a  velocity  of  10  feet  per 
second,  and  returns  with  a  velocity  of  30  feet  per  second.     If  the  duration 
of  the  blow  be  -^  second,  find  the  average  (time)  force  exerted  by  the 
striker.  Ans.,  9'32  Ibs. 

6.  A  body  weighing  50  Ibs.  moving  at  the  rate  of  10  feet  per  second 
overtakes  another  body  of  25  Ibs.  moving  at  the  rate  of  6  feet  per  second. 
If  both  masses  be  perfectly  elastic,  find  their  velocities  after  the  shock. 

Ans.,  7  £  and  11£  feet  per  second. 

7.  A  hammer-head  of  2|  Ibs.,  moving  with  a  velocity  of  50  feet  per 
second,  is  stopped  in  0-001  second.     What  is  the  average  force  of  the 
blow  ?  Ans.,  3,882  Ibs. 

8.  A  shell  bursts  into  two  fragments,  whose  weights  are  12  and  20  Ibs. 
The  former  travels  onward  with  a  A^elocity  of  700  feet  per  second,  and 
the  latter  with  a  velocity  of   380  feet  per  second.     What   was  the 
momentum  of  the  shell  when  the  explosion  took  place  ?        Ans.,  496-8. 

9.  A  shell  weighing  20  Ibs.  explodes  when  in  motion  with  a  velocity 
of  600  feet  per  second.     At  the  moment  of  explosion  one-third  of  the 
shell  is  reduced  to  rest.     Find  the  momentum  of  the  other  two-thirds. 

Ans.,  372-7. 

10.  Water  is  flowing  through  a  service  pipe  at  the  rate  of  60  feet  per 
second.     If  the  water  be  brought  to  rest  uniformly  in  one-tenth  second 
by  closing  the  stop-valve,  what  will  be  the  increase  of  the  pressure  of  the 
water  near  the  valve,  the  pipe  being  taken  as  50  feet  long,  the  resistance 
of  the  pipe  and  the  compressibility  of  the  water  being  neglected  ? 

Am.,  403  Ibs.  per  sq.  in. 

394.  As  for  the  way  in  which  the  vibrations  take  place  in 
two  colliding  bodies,  if  mathematically  treated  it  is  generally 
difficult;    but  very  good  working  notions  for   the   engineer 
are  derivable   from   the   results   of  experimental   study,  the 
subject  being  taken  up  in  books  on  physics  under  the  head 
"  Acoustics."     The  best  mathematical  treatise  is  Lord  Ray- 
leigh's  "  Sound."     The  time  of  an  impact  is  known  in  several 
cases  in  which  it  has  been  calculated  from  theory.     (See  Art. 
404.) 

395.  Let  us  consider  what  takes  place  when  two  ivory  balls 
come  together.     There  is  a  certain  instant  after  they  first 
touch  when  their  centres  move  together  just  as  if  they  were 
composed  of  soft  clay — then  they  act  on  each  other  with  their 
greatest  pressure ;   they  are  in  their  most  strained  condition, 


APPLIED    MECHANICS.  495 

and  supposing  no  loss  by  internal  friction,  the  strain  in  the 
balls  represents  an  amount  of  stored-up  energy  (see  Arts.  259, 
267)  equal  to  the  kinetic  energy  which  the  bodies  have  lost. 
It  is  very  important  to  remember  this  fact,  that  if  bodies  are 
to  return  to  their  old  states  after  the  collision,  we  must 
suppose  that  during  the  collision  there  is  a  storage  of  kinetic 
energy  in  the  form  of  strain.  All  the  kinetic  energy  will  not 
be  given  out  again,  nor  can  we  say  that  it  is  all  stored,  because 
there  is  a  sort  of  internal  friction  causing  part  of  the  strain 
energy  to  be  converted  into  heat  when  any  change  occurs. 
Now,  if  the  whole  of  the  stored-up  energy  is  confined  to  one 
portion  of  the  body,  the  strain  may  be  too  great.  Thus,  a 
Steel  rod  1  square  inch  in  section,  1  foot  long,  will  store  up 
167  foot-pounds  of  strain  energy  in  its  stretched  condition 
before  it  breaks.  For  suppose  breaking  stress  to  be  100,000 
Ibs.  per  square  inch.  This  will  occur  when  there  is  a  length 
ening  or  shortening  of  -0033  foot,  so  that  the  energy  stored 
up  is  the  work  done  by  a  force  whose  average  amount  is 
50,000  Ibs.  acting  through  -0033  foot,  or  167  foot-pounds.  If 
2  feet  of  the  same  rod  stored  up  this  same  amount  of  energy, 
there  would  only  be  83  foot-pounds  in  each  foot  of  its  length ; 
and  it  is  easy  to  see  that  the  stress  is  no  longer  the  breaking 
stress  of  100,000  Ibs.  per  square  inch,  but  only  70,700  Ibs. 
per  square  inch.  As  we  store  the  same  amount  of  energy  in 
smaller  and  smaller  portions  of  a  body,  it  is  evident  that  we 
must  approach  a  condition  of  fracture. 

396.  We  see,  then,  that  at  the  place  where  contact  occurs, 
two  bodies,  A  and  B,  are  strained ;  but  if  A  is  of  some  very 
elastic  material,  such  as  tempered  steel,  the  strain  energy  is 
conveyed  very  rapidly  to  every  part  of  the  body ;  whereas  if  B 
is  a  feebly  elastic  body,  the  strain  accumulates  at  one  place, 
leaving  the  rest  of  the  body  unstrained,  whilst  at  this  place 
the  strain  may  produce  fracture.  This  slowness  to  communi- 
cate strain  to  the  rest  of  the  body  may  also  be  produced  by 
the  shape  of  the  body.  For  instance,  a  rod  struck  sidewise  or 
a  thin  plate  struck  in  the  middle  does  not  so  immediately  com- 
municate its  strain  to  the  remote  parts  as  a  rod  struck  end- 
wise. Again,  the  nature  of  the  parts  of  A  and  B  in  contact 
may  be  such  that  not  only  does  the  strain  energy  leave  this 
part  of  A  rapidly,  but  immediately  in  the  neighbourhood  of 
the  place  of  contact  there  is  a  greater  capacity  to  bear  strain 
energy  without  rupture  than  is  the  case  with  B.  Thus,  when 


496  APPLIED    MECHANICS. 

ship  A  ranis  the  broadside  of  ship  B,  the  side  of  B  is  bent 
inwards  and  the  strain  energy  produced  is  accumulated  near 
the  place  of  contact  till  fracture  occurs ;  whereas,  not  only  is 
A'S  stem  able  to  transmit  to  all  parts  of  A  with  great  rapidity 
the  strain  energy  which  must  be  stored  up  in  the  whole  mass, 
but  at  the  stem  itself  the  material  of  A  is  capable  of  with- 
standing greater  stresses  than  the  material  of  B'S  side. 
Suppose,  however,  that  A'S  stem  is  not  of  steel,  still  B'S  iron 
or  wooden  side  will  be  perforated  if  A  has  enough  velocity ; 
A'S  stem  may  also  be  damaged  in  the  impact  in  such  a  case. 

397.  A  candle  may  be  fired,  it  is  said,  through  a  thin 
deal  board  with  very  little  injury  to  its  shape,  and  the  usual 
explanation  of  this  phenomenon  given  in  books  is  that  the 
candle  has  not  time  to  get  broken.  This  explanation  is  not 
eatisfactory  ;  it  is  a  little  too  vague.  If  we  had  the  board  in 
rapid  motion,  and  striking  the  candle  in  the  same  relative 
position,  the  candle  having  previously  been  at  rest,  would  the 
candle  perforate  the  board  1  There  cannot  be  any  doubt  that 
it  would.  Hence  it  is  not  the  body  struck  which  must  in 
every  case  get  hurt ;  the  pressure  on  one  is  equal  to  that  on 
the  other.  Suppose  ship  A  rushes  at  ship  B  when  B  is  broad- 
side on,  and  rams  her,  B  will  probably  be  sunk,  even  if  she  is 
a  much  larger  and  better  ship  than  A.  But  suppose  that  B  is 
able  to  meet  her  adversary  stem  to  stem,  if  they  are  equally 
strong  they  will  equally  injure  one  another,  and  if  B  is  the 
stronger  A  will  suffer  the  most.  This  case  differs  very  much 
from  that  of  the  candle,  because  we  can  assume  greater 
strength  even  for  slowly  applied  pressures  from  stem  to 
stern  of  the  ship  A  than  from  side  to  side  of  B ;  whereas  the 
strength  of  the  candle  for  slowly  applied  pressures  cannot  be 
compared  with  that  of  the  wood  which  it  punches  from  the 
board.  What  is  meant  by  the  usual  explanation,  "  The  candle 
has  not  time  to  get  deformed  "  ?  Why  has  not  the  soft  candle 
time  to  get  broken,  and  yet  the  wood  has  time  to  get  torn 
asunder  1  The  fact  is,  the  wood,  if  it  were  slowly  pressed, 
would  communicate  its  strain  energy  to  every  part  of  the 
board  and  its  supports ;  but  this  communication  takes  an 
appreciable  interval  of  time,  however  suddenly  the  pressure 
may  be  applied,  or  however  great  it  may  be.  As  the  strain 
energy  is  rapidly  produced  it  becomes  accumulated  near  the 
place  of  contact  to  such  an  extent  as  to  produce  fracture  of  the 
wood.  Now  the  point  of  the  candle  is  subjected  to  the  same 


APPLIED    MECHANICS.  497 

pressure  as  the  wood,  and  begins  to  get  spoiled  in  shape — that 
is,  it  is  compressed — and  this  compression  produces  a  lateral 
spreading.  In  the  meantime,  however,  the  compressive  strain 
energy  is  communicated  very  rapidly  backwards  along  the 
candle,  and  the  spreading  and  spoiling  goes  on  along  its  entire 
length,  but  is  small  at  any  point,  since  it  is  distributed  over 
the  whole  mass.  Practically,  therefore,  the  spoiling  occurs 
only  at  the  point  of  the  candle,  since  time  is  needed  for 
fracture  of  the  material. 

398.  An  earthquake,  when  it  acts  on  a  house,  usually 
tends  to  move  it  through  a  distance  of  probably  a  very  small 
fraction  of  an  inch-,  but  it  does  this  in  a  very  short  time — that 
is,  the  house  gets  a  considerable  velocity.     The  mass  of  the 
house  multiplied  by  the  greatest  velocity,  and  divided  by  the 
short  time  during  which  the  momentum   is  being  communi- 
cated, gives  the  pressure  which  the  foundations  of  the  house 
are  subjected  to.     Now,  when  the  foundations  are  not  very 
rigidly  connected  with  the  ground,  the  time  of  communication 
of  the  momentum  is  lengthened,  and  the  pressure  is  conse- 
quently   diminished.       This   is   the   usual   Japanese   plan   of 
providing   for   earthquake   effects.      Unfortunately,  the  very 
means  taken  to  diminish  the  pressure  on  the  foundations  also 
diminishes  their  capability  of  withstanding  forces,  and  it  has 
not  yet  been  decided  what  sort  of  a  house  is  best  fitted  to  with- 
stand destructive  earthquakes.     Want  of  rigidity,  combined 
with  strength  or  toughness  in  the  materials,  and  especially  the 
quality  of  internal  friction  in  the  materials,  so  that  vibrations 
may  rapidly  die  away — these  are  the  qualities  needed.     They 
are  found  in  steel,  wrought  iron,  and  wood,  and  especially  in 
wicker-work,  in  a  less  degree  in  cast  iron  and  in  brick  or  stone 
set  in  cement,  and  less  still  in  brick  and  stone  set  in  bad 
mortar. 

399.  When  you  in  some  way  understand  the  possibility  of 
a  candle  perforating  a  board,  you  will  be  able  to  comprehend 
how  sand,  when  blown  in  air  against  tempered  steel,  is  able  to 
abrade  it ;  how  the  emery  wheel  and  grindstone  going  at  great- 
velocities  are  able  to  cut  into  hard  metals  ;  and  how  in  Cali- 
fornia a  jet  of  water  going  with  very  great  velocity  is  used  for 
mining  purposes  instead  of  iron  tools. 

400.  Quasi-Rigidity  Produced  by  Rapid  Motion. — A  top 
when  not  spinning  can  with  difficulty  be  balanced  on  its  point, 
and  if  left  to  itself  it  almost  instantly  falls ;  whereas  when  it 


498  APPLIED    MECHANICS. 

is  spinning  the  effect  of  slightly  tilting  it  out  of  the  perpen- 
dicular is  not  to  make  it  fall,  but  to  make  it  take  a  slow 
precessional  motion. 

There  is  a  piece  of  apparatus  called  a  g'yrostat,  which  is, 
in  a  more  or  less  perfect  form,  to  be  found  in  every  mechanical 
laboratory,  and  the  student  ought  to  experiment  for  himself 
with  this  apparatus  on  the  curious  effects  of  quasi  rigidity 
which  manifest  themselves  in  tops  and  other  spinning  bodies. 
If  he  has  a  slight  acquaintance  with  astronomy  he  will  be 
interested  in  tracing  the  connection  between  the  behaviour  of 
a  tilted  top  and  the  precession  of  the  earth's  equinoxes. 

When  a  circular  sheet  of  drawing-paper  is  mounted  like 
a  very  thin  grindstone  on  an  axis,  and  is  gradually  made  to 
rotate  rapidly,  it  is  found  to  have  become  quite  rigid — that  is, 
it  greatly  resists  bending  as  if  it  were  made  of  steel.  In  the 
same  way  a  long  loop  of  rope,  hanging  round  a  high  pulley, 
which  gives  it  a  quick  motion,  takes  a  certain  form  which  it  is 
very  difficult  to  alter,  as  may  be  shown  by  striking  it  with  the 
hand  or  with  a  stick  :  it  resembles  more  a  rigid  rod  than  a 
flexible  rope. 

Again,  in  the  well-known  lecture-experiments  on  smoke 
rings,  we  see  that  these  little  whirlpools  of  air  have  many 
properties  in  common  with  elastic  solid  bodies  on  account  of 
the  partial  rigidity  which  is  due  to  their  rapid  motion. 

In  objects  which  spin  and  rub  on  a  level  surface,  like  tops, 
we  have  the  interesting  general  rule,  "Positions  which  are 
stable  when  there  is  no  spin,  are  unstable  when  there  is  spin, 
and  vice-versd."  Students  of  statics  insist  on  a  low  centre  of 
gravity  in  vehicles ;  students  of  dynamics  sometimes  insist  on 
a  high  centre  of  gravity,  as  giving  greater  stability  when  there 
is  rapid  motion. 

It  would  be  beyond  the  scope  of  a  book  like  this  to  explain 
these  curious  phenomena,  and  I  merely  direct  the  attention  of 
students  to  these  instances  in  order  to  incite  them  to  make 
experiments,  and  to  seek  for  the  explanation  of  what  they 
observe.  My  popular  lecture  on  Spinning  Tops  may  be  worth 
reading. 

401.  Motion  Produced  by  a  Blow. — When  a  body  sub- 
jected to  a  blow  is  quite  free  to  move  in  any  way,  unless  the 
blow  acts  through  its  centre  of  gravity,  the  body  will  not 
merely  move  as  a  whole,  but  it  will  revolve.  When  the  blow 
acts  in  a  direction  through  the  centre  of  gravity  there  is  no 


APPLIED    MECHANICS. 


499 


rotation  produced.  It  is  usual  in  such  a  case  to  consider  the 
motion  of  the  centre  of  gravity  of  the  body,  and  the  motion  of 
the  body  about  an  axis  through  its  centre  of  gravity,  for  it  is 
known  that  any  motion  whatsoever 
of  the  body  consists  of  a  combination 
of  two  such  motions.  It  is  found 
that  the  kinetic  energy  communicated 
to  a  body  by  means  of  a  blow  it. 
best  calculated  in  the  following  way. 
if  we  know  the  nature  of  the  motion  ; 
— First,  find  the  kinetic  energy,  a* 
if  every  portion  of  the  body  had  the 
motion  of  the  centre  of  gravity. 
Secondly,  find  the  kinetic  energy  of 
rotation  as  if  the  axis  of  rotation 
through  the  centre  of  gravity  were 
fixed  in  space.  Add  these  two  re- 
sults together.  We  may  regard  from 
another  point  of  view  the  instan- 
taneous motion  of  a  body  when  it  is 
struck — namely,  as  a  rotation  about 
some  axis  which  does  not  itself  move. 
This  is  only  the  case  for  an  instant ; 
immediately  afterwards  we  must  re- 
gard the  body  as  moving  about  another 
fixed  axis.  If  a  body  is  hinged  so 
that  it  can  only  move  about  a  fixed 
axis,  it  is  always  possible  to  find  the 
point  at  which  the  body  may  be 
struck,  and  the  direction  of  the  blow, 
which  will  tend  to  produce  an  instan- 
taneous rotation  about  this  particular 
axis,  and  therefore  to  produce  no 
pressure  at  the  hinge.  Thus  the 
ballistic  pendulum  of  Fig.  278  is 
always  struck  in  such  a  way,  and  Fig.  278. 

the  point   in   which   it   is  struck  is 

called  the  "  centre  of  percussion."  An  easy  way  to  find  the 
centre  of  percussion  is  as  follows : — Make  the  body  vibrate 
like  a  pendulum,  about  its  axis  of  suspension,  under  the  action 
of  gravity.  Now  find  the  length  of  the  equivalent  simple  pen- 
dulum. This  is  the  distance  of  the  centre  of  percussion  from 


500  APPLIED    MECHANICS. 

the  axis.  In  a  tilt  hammer  all  blows  ought  to  be  delivered 
from  this  centre  of  percussion  if  we  wish  to  have  no  pressure 
on  the  bearings.  A  cricket-bat  or  a  rod  of  iron  tingles  the 
hand  when  we  strike  a  blow  with  it,  unless  we  happen  to 
strike  at  the  centre  of  percussion.  For  a  rod  of  iron  free  to 
move  about  one  end,  the  centre  of  percussion  is  at  two-thirds 
of  the  way  towards  the  other  end.  (See  Art.  454.) 

402.  The  Ballistic  Pendulum  of  Fig.  278  is  a  contrivance 
\*hich  enables  us  to  measure  the  velocity  of  a  bullet.  It  con- 
sists of  a  mass  of  wood,  A,  forming  part  of  a  pendulum.  The 
bullet  is  fired  into  it,  and  the  wood  swings  backwards  in  con- 
sequence. The  bullet  is  fired  in  such  a  way  that  it  will 
cause  no  jar  to  be  given  to  the  pivot  B.  The  momentum 
existing  in  the  bullet  before  it  enters  the  wood  belongs  now  to 
the  whole  mass  of  which  it  becomes  a  part.  C  is  a  silk  ribbon 
which  is  pulled  through  a  moderately  tight  hole  or  over  the 
edge  of  a  table  by  the  swing  of  the  pendulum,  and  the  length 
of  ribbon  pulled  through  is  found  to  be  proportional  to  the 
momentum  of  the  bullet  before  entering  the  wood. 

If  w  is  the  weight  of  the  bullet,  its  horizontal  velocity  in  the 
plane  of  swinging  heing  v,  its  momentum  is  —  v.  The  moment  of 

momentum  x  -  v  (if  x  is  the  perpendicular  distance  from  B  to  the 

axis  of  the  bullet)  "before  impact  is  equal  to  the  moment  of 
momentum  after  impact ;  and  if  i  is  the  moment  of  inertia  of  A 
and  the  bullet  ahout  B  after  collision,  and  a  is  its  initial  angular 

velocity,  then  i  a  =  z  —  v  .....  (I}.     A  certain  part  of  the  whole 

kinetic  energy  has  "been  converted  into  heat ;  the  mechanical 
energy  now  in  the  system,  3  i  a2,  will  be  converted  into  potential 
energy,  w  being  the  total  weight,  the  centre  of  gravity  will  be 
raised  through  the  distance  h  such  that  w  h  =  ^  i  a2  .  .  .  .  (2) ;  so 
that,  as  a  is  known,  h  is  known.  Knowing  h,  we  can  either  find  0, 
the  total  swing,  or  we  can  calculate  or  find  experimentally  the 
length  of  the  ribbon  which  may  "be  drawn  up,  and  from  the  length 
of  the  ribbon  we  can  calculate  v. 

In  the  short  time  when  the  bullet  is  being  lodged  it  is  exercising 
horizontal  force  at  each  instant,  and  there  may  be  horizontal  force 
at  B  in  the  same  direction.  AVe  may  say  that  these  external  forces, 
acting  at  the  centre  of  gravity,  arc  balanced  by  the  mass  of  the 
whole  body  multiplied  by  the  acceleration  of  the  centre  of  gravity 
at  each  instant,  and  hence  their  integral  effects  balance ,  that  is, 

then,  the  impulse  —  v  +  p  (where  P  is  the  horizontal  impulse  from 


APPLIED    MECHANICS.  501 

the  point  of  support  in  the  same  direction)   =     -  a  .  B  o  if  o  is 

the  centre  of  gravity,  "because  a  .  B  o  is  the  velocity  produced  in 
o,  and  w  is  the  weight  of  the  pendulum  and  bullet-  Hence 

p=-«.BQ--t; (3), 

9  9 

This  is  0,  therefore,  if 

w.a.BQ  =  «w....  (4). 

But  by  (1),  (4)  means  that  w  .  a  BG  =  w  JPa/x,  or  x  =  &2/B  G. 
Now  this  is  the  rule  (Art.  454)  by  which  we  find  the  length 
of  the  equivalent  simple  pendulum  or  the  distance  to  the  point 
p  of  oscillation  (Art.  454).  It  also  for  this  reason  gets  the  name 
"  point  of  percussion." 


APPENDIX   TO    CHAPTER  XXIII. 

403.  In  the  Example,  Art.  385a,  we  had  an  example  of  stop- 
page of  water  in  a  pipe  as  if  all  the  water  had  exactly  the  same 
motion.  It  was  an  interesting  academic  exercise.  The  following 
exercise  is  also  interesting,  and  will  give  us  more  information. 

Sudden  Stoppage  of  Water  in  a  Pipe  of  Uniform  Section. — 

Suppose  the  pipe  to  be  infinitely  rigid.  It  will  be  found  that  the 
effects  are  independent  of  whether  the  pipe  is  vertical  or  horizontal, 
but  we  shall  consider  it  to  be  vertical.  It  will  be  found  on  closer 
examination  that  if  we  know  the  pressure  we  need  only  con- 
sider, in  our  problem,  j9,  the  pressure  above  the  normal  pressure, 
and  this  is  what  we  shall  do.  Let  v  be  the  axial  velocity 
of  a  particle  of  water,  w  the  weight  per  unit  volume,  so  that 
w/ff  is  its  mass  per  unit  volume,  often  called  p;  if  K  is  the 
cubical  elasticity  (during  quick  changes  of  volume)  and  t 
is  time ;  if  x  is  distance  to  a  point  o,  measured  along  the  pipe  in 
the  direction  of  motion,  and  if  the  pipe  is  taken  of  unit  cross- 
sectional  area;  if  x  is  the  total  distance  already  travelled  by 

a  particle  at  Q,  so  that  at  x  +  1,  x  becomes  x  +  -^-,  fluid 
which  once  occupied  length  1  of  the  pipe  now  occupies  length 

1  +  -r-,  so  that  its  increased  volume  is  ^— .     Hence,  as  pressure- 
dx  ax 

producing  volumetric  strain  =  K.  x  compressive  strain, 
_  K^X 

Now  velocity  v  =  -^,  and  as  the  mass  between  x  and  x  +  5x  is 
p  .  tix  and  the  pressures  are  p  and  p  +  $p,  we  have  the  force  -  9p 


502  APPLIED    MECHANICS. 

d2  X  . 

causing  acceleration  —  Q   in  the  mass  p  .  8#,  and  hence 


d2  x  dp  d?  x 

-if       - 

^X 

—  ^ 


rf2  X  ^X 

or,  from  (1),  K          =  p  —  ^  .     If  K/p  or  Kff/tv  be  called  a2,  we  have 


2x  = 

^  ^2     ' 

and  we  recognise  this  as  the  equation  of  wave  propagation  with 
the  velocity  a.     Differentiate  (3)  with  regard  to  t  and  use  v  for 

„  cPv       cPv 
dxfdt,  and  we  have  a  2—  2  =  ^-2. 

Let  our  problem  be  this.  When  water  is  flowing  with  a 
uniform  velocity  -  v0,  let  an  infinitely  rigid,  thin  diaphragm 
suddenly  produce  a  stoppage  at  x  =  0  ;  what  will  happen  ?  Con- 
sider the  pipe  only  for  positive  values  of  x.  We  have  v  =  —  v« 
everywhere  at  t  =  0,  and  at  x  =  0  we-  have  v  such  a  function  ol 
the  time  that  it  was  -  v0  till  t  =  0,  and  ever  afterwards  v  there 
isO. 

To  make  things  simpler,  let  v1  be  a  new  variable  &uch  that 
v1  =  v  +  v0,  and  therefore  the  conditions  are  that  v1  =  0  every- 
where at  t  =  0,  and  at  x  =  0  we  have  v1  such  a  function  of  the  time 
that  it  was  0  till  t  =  0,  and  ever  afterwards  v1  —  t>0.  We  see  that 


If  we  let  ^-  be  denoted  by  9  in  the  usual  symbolical  way,  (4)  is 


Solving  this  as  the  simplest  linear  equation  would  be  solved  if  6 
were  a  quantity  independent  of  x,  we  have 


if  A  and  B  are  functions  of  the  time  corresponding  to  x  =  0.     As 
x  may  be  infinite,  B  ~  0  ;  so  that 


where  /  (t)   is   0  till  t  =  0.  and  then   is  t;^   and  remains  of  tho 
value  v0  ;  so  that 

*  =   ~  •„  +/('  -  */«)  ----  (6>-* 

We  see  now  what  occurs.     Until  t  -  x\a  =  0  or  t  =  xja,  v  =  -  v0 

at  any  place  ;   and  after  t  =  x/a,  v  =  0.     Hence  we  have  from 

=  0  a  state  of  no  velocity  spreading  along  the  pipe  with  the 

*  The  symbolic  f™f(t)  means  f(t  +  6) 


APPLIED   MECHANICS.  603 

velocity  «  to  infinity.     Since  v  =  -^t  if  we  integrate  (6)  in  regard 

to  time,  we  have 

x  =   -  V  +  *  (*  -  */«)  ----  (7), 

where  F  is  such  that  —jjp  =  /  (y\  whatever  y  may  bo.     Hence, 

using  (I),  p  =  —  f  (t  -  xfa);  or,  since 

K/a  —  K  tt>  i  /  K  i  g\  —  K  £  w$  /  ffv  , 
we  have  p  =    //—  /  (*  -*/«)....  (8). 

° 


This  tells  us  that,?  =  0  till  t  -  xja  =  0,  and  then  is  t>0  A      —  ,  and 

remains  of  this  value.  This  is  the  state  of  pressure  produced  every- 
where with  perfect  rest  accompanying  it,  at  the  velocity  a  from  the 
place  of  the  sudden  stoppage  if  the  pipe  is  infinitely  rigid. 

If  at  x  =  I  there  is  a  place  where  the  pressure  is  kept  constant 
^we  say  p  =  0  there),  at  zero,  our  wave  is  reflected  as  a  wave  of 
velocity  +  r0  and  no  pressure,  till  on  reaching  x  =  0  it  is  again 
reflected  as  a  wave  of  no  velocity,  and  so  on.  As  the  pipe  is  not 
infinitely  rigid,  and  as  there  is  friction,  the  wave  really  diminishes 
in  its  values  of  velocity  and  pressure  as  it  travels,  but  we  may  take 
a  pressure  approaching  the  value  v0  */Tnw\g  as  being  instantly  pro- 
duced by  a  sudden  stoppage.  Experiments  with  suddenly  closed 
valves  give  measured  pressures  in  the  laboratory  considerably  less, 
partly  because  the  stoppage  is  not  one  of  infinite  suddenness,  partly 
because  of  the  inertia  of  the  pressure-measuring  apparatus.  In  the 
case  of  water  the  pressure  in  pounds  per  square  inch  is  20  v0  if  v0  is 
the  velocity  in  feet  per  second. 

This  great  pressure  produced  by  stoppage  is  taken  advantage  of 
in  the  hydraulic  ram.  If  the  bottom  of  the  sea  were  smooth,  and 
a  sea  with  a  translational  velocity  v0  were  suddenly  stopped  on  its 
motion  by  a  vertical  wall,  the  pressure  on  the  wall  would  be  what 
we  have  given  above,  for  a  very  short  time,  if  the  wall  were  per- 
fectly rigid  ;  what  would  occur  subsequently  I  do  not  know,  be- 
cause there  is  atmospheric  pressure  at  the  surface  of  the  sea.  .  We 
know  enough,  however,  to  see  the  necessity  for  some  springiness  at 
the  wall  surface.  It  is  for  the  same  sort  of.  reason  that  heavy  seas 
produce  so  much  damage  sometimes  when  objects  are  struck  by 
them  on  board  ship.  Curious  stories  are  told  by  sailors  of  half  -inch 
bars  of  iron  being  bent  and  broken  by  seas  coming  over  the  bul- 
warks, and  it  is  just  possible  that  they  may  be  true,  although  it  is 
more  probable  that  such  fractures  are  due  to  blows  from  passing 
wreckage.  Anyone  who  has  seen  the  ruined  breakwater  at  Wick 
will  believe  in  the  greatness  of  the  forces  due  to  blows  from  ocean 
waves. 

404.  Students  will  find  it  an  excellent  easy  mathematical  exercise 
to  assume  that  in  an  infinite  length  of  pipe  filled  with  water  there  is 
a  piston  whose  displacement  is  a  pure  function  of  the  time.  The 


504  APPLIED   MECHANICS. 

problem  is  identical  with  the  simplest  problem  in  Telephonic 
Signalling.  If  the  pipe  is  supposed  to  yield  and  to  be  leaky,  and 
if  the  viscosity  of  the  liquid  is  considered,  we  have  the  same 
problem  as  that  of  the  Philosophical  Magazine,  page  223,  August, 
1893. 

When  a  prismatic  bar  of  length  ^  and  velocity  vl  in  the  direc- 
tion of  the  common  axis  overtakes  a  bar  of  length  72  (greater  than 
*7),  moving  with  a  velocity  v2,  the  bars  being  of  the  same  material 
and  cross  section,  Mr.  Love  says  that  the  ends  at  the  junction 
move  with  a  common  velocity  ^  (^  +  v2),  and  a  compressive  strain 
i  (vi  ~  V2)la^  produced,  a  is  the  velocity  of  a  longitudinal  wave 
of  sound,  or  E  =  a*p  where  E  is  Young's  modulus  and  p  is  the  mass 
of  the  bar  per  unit  volume,  or  our  w/g  above.  Waves  of  compres- 
sion run  from  the  junction  along  both  bars,  and  each  element  of 
either  bar,  as  the  wave  passes  over  it,  takes  suddenly  the  velocity 
£  (#j  +  vz)  and  the  compression  \  (v\  -  v2~)/a. 

When  the  wave  reaches  the  free  end  of  the  shorter  bar  it  is 
reflected  as  a  wave  of  extension ;  each  element  of  the  bar  as  the 
wave  passes  over  it  takes  suddenly  the  velocity  which  initially 
belonged  to  the  longer  bar  and  zero  extension.  After  a  time,  equal 
to  twice  that  required  by  a  wave  of  compression  to  travel  over  the 
shorter  bar,  this  bar  has  uniform  velocity,  equal  to  that  which 
originally  belonged  to  the  longer  bar,  and  no  strain. 

The  impact  now  ceases,  and  there  is  in  the  longer  bar  a  wave  of 
compression  of  length  equal  to  twice  that  of  the  shorter  bar.  The 
wave  at  this  instant  leaves  the  junction,  and  the  junction  end  of 
the  longer  bar  takes  a  velocity  equal  to  that  which  it  had  before 
the  impact.  The  ends,  therefore,  remain  in  contact  without  pres- 
sure. This  state  of  things  continues  until  the  wave  returns 
reflected  from  the  further  end  of  the  longer  bar.  When  the  time 
from  the  beginning  of  the  impact  is  equal  to  twice  that  required  by 
a  wave  of  compression  to  travel  over  the  longer  bar,  the  junction 
end  of  the  latter  suddenly  acquires  a  velocity  equal  to  that  origin- 
ally possessed  by  the  shorter  bar  and  the  bars  separate.  The 
shorter  bar  rebounds  without  strain  and  with  the  velocity  of  the 
longer,  and  the  longer  bar  rebounds  vibrating. 


605 


CHAPTER   XXIV. 

FLUIDS        IN       MOTION. 

405.  WE  tried  in  Art.  145  to  give  exact  notions  on  the 
subject    of    pressure   in   fluids    not  in    motion.      When  we 
supposed  that  the  weight  of  the  fluid  itself  was  insignificant, 
we  found  that   the  pressure  on  each  square  inch  of   surface 
touched  by  the  water,  and  on  each  square  inch  of  interface 
separating  two  portions  of  the  fluid,  was  everywhere  the  same. 

One  of  the  best  methods  of  observing  the  pressure,  any- 
where, is  by  inserting  a  pressure  gauge. .  The  gauge  indicates 
the  pressure  per  square  inch  at  the  part  of  the  liquid  to  which 
a  communicating  little  gauge  tube  penetrates.  It  always  does 
so  when  there  is  no  motion  of  the  fluid  at  the  point  in  ques- 
tion ;  but  unfortunately,  when  there  is  motion  there,  the 
introduction  of  the  tube  alters  the  motion  there  and  more 
or  less  falsifies  our  measurement. 

We  are  now  about  to  consider  pressure  in  fluids  in  motion, 
and  we  approach  our  subject  by  first  speaking  of  pumps. 

406.  A  pump  is  a  machine  which  gives  energy  to  water — 
that   is,  it  can   raise  water  to   a   height,  giving  it  potential 
energy.     It  can  force  water  into  a  vessel  under  great  pressure, 
moving  pistons — that  is,  it  can  give  to  water  pressure  energy. 
It  can  set  water  in  motion — that  is,  give  it  kinetic  energy. 

Reciprocating-pumps  are  either  lift-pumps,  or  force-pumps, 
or  combinations  of  both. 

In  Art.  137  we  described  the  force-pump  used  with  hydrau 
lie  presses.  In  principle  this  is  the  same  as  all  other  force- 
pumps.  The  feed  and  other  pumps  of  steam-engines,  and  the 
pumping  parts  of  pumping  engines,  are  described  in  detail  in 
works  on  those  subjects.  As  to  Lifting  tumps,  in  Fig.  279 
we  have  a  diagrammatic  representation  of  the  common  village 
or  house  pump.  The  rod  R,  usually  worked  by  means  of  a 
handle  or  lever,  pulls  the  bucket,  B,  up  and  down.  The 
bucket  has  openings  arranged  with  flaps  or  other  valves  in 
various  ways,  so  that  fluid  may  pass  upwards  but  not  down- 
wards. At  w  there  is  another  valve  which  opens  upwards. 
First  suppose  there  is  only  air  between  B  and  the  well,  J. 
1  Sometimes  in  the  morning  it  is  necessary  to  throw  in  some 


506 


APPLIED    MECHANICS. 


water  on  the  top  of  B  to  make  it  air-tight.     As  B  descends, 

much  of  the  air  between  B  and  w  passes  through  B.  As  B  is 
lifted  it  produces  a  partial  vacuum  below 
it ;  air  passes  from  H  to  the  barrel,  and 
water  rises  in  H.  As  the  pumping  pro- 
ceeds there  is  more  of  a  vacuum  until 
water  fills  H  and  the  barrel,  in  case  B  is 
not  nearly  34  feet  above  the  level  in  the 
well.  25  feet  is  probably  the  maximum 
height  in  house  pumps.  After  this  it  is 
water  that  pasaes  through  B  in  every  down 
stroke,  and  is  lifted  in  every  up  stroke. 

When  a  higher  lift  is  needed,  we 
sometimes  place  a  valve  in  the  upper  part 
of  the  barrel,  or  in  the  delivery  pipe,  to 
prevent  the  return  of  the  lifted  water. 
Fig.  280  is  a  dia- 
grammatic repre- 
sentation of  a 
force  -pump. 
Sometimes  a  piston 
is  used  instead  of 
the  plunger,  p.  A 
is  an  air-vessel. 

As  more  and  more  water  enters  at  B, 

the  pressure  of  the  air  becomes  great 

enough    to    force    the  water   up    the 

delivery   pipe,    E.      In   this   case  the 

stream   is  not   so  intermittent.     Un- 
fortunately the  air  is  apt  to  dissolve 

in  the  water  and  get  carried  off  at  a 

rate  which  depends  upon  how  much 

air  is  already  in  the  water.     Fig.  281 

shows   a   double-acting   force-pump. 

When  the  piston  B  moves  to  the  right, 

water  passes  through   the   valves,    A, 

to  the  delivery  pipe,  D,  and  enters  the 

barrel   through   D    from    the    suction 

pipe,  s,  and  well.     When  B  moves  to  the  left,  c  is  the  delivery 

valve,  and  E  is  the  suction  valve.     Fig.  282  shows  a  combined 

force  and  lift  pump  whose  action  is  very  easy  to  understand. 

The  delivery  valve  is  not  shown. 


Fig.  280. 


UNIVERSITY   1 


APPLIED    MECHANICS. 


507 


407.  A  pump  must  be  efficient  —  that  is,  it  must  do  nearly 
as  much  work  on  water  as  the  pump  itself  receives  from  an 
engine  or  labourers.     But  it  must  be 
remembered  that  the  best  pumps  used 
for  different  purposes  have  very  differ- 
ent efficiencies.     Forty  per  cent,  would 
be  regarded  as  a  reasonable  efficiency 
for    reciprocating   pumps    of   low   lift, 
D 


Fig.  282. 


S  I       Fig.  281. 

whereas  it  would  be  regarded  as  yathei 
poor  for  pumps  of  high  lift. 

Example. — (1)  Suppose  that  when  a 
pump  is  delivering  a  certain  quantity 
Q,  =  1,000  gallons  per  minute,  there  is 
a  loss  of  35  foot-pounds  per  pound  of 
water  in  the  pump  and  horizontal  pipes, 


and  0-05  foot-pound  per  pound  for  each  foot  of  vertical  pipe. 
Calculate  the  total  loss  per  pound  in  each  of  the  following  cases, 
the  vertical  height  of  pipes  and  of  delivery  being  called  h  feet : — 


Useful  work 

h 

per  pound  of 

Lost  energy. 

Total  energy. 

Efficiency. 

water. 

20 

20 

36 

56 

0-36 

50 

50 

37'5 

87-5 

0-57 

100 

100 

40 

140 

0-71 

200 

200 

45 

245 

0-82 

300 

300 

50 

350 

0-86 

400 

400 

55 

455 

0-88 

(2)  If  the  loss  of  the  pump  is  20  +  15  x  10~6  Q    per  pound  of 
water,  and  of  the  vertical  pipe  is  5  x  10~8  Q2  per  pound,  Q  "boing 


508 


APPLIED    MECHANICS. 


gallons  per  minute,  work  out  two  more  tables  like  the  above — 
one  when  the  delivery  is  500,  and  the  other  when  it  is  1,500  gallons 
per  minute. 

408.  The  peculiarity  of  reciprocating-pumps  is  that,  when 
there  is  no  slip,  there  is  the  same  quantity  of  water  passed  through 

the  pump  at  every  stroke.  If  we 
know  the  size  of  everything,  then 
the  speed  of  the  pump  tells  how 
much  water  is  delivered,  and  if  we 
know  the  height  to  which  it  is 
delivered,  we  know  the  work  done. 
Now,  in  a  centrifugal  -  pump 
things  are  somewhat  different ;  with 
a  given  speed  of  pump  we  may 
have  very  different  quantities  of 
water  passing.  We  may  have  the 
pump  running  at  a  certain  speed, 
and  no  water  being  delivered,  and 
very  little  work  being  done,  only 
frictional  work,  in  fact.  Now  a 
slight  increase  of  speed  may  cause 
an  abundant  flow  of  water,  and  a 
tremendous  increase  in  the  work 
usefully  done.  Strange  to  say,  if 


Fig.  283. 


this  speed  be  now  diminished  to  what 
it  was  at  first,  it  does  not  follow 
that  the  water  will  cease  to  flow. 
In  any  centrifugal-pump  such  as 
Fig.  283  we  observe  that  there  is  a 
central  wheel,  A  B,  with  vanes,  which 
can  be  rotated  very  rapidly.  Water 
can  enter  the  wheel  on  both  sides, 
C  C,  at  its  centre  from  two  supply 
pipes,  D  D,  which  meet  in  one  pipe, 
E,  below.  The  water  fills  the  wheel 
or  space  bet  ween  the  vanes,  and,  being 
whirled  round  with  great  velocity, 
tries  to  get  away  at  the  circumfer- 
ence of  the  wheel,  because  of  the  cen- 
trifugal force,  and  it  flows  out  into  the  casing,  F  F,  which 
gradually  becomes  the  discharge-pipe  of  the  pump,  as  shown 
in  the  small  scale  drawing,  Fig.  284.  This  is  a  popular 


\ 

V 

1 

t 

- 
1 

1 

Fig.  284. 


APPLIED    MECHANICS.  509 

explanation  of  what  occurs,  but  we  must  examine  the  matter 
more  carefully. 

409.  What  occurred  when  the  pressure  in  the  pump  of  the 
hydraulic  press  became  greater  than  the  pressure  of  fluid  in 
the  press  ?  There  occurred  a  flow  of  fluid.  The  fluid  was  set 
in  motion  from  pump  to  press.  A  difference  of  pressure 
between  two  places  which  communicate  with  one  another 
usually  means  a  tendency  to  produce  motion. 

In  the  hydraulic  press  the  flow  is  intermittent.  Why? 
Because  the  pressure  is  intermittent.  We  may  be  sure  that 
when  we  have  a  certain  difference  of  pressure  between  two 
places,  and  this  is  always  the  same,  the  flow  of  fluid  is  per- 
fectly steady ;  and  we  saw  in  Art.  408  that  work  is  then  done 
on  the  fluid  with  perfect  uniformity.  We  also  saw  how  to  tell 
what  work  was  done.  The  work  done  on  a  fluid  in  a  minute 
is  simply  the  difference  of  pressure  per  square  foot  which  causes 
the  flow,  multiplied  into  the  number  of  cubic  feet  of  water 
which  flow  per  minute.  This  gives  the  answer  in  foot-pounds. 

Suppose  that  with  pumps,  or  in  any  other  way,  we  establish 
A  difference  of  pressure  between  a  place  A  and  a  place  B,  and 
suppose  we  know  that  our  pumps  and  other  arrangements  will 
not  break  down — in  fact,  that  the  difference  of  pressure 
between  A  and  B  is  really  a  fixed  thing  on  which  we  can 
depend — we  know  that  the  flow  of  water  from  A  to  B  will  be 
the  same  at  all  times,  and  that  the  same  amount  of  work  will 
be  done  upon  it  every  minute.  If,  now,  we  leave  out  of  our 
minds  all  consideration  of  how  that  constant  difference  of 
pressure  has  been  produced — merely  think  of  the  two  vessels 
A  and  B — we  know  that  this  difference  of  pressure  which  has 
been  established  is  really  a  store  of  energy.  What  enables  us 
to  call  it  a  store  of  energy  ?  The  fact  that  we  know  it  will 
not  be  suddenly  destroyed. 

Suppose  we  know  that  a  man  has  a  certain  income  paid, 
say,  by  Government,  and  suppose  we  are  perfectly  certain  that 
this  income  is  constant,  we  can  regard  the  certainty  of  the 
man's  income  as  a  store.  To  say  that  a  man  makes  a  sovereign 
in  a  day  is  not  of  much  importance,  but  to  say  that  the  man 
has  a  regular  income  of  one  pound  a  day  makes  him  a  respect- 
able member  of  society,  and  a  store  of  social  energy. 

A  man  may  be  sitting  in  Parliament,  but  this  in  itself  does 
not  make  him  a  store  of  political  energy ;  whereas  if  we  know 
that  he  is  certain  to  sit  there  for  a  length  of  time — that  a 


510  APPLIED    MECHANICS. 

general  election  or  general  vote  of  the  House  is  unlikely  to 
unseat  him — we  can  regard  him  as  possessing  a  store  of 
political  energy. 

Similarly,  a  pound  of  water  in  the  vessel  A,  at  rest, 
possesses  more  energy  than  a  pound  of  water  in  the  vessel 
B,  at  rest.  (We  must  remember  that  there  are  some  means  of 
keeping  the  pressures  in  A  and  B  what  they  were.)  How  much 
more  energy  has  it  ?  If  the  difference  of  pressure  is  p  Ibs.  per 
square  inch,  then  it  has  2*3  p  foot-pounds  of  energy ;  not  in 
virtue  of  its  own  intrinsic  worth,  but  because  it  is  where  it  is, 
and  because  we  know  that  if  it  flows  into  the  vessel  B,  nothing 
will  alter  in  the  pressure  conditions  of  the  two  vessels  till  it 
gets  into  the  vessel  B.  We  understand,  then,  that  a  pound 
of  water,  subjected  to  a  pressure  of  p  Ibs.  per  square  inch,  may 
be  said  to  have  a  store  of  2-3  p  foot-pounds  of  energy,  if  we 
know  that  the  motion  which  is  occurring  in  the  water  is  steady, 
and  is  not  altering  capriciously. 

Thus  a  pound  of  water  at  the  pressure  of  the  atmosphere, 
14*73  pounds  per  square  inch,  possesses,  in  virtue  of  this  pres- 
sure, 2*3  x  14 '7  3  or  34  foot-pounds  of  energy.  It  would 
possess  the  same  energy  if  it  were  at  the  pressure  O,  the 
pressure  in  a  vacuum,  but  were  34  feet  higher  than  it  is  in 
position. 

Suppose  that  A  is  a  closed  box,  and  that  it  is  filled  with 
water,  under  great  pressure.  Now,  suppose  we  open  a  valve, 
and  let  this  water  escape.  Although  there  was  a  great  pres- 
sure, p,  just  for  an  instant,  and  therefore  a  rapid  flow  of  water 
just  for  an  instant,  this  almost  instantly  dies  away,  because 
the  pressure  in  A  is  almost  instantly  diminished.  Every 
pound  of  the  water  did  not  then  have  a  store  of  2'3  p  foot- 
pounds of  energy,  and  yet  it  was  at  the  pressure  of  p  pounds 
per  square  inch.  It  is  the  certainty  that  the  state  of  pressure 
in  A  will  continue  constant  that  gives  to  pressure  its  signifi- 
cance, and  gives  to  us  the  liberty  of  regarding  pressure  as  a 
store  of  energy. 

410.  Suppose  that  we  have,  anywhere,  steady  motion  of 
water.  Consider  a  pound  of  the  water.  What  is  its  total 
store  of  energy  ? 

1.  It  is  A  feet  above  some  datum  level.  Then  if  one 
pound  of  water  ever  were  allowed  to  fall  to  the  datum  level, 
it  would  do  h  foot-pounds  of  work  in  falling.  The  mechanical 
energy  stored  up  in  a  miller's  dam  is  simply  the  weight  of  the 


APPLIED    MECHANICS. 


fill 


water,  multiplied  by  the  height  through  which  it  can  fall.  Of 
course,  if  any  other  volumetric  force  acts  on  the  pound  of 
water,  this  will  constitute  another  store  of  potential  .energy. 
"We  are  supposing  that  the  weight  of  the  water  is  the  only 
volumetric  force. 

2.  As  the  motion  is  steady,  if  the  water  is  at  a  place 
where  pressure  is  p  pounds  per  square  inch,  it  possesses,  in 
virtue  of  the  fact  that  the  motion  is  of  a  steady  character,  the 
store  2-3  p  foot-pounds  of  energy. 

3.  As  the  water  is  in  motion,  if  v  is  its  velocity  in  feet 
per  second,  as  its  mass  (the  mass  of  one  pound)  is  1-4-32-2,  we 
know  that  its  kinetic  energy,   or   energy  of   motion,   is  the 
square  of  the  velocity  divided  by  64-4. 

Exercise. — Calculate  the  numbers  in  the  following  table, 
which  shows  the  relative  values  of  h  and  p  and  v,  if  we  wish 
to  convert  one  of  these  forms  of  energy  into  another.  Thus, 
the  energy  due  to  a  difference  of  level  of  2*3  feet  is  equiva- 
lent to  that  due  to  a  difference  of  pressure  of  1  Ib.  per  square 
inch,  or  to  that  due  to  a  velocity  of  12-18  feet  per  second. 


Difference  of  level. 

Pressure. 

Velocity. 

2-3  feet. 
16-1     „ 
34        „ 
64-4     „ 

1       Ib.  per  square  in. 

14-73 

28 

12-18  feet  per 
32-2 
45-7 
64-4 

second 
» 

411.  Now,  we  shall  not  suppose  that  the  pound  of  water  has 
any  other  stores  of  energy  than  these.  We  know  that,  as  it 
is  compressible  to  some  extent,  it  may  have  a  store  as  the 
mainspring  of  a  clock  has.  This  store  must  always  be  taken 
into  account  when  the  fluid  is  air  or  any  other  gas.  It  may 
•also  be  electrified  or  at  a  high  temperature,  or  it  may  have 
other  stores  of  energy  which  we  are  neglecting.  Merely  think 
of  these  three  stores : — Potential  energy  due  to  height  above 
a  datum  level ;  pressure  energy  due  to  the  unchangeableness 
of  tilings ;  kinetic  energy  due  to  its  actual  motion. 

What  we  must  remember  carefully  is  the  fact  that  this 

pound   of  water  retains   all    of   this   energy  except  what  it 

loses   in   friction.      Suppose  that  a  man  has  capital  in  the 

shape  of  gold,  capital  in  the  shape  of  shares  which  never  alter 

R 


512 


APPLIED    MECHANICS. 


in  price,  and  capital  in  the  shape  of  a  pet  manufactory,  which 
wastes  money  just  in  proportion  to  the  amount  of  capital 
invested  in  it.  He  may  buy  more  shares  or  sell  them  out,  in- 
vest more  or  less  money  in  the  pet  manufactory,  but  all  the 
time  his  only  loss  is  the  loss  from  the  manufactory.  He  may 
have  no  gold,  or  no  shares,  or  very  little  money  in  the  factory, 
but  all  the  time  his  total  capital  is  unchanging,  except  that  he 
loses  in  proportion  to  the  value  of  his  factory.  It  is  only  when 
water  has  part  of  its  energy  in  the  shape  of  kinetic  energy,  only 
when  it  is  in  motion,  that  it  loses  any  part  of  its  total  store. 

412.  Consider  a  lake  of  water  at  rest.  Consider  a  point  A, 
and  somewhat  below  its  level  another  point  B.  A  pound  of  water 
at  A  has  the  same  store  of  energy  as  if  it  were 
at  B.  In  neither  case  is  there  any  energy  of 
motion.  The  store  of  energy  at  A  is  merely  due 
to  height  above  some  datum  and  the  pressure 
per  square  inch  at  A.  Now,  if  a  pound  of  water 
gets  to  A  from  B,  it  loses  potential  energy  h  foot-pounds,  if  the 
difference  of  the  level  is  h  feet,  and  it  ought  to  gain  an  equiva- 
lent of  pressure  energy,  and  the  gain  of  pressure  is,  as  we  have 

already  seen,  simply  ^  pounds  per  square  inch.     Thus,  if  h  is 

34  feet,  and  p  is  the  gain  of  pressure,  then  there  is  a  gain  of 
pressure  energy  of  34  foot-pounds — that  is,  there  is  a  pressure 
at  B  of  34-i-2'3,  or  14-73  Ibs.  per  square  inch  greater  than  the 
pressure  at  A.  This  is  an  increase  of  pressure  called  one 
atmosphere.  In  still  water  there  is  an  increase  of  pressure  of 

one    atmosphere   for   every   34    feet    of 

descent.     (See  Art.  173.) 

413.  To  further  familiarise  us  with 

the  idea,  consider  the  flow  of  water  from 

an  orifice. 

In  Fig.  286  we  see  the  stream  lines 

along  which  water  flows  out  of  the  orifice. 

The  shapes  of  these  stream  lines  will  depend 

on  the  shape  and  position  of  the  orifice. 

Now,  a  pound  of  water  at  the  upper 
still  surface  A  is  at  atmospheric  pressure,  and  when  it  reaches 
c  it  is  also  at  atmospheric  pressure,  so  that  its  pressure 
energy  remains  the  same. 

But  at  c  it  has  fallen  h  feet ;  it  has  lost  h  foot-pounds  of 
potential  energy,  and  it  must  therefore  have  gained  h  foot-pounds 


Fig.  286. 


APPLIED    MECHANICS. 


513 


Fig.  286. 


of  kinetic  energy.  If  v  is  its  velocity,  then  v2  -4-  64-4  must  be 
equal  to  h,  so  that  v  may  be  calculated;  hence  its  velocity  is  just 
the  same  as  that  of  a  stone  which  had  fallen  freely  from  A  to  c. 

To  illustrate  this,  a  high  vessel  may  be  used,  from  which  at 
different  levels  tubes  come  out,  ending  in  nozzles,  throwing 
jets  vertically  upwards.  These 
jets  do  not  rise  to  the  same  height, 
because  there  is  a  loss  of  energy 
due  to  friction,  and  this  friction 
occurs  principally  at  the  nozzles. 
If  the  jet  reaches  within  a  dis- 
tance Ax  of  the  level  of  still  water 
inside,  and  if  the  nozzle  is  at  the 
depth  h  below  still  water  level, 
then  if  we  may  say  that  all  the 
friction  occurs  at  the  nozzles, 
h^h  expresses  the  loss  as  a  frac- 
tion of  the  whole  kinetic  energy  at  the  nozzle. 

When  the  measurements  and  calculations  are  made  for 
different  levels  of  the  water  in  the  vessel,  we  find  nearly  the 
same  result  in  every  case,  showing  that  the  frictional  loss  of 
energy  seems  to  be  proportional  to  the  kinetic  energy  there. 

There  is  very  little  friction  in  the  case  we  are  considering 
in  Fig.  286,  where  the  orifice  is  sharp-edged,  and  we  have  a 
very  simple  statement  of  the  velocity  at  c. 

Can  we  say  the  same  about  the  velocity  at  B  ?  Certainly 
not.  If  we  knew  the  pressure  energy  at  B,  we  could  say  how 
much  of  the  lost  potential  energy  has  been  invested  in  this 
shape,  and  therefore  how  much  has  been  invested  in  the 
shape  of  kinetic  energy ;  but  without  knowing  the  pressure  at 
B,  we  cannot  tell  what  is  the  velocity  there. 

414.  In  this  subject  of  flowing  water  there  are  more  mislead- 
ing hypotheses,  due  to  perverted  ingenuity,  than  in  almost  any 
other ;  and,  unfortunately,  the  logical  conclusions  drawn  from 
these  hypotheses,  when  known  to  be  untrue,  are  said  to  be  the 
statements  of  theory  as  opposed  to  practice.  We  often  hear 
the  statement,  "the  theoretical  velocity  at  an  orifice  is  the 
velocity  which  the  water  would  have  acquired  if  it  had  fallen 
freely,  as  in  a  vacuum,  from  still  water  level ; "  whereas  it  is 
evident  that  we  cannot  tell  the  velocity  at  any  point  in  the  jet 
unless  we  know  the  pressure  there,  and  we  only  know  the 
pressure  on  the  very  outside  of  the  jet. 


514 


APPLIED    MECHANICS. 


Now,  although  we  do  not  know  the  pressure  or  velocity  at 
every  point  of  water  flowing  from  an  orifice,  the  studies  of 
Professor  James  Thomson  enable  us  to  make  certain  im- 
portant statements  which  agree  with  experiment.  One  of 
these  is  this  : — 

When  frictionless  liquid  flows  from  two  similar  vessels 
through  similar  orifices  similarly  situated  with  regard  to  free 
water  level,  the  lines  of  flow  are  exactly  of  similar  shape ;  the 
velocities  at  similar  points  are  exactly  as  the  square  roots  of 
the  dimensions  of  the  vessels,  and  the  total  quantities  of  liquid 
which  flow  are  proportional  to  the  square  roots  of  the  fifth 
powers  of  the  dimensions. 

Thus,  if  we  have  water  flowing  similarly  from  three  similar 
vessels,  all  made  from  the  same  drawings,  but  to  different 
scales — say  one  1  foot,  another  4  feet,  and  another  9  feet  deep 
• — the  velocities  at  similar  points  are  as  1  to  2  to  3,  the 
sections  of  stream  tubes  are  as  1  to  16  to  81,  and  the 
quantities  of  liquid  flowing  from  the  three  vessels  are  as  1 
to  32  to  243  ;  thus  we  are  quite  sure  that  243  times  as  much 
liquid  flows  from  the  third  vessel  as  from  the  first. 

Hence,  suppose  we  want  to  know  how  much  water  is  flow- 
ing in  a  small  stream,  we  dam  the  water  up  somewhere,  and 
let  it  flow  out  of  our  dam  through  a  notch  like  Fig.  287  (a 

right-angled  isosceles  notch)  in  a 
wooden  board  with  sharp  edges. 
Indeed,  we  prefer  to  have  the 
edges  of  the  notch  made  of  metal, 
so  that  we  shall  be  sure  that 
they  are  straight  and  sharp. 
Measure  the  height  of  the  still 
water  in  the  dam  above  the 
lowest  point  of  the  notch  (D). 
A  graduated  post,  rising  from  the  bottom  of  the  dam 
some  feet  away,  its  zero  being  on  the  level  of  D,  is  very 
convenient  for  this.  This  one  measurement  tells  us  how 
much  water  is  tumbling  over.  For  we  know  that,  sup- 
pose there  is  rather  a  drought  one  day,  and  a  shows  the 
appearance  of  the  notch ;  and  on  another,  a  rainy  day,  A  shows 
its  appearance,  we  observe  that  the  orifices  through  which  the 
water  is  flowing  are  similar  and  similarly  situated  with  regard 
to  the  still  water  level ;  and  Thomson's  theory  enables  us  to 
say  that,  if  on  one  day  the  height  is  1  foot,  and  on  another  ifc 


Fig.  287. 


APPLIED   MECHANICS.  615 

is  4  feet,  then  32  times  as  mucn  water  comes  over  on  the  second 
day  as  on  the  first.  The  flow  of  water  is  exactly  proportional 
to  the  square  root  of  the  fifth  power  of  the  height  D  A. 

Now,  Thomson  measured  very  accurately  how  much  water 
is  flowing  when  the  height  is  1  foot,  and  he  found  it  to  be  2 -6  35 
cubic  feet  per  second.  Hence  we  have  the  rule :  measure  the 
vertical  height  D  A,  at  any  instant,  in  feet ;  raise  this  to  the 
fifth  power,  and  extract  the  square  root,  and  multiply  by 
2-635,  and  we  know  how  much  water  is  flowing  in  cubic  feet 
per  second. 

If  we  know  the  cubic  feet  of  water  flowing  per  second,  we 
know  the  weight  of  the  water,  since  a  cubic  foot  of  water 
weighs  62-3  Ibs.,  and  the  weight  of  water,  multiplied  by  the 
number  of  feet  through  which  we  can  let  it  fall,  tells  us  the 
foot-pounds  per  second — the  available  power  of  the  stream. 
The  foot-pounds  per  minute,  divided  by  33,000,  is  the  horse- 
power of  the  stream. 

415.  Gauge  notch  observations,  made  from  day  to  day  on 
a  stream,  enable  a  person  to  make  very  exact  calculations  as 
to  the  power  available  for  the  driving;  of  mills  by  turbines,  for 
the  working  of  hoists,  cranes,  and  lifts,  and  for  hundreds  of 
other  purposes. 

How  Thomson  used  his  theory  in  proving  that  the  famous 
Lowell  empirical  formula,  for  rectangular  gauge  notches,  is 
really  a  rational  one,  will  be  found  in  Art.  436.  Students 
ought  to  treasure  anything  published  by  Thomson  on  fluid 
motion. 

416.  Many  rather  abstruse-looking  questions  are  easily  an- 
swerable when  we  fully  grasp  the  significance  of  the  energy  law. 

The  fundamental  fact  which  makes  any  hydraulic  problem 
clear  to  you  is  this.  If  we  may  neglect  friction,  then  a  pound 
of  water  at  any  place  has  its  total  energy  in  three  shapes.  It 
has  h  foot-pounds  of  energy,  because  it  is  A* feet  above  a  datum 
level.  It  has  2-3  p  foot-pounds  of  energy,  because  its  pressure 
is  p  pounds  per  square  inch  ;  and  it  has  v2  -^  64 -4  foot-pounds 
of  energy,  because  its  velocity  is  v  feet  per  second. 

To  take  another  example  : — 

Suppose  we  have  a  pipe  which  is  in  the  main  horizontal, 
so  that  we  may  neglect  differences  of  level.  Then  we  have  to 
remember  that  the  pressure  energy,  plus  the  kinetic  energy,  of 
a  pound  of  water  does  not  alter.  When  water  flows  along  a 
pipe  which  is  full,  there  must  be  the  same  quantity  flowing 


518 


APPLIED    MECHANICS. 


everywhere.  We  are  sure,  therefore,  that  there  must  be 
greater  velocity  wherever  the  pipe  is  contracted.  But  greater 
velocity  meafiS  greater  kinetic  energy,  and  if  this  water  invests 


J 

K 

L          Ml 

,-^_-_---I 

s 

•l-D 

1 

"C 

^F 
- 

- 

••-•% 

Fig.  288. 

more  of  its  energy  kinetically,  it  must  have  less  in  the  shape 
of  pressure  energy.  That  is,  the  pressure  of  the  water  at  B 
(Fig.  288)  is  less  than  at  A  or  at  c.  The  pressure  at  B  may 
become  very  small  indeed.  It  is  easy,  in  this  way,  to  reduce 
the  pressure  to  much  less  than  the  atmospheric  pressure — 
merely  contracting  the  cross  section  of  the  pipe  is  sufficient. 
We  cannot  make  the  pressure  as  small  as  that  of  a  vacuum, 
for  before  that  limit  is  reached  vapour  forms. 

417.  Suppose  a  conduit  of  this  kind  were  carried  over  our 
fields,  and  that  a  quantity  of  water  lay  in  our  fields,  at  not  too 
great  a  depth  below  the  conduit.  If  we  bring  a  pipe  to  the  point 
B  from  the  field-water,  this  becomes  a  suction-pipe,  and  we  get 
our  fields  drained  at  the  expense  of  the  conduit  owners.  We 
spoil  their  water,  if  it  is  clean,  but  at  all  events  we  get  our 
fields  drained. 

In  this  lies  the  theory  of  jet  pumps,  and  much  of  the 
theory  of  injectors,  etc.  The  jet  pump  of  Professor  James 


Fig.  289. 

Thomson  simply  consists  of  a  small  pipe,  A  (Fig.  289),  which 
ends  in  a  nozzle.      Through  this,  let  us  suppose,  we  have  a 


APPLIED   MECHANICS.  6l7 

small  supply  of  water  flowing  from  some  pretty  high  reservoir. 
Suppose  that  this  water  flows  into  the  atmosphere  at  c. 
Evidently  the  pressure  at  a  is  much  less  than  the  pressure  at 
c.  It  is  less  then  than  the  atmospheric  pressure,  and  hence 
the  neighbourhood  of  a  is  a  partially  vacuous  space,  so  that 
the  pipe  B  E  becomes  a  suction-pipe,  and  water  tends  to  flow 
from  a  point  at  E,  to  B,  and  on  to  c.  Thus,  if  we  have  a  small 
supply  of  water  from  a  high  reservoir,  we  are  able  to  drain  a 
marsh  with  it. 

418.  There  is  not  space  here  to  speak  of  the  hundred  other 
ways  in  which  our  principle  comes  in  to  simplify  all  sorts  of 
puzzling  phenomena.  We  may  refer  to  Mr.  James  Perry's 
siphon  for  the  discharge  of  flood- waters  at  the  weirs  in  rivers.  It 
has  no  moving  parts  in  the  water,  being  simply  a  wide,  open  pipe 
of  varying  sectional  area,  through  which  an  object  as  large  as  a 
bullock  might  pass,  without  injury  to  the  sluice.  It  has  been 
found,  by  actual  trial,  that  the  quantity  of  water  passing  per 
minute  through  such  a  sluice  is  independent  of  the  fall,  so 
long  as  there  is  sufficient  fall  to  balance  the  waste  by  friction 
in  the  pipe,  a  few  inches  being  enough  in  the  case  of  wide 
pipes ;  and  a  velocity  of  45  feet  per  second,  at  the  smallest 
section,  may  be  calculated  upon  when  the  sluice  is  working 
full  power.  The  cross-section  of  the  siphon  is  like  the  letter 
D  at  the  tail ;  the  masonry,  or  concrete  basin,  is  actually  a 
portion  of  the  siphon,  and  the  horse-shoe  shaped  space  between 
the  lip  of  the  basin  and  the  iron  edge  of  the  siphon  proper  is 
the  actual  opening.  The  quantity  of  water  passing  at  any 
time  is  regulated  by  the  admission  of  air  to,  or  its  exclusion 
from  the  siphon.  There  is  a  throttle-valve  arrangement  by 
means  of  which  the  siphon  may  be  adjusted  at  any  time,  so  as 
to  vary  what  may  be  called  the  normal  water  level  in  the 
reach  above  the  sluice,  and  to  vary  the  opening  through  which 
a  constant  stream  of  water  falls  on  a  ridge-shaped  portion  of 
the  siphon.  This  stream  of  water  is  needed  to  exhaust  the 
siphon  of  air,  so  that  action  may  be  set  up  at  any  time,  even 
when  the  water  is  not  passing  over  the  top  of  the  throttle- 
valve.  The  manipulation  of  such  sluices  is  perfectly  simple — 
they  may  be  made  self-acting ;  but  it  is  proposed  in  important 
places  to  work  all  the  sluices  on  a  river  from  a  single  station 
electrically. 

Example. — There  is  a  circular  sharp-edged  orifice  in  a  tank 
in  which  there  is  liquid  kept  standing  to  a  certain  height.  A 


518  APPLIED    MECHANICS. 

man  is  told  that,  without  interfering  with  the  actual  edge  cf 
the  hole,  he  is  allowed  to  do  what  he  pleases  to  increase  the 
How.  How  may  he  do  so1?  Evidently  by  fitting  on  a  tube 
which  quickly  but  gradually  gets  to  be  of  much  larger 
diameter;  He  takes  care  that  the  tube  shall  run  full. 

Example. — A  number  of  mill-owners  receive  water  from 
the  same  lake.  Each  has  a  rectangular  opening  the  depth  of 
which  below  the  lake  and  its  breadth  are  supposed  to  fix  his 
supply.  Show  that  if  a  mill-owner  is  allowed  to  do  anything 
he  pleases  on  his  own  side  of  the  opening,  he  may  procure  a 
very  much  greater  supply  of  water. 

419.  We  know  that  loss  of  energy  per  pound  by  friction, 
in  water,  is  proportional  to  the  square  of  the  velocity,  at  such 
speeds  as  are  common  in  pumps.  In  Art.  48  we  gave  the 
rules  by  which  we  are  able  to  calculate  the  loss  of  energy 
which  a  pound  of  water  experiences  in  going  along  pipes.  But 
we  wish  to  impress  on  you  the  fact  that  this  loss  always  be- 
comes very  great  when  the  flow  of  the  water  has  to  occur  ab- 
normally. In  the  pipe  (Fig.  288)  the  wide  and  narrow  parts 
gradually  change  into  one  another  by 
continuous  curves.  It  is  practically 
impossible  for  a  liquid  to  flow  in  a 
discontinuous  curve.  Suppose  we  try, 
then,  to  make  it  flow  along  the  pipe 
shown  in  Fig.  290.  What  the  water 
does  is  this  :  when  it  comes  to  the  corner 
Fig.  290.  it  produces  for  itself  wheels,  little 

eddies  or  whirlpools,  as  we  might  put 
rollers  under  a  log   of  wood  that  we  wanted    to  get  along 
easily ;  and  there  is   great  loss  of  energy   due  to   this,   for 
the   eddies   have  not  only  to  be  in  the  corners,  but  there 
are  smaller  eddies  carried  along  by 
the  water  itself,  maintained  so  long 
as  they  are  needed.     We  have  been 
speaking  of  actual  discontinuity  in 
Fig.  291.  the  flow.     But  there  is  a  fact  which 

many  hydraulic  engineers  seem  to 

be  quite  ignorant  of — namely,  that  a  liquid  cannot  flow  along  a 
path  which  suddenly  changes  in  curvature.  A  liquid  cannot,  for 
example,  flow  along  this  path  (Fig.  291).  At  A  it  changes 
from  a  straight  line  suddenly  to  the  arc  of  a  circle,  and,  con- 
sequently, the  water  digresses  at  A,  and  creates  little  castors, 


APPLIED    MECHANICS.  519 

little  eddies  to  carry  it  by  a  path  of  continuous  change  of 

curvature  from  B  to  c.     Now  this  analogy  can  be  shown  to  be 

true  experimentally.      Thus  it  has  been  found  that  if  water 

flows  along  the  bend  (Fig.  291)  it  loses  a  certain  amount  ol 

energy  on  account  of   the  bend  ;    but  if 

we  make  the  pipe  bend  as  much  again  in 

the    same  direction  as  in  A  B  (Fig.  293), 

we   do    not   get    again   the    same   loss  j 

indeed,  there  is  comparatively  very  kittle 

loss   at   the   second    bend.      But   if    we 

bend  the  pipe  in  the  opposite  direction, 

as  in  c  D    (Fig.  292),  there  is  as   much 

loss  at  the  second  bend  as  at  the  first. 

The    little    wheels,     or    castors,  •  or 
eddies  which  the  fluid  creates  for  itself  to       Pig.  292.      Fig.  293. 
carry  it  through  the  bend  A  are  available 

when  the  water  needs  them  again  at  B,  whereas  the  wheels  or 
eddies  produced  at  c  have  to  be  destroyed,  and  new  ones 
created,  rotating  just  in  the  opposite  direction  to  carry  the 
fluid  through  the  bend  D. 

We  see  now  how  necessary  it  is  that  all  curved  vanes  or 
other  surfaces  along  which  water  flows  should  be  drawn,  not 
with  a  pair  of  compasses,  but  rather  with  a  batten,  a  thin 
Strip  of  wood,  which  bends  gradually. 

420.  We  regard  a  pump  as  a  contrivance  which  gives  to 
every  pound  of.  water  passing  through  it  an  additional  store  of 
energy.     From  the  pond  to  the  entrance  to  the  pump,  every 
pound  of  water  has  just  the  energy  it  has  in  the  pond,  barring 
frictional  loss.      From  pump  to  cistern,  every  pound  of  water 
has  an  additional  store  of  energy,  and  it  is  the  pump  which 
gives  it  this  store. 

Suppose  we  have  a  centrifugal  pump  going  at  a  regular 
speed,  and  discharging  a  regular  quantity  of  water.  In  the 
supply-pipe  a  pound  of  water  has  a  certain  total  amount  of 
energy  which  we  know,  if  we  know  its  height  above  datum,  its 
pressure,  and  its  velocity.  But,  in  passing  through  the  wheel, 
it  receives  a  supply  of  energy.  The  total  energy  of  a  pound 
of  water  in  the  discharge-pipe  is  greater  than  what  it  is  in  the 
supply-pipe.  We  can  make  all  sorts  of  calculations,  if  we 
know  what  is  the  clear  gain,  the  clear  gift  of  energy  it  gets  in 
passing  through  the  pump. 

421.  Suppose  a  man  jumps  in  to  an  American  railway  train 

T>A" 


520  APPLIED    MECHANICS. 

anywhere,    and   after    wandering  about,  fore  and  aft,  jumps 
out  again.      Find  the  man's  momentum  in  the  direction  of 
the  train's  motion  just  before  he  alights  on  the  train.     Find 
his  momentum  in  the  same  direction 
when   he   lias   just   sprung   from   the 
train ,  the  difference  of   these  is  the 
total   impulse  with  which  he  acts  on 
the  train.     It  is  the  momentum  which 
he  gives  to  the  train.     Suppose  that  a 
Fig.  294.  number  of  people  could  perform  this 

acrobatic  feat  every  second  with  the 

greatest  regularity,  then  the  momentum  given  in  one  second  to 
the  train  could  be  calculated.  But  momentum  given  per  second 
is  what  we  call  force ;  hence  we  have  found  the  force  acting 
on  the  train  due  to  these  jumping  individuals,  and  this  force, 
multiplied  into  the  space  passed  through  by  the  train  in  one 
second,  gives  the  propelling  work  done  upon  the  train  per 
second.  We  have  nothing  to  do  with  whether  it  is  a  pro- 
pelling force  or  a  retarding  force.  In  the  one  case,  the 
acrobats  give  momentum  to  the  train ;  in  the  other,  the  train 
gives  momentum  to  them.  The  loss  of  momentum  per  second 
in  a  regular  stream  of  people,  going  on  and  off  the  train,  is  a 
force  which  is  applied  to  the  train.  We  only  have  to  do  with 
their  momentum  in  the  direction  of  the  train's  motion. 

Now  suppose  that,  instead  of  its  being  a  train,  it  were  a 
sort  of  circular  turntable,  or  a  merry-go-round,  and  that  a 
regular  stream  of  people  jumped  on  and  off.  In  this  case,  the 
place  where  a  man  jumps  on  may  be  going  at  a  different  speed 
from  the  place  where  a  man  jumps  off;  but  our  rule  is  not 
very  different.  Find  how  much  is  added  per  second  to  the 
momentum  of  the  wheel  at  the  point  where  people  leap  on, 
and  regard  this  as  a  force.  Multiply  by  the  speed  of  the 
wheel  there,  and  this  is  the  work  done  by  the  mere  leaping 
on,  and  staying  on  the  wheel.  Now,  find  how  much  per 
second  is  taken  from  the  momentum  of  the  wheel  at  the  place 
where  leaping  off  occurs,  and  regard  this  as  a  force  opposing 
the  motion.  Multiplied  into  speed  at  the  leaping-off  place,  we 
have  the  work  taken  by  the  stream  of  people  from  the  wheel, 
per  second,  because  they  leap  off.  The  difference  in  these 
two  things  is,  of  course,  the  work  done  in  the  jumping 
on  and  off. 

Now,  the  water  moves  from  the  centre  towards  the  vanes 


APPLIED    MECHANICS.  521 

of  the  centrifugal  pump,  merely  radially,  and  hence  the  water 
entering  the  vane  cannot  add  to  or  diminish  the  momentum  of 
the  wheel  just  there.  It  had  no  momentum  of  its  own 
previously  in  this  direction.  What  occurs  inside  the  wheel 
now  we  have  nothing  to  do  with,  excepting  that  we  know  that 
frictional  loss  occurs  there.  We  are  only  concerned  with  how 
the  water  leaves  the  wheel.  The  way  in  which  it  is  made  to 
leave  the  wheel  determines  how  much  energy  it  takes  from 
the  wheel. 

Take  the  simplest  case.  Suppose  the  vanes  to  be  radial  at 
B  (Fig.  294).  That  is,  besides  moving  outwards  radially  at  B, 
the  water  leaves  the  vane  with  the  same  tangential  velocity  as 
the  vane  at  B  has.  Suppose  this  tangential  velocity  to  be  v. 
Then  w  Ib.  of  water  leaving  B  per  second  leaves  with  a  tan- 
gential momentum,  wv  -r-  32 -2,  and  retards  the  wheel  with  a 
force  of  this  amount  acting  at  B.  This  force  x  v  is  the  energy 
which  it  receives  per  second  from  wheel,  or  wv2  -f-  32*2.  One 
pound  of  water,  therefore,  receives  the  energy  v2  -f-  32-2  from 
the  wheel  in  passing  through  it. 

We  understand,  then,  that  a  pound  of  water  in  the  dis- 
charge-pipe of  the  centrifugal  pump  has  this  greater  store  of 
energy  than  a  pound  of  water  in  the  supply-pipe,  except  for 
frictional  losses.  If  we  make  the  water  go  out  from  the  wheel 
as  backward  bent  vanes  make  it  go  (Fig.  296),  and  as  it 
would  be  dangerous  for  us  to  do  from  a  railway  train,  less 
work  has  been  done  upon  it  by  the  wheel.  If  it  goes  out  in 
the  direction  of  motion  relatively  to  the  wheel,  more  work  is 
done  upon  it  than  we  are  now  supposing. 

422.  The  wheel  with  radial  vanes  gives  -y2  -f-  32 '2  foot- 
pounds of  energy  to  one  pound  of  water.  If  we  know  how 
many  pounds  of  water  pass  through  the  wheel,  we  know  then 
the  total  amount  of  work  done  by  the  wheel. 

The  water  gets  this  energy  to  squander  or  store  as  it 
pleases,  and  it  does  squander  it  in  friction  to  a  large  extent. 
But  suppose  it  squandered  none  of  it,  but  converted  it  all  into 
potential  energy  in  lifting  itself  up  to  a  cistern,  it  would  lift 
itself  i?2_i_32-2  feet  high;  that  is,  it  would  lift  itself  above 
the  pond  to  twice  the  height  due  to  the  velocity  of  the  rim 
of  the  wheel.  Suppose  the  rim  of  the  wheel  has  a  velocity  of 
457  feet  per  second,  a  stone  would  have  to  fall  freely  34  feet 
to  acquire  this  velocity,  and  hence  the  total  rise  of  water 
would  be  68  feet,  twice  the  height  due  to  the  velocity  of  the 


522  APPLIED    MECHANICS. 

rim  of  the  wheel.     In  this  case  we  should  say  that  the  pump 
was  perfect. 

The  wheel  itself  receives  energy  from  the  engine,  else  it 
could  not  give  energy  to  the  water.  It  gives  out  all  the 
energy  that  leaves  the  engine,  except  what  is  wasted  in  bear- 
ings everywhere,  and  what  is  wasted  in  friction  with  the 
water. 

The  energy  given  out  by  the  engine  per  pound  of  water, 
divided  into  v2  -f-  32 -2,  is  the  efficiency  of  the  shafting,  belt- 
ing, and  wheel.  Again,  the  real  height  to  which  water  is 
lifted  by  the  pump,  divided  by  the  ideal  height,  v2  -r  32 '2,  is 
the  efficiency  of  the  water  passages  from  pond  to  wheel  and 
from  wheel  to  cistern. 

The  loss  in  these  places  is  due  to  friction.  Make  the 
supply-pipe  wide,  bell-mouthed  at  the  bottom,  where  water 
enters  it,  so  that  it  may  enter  by  gradual  curves ;  make  the 
approach  to  the  wheel  as  gradual  as  possible  ;  let  the  vanes  of 
the  wheel  make  the  calculable  angle  with  the  central  circle, 
which  will  reduce  the  shock  there  (see  Art.  428) ;  make  the 
discharge-pipe  wide,  and  let  the  velocity  with  which  the  water 
enters  the  upper  cistern  be  as  small  as  possible,  and  we  greatly 
reduce  the  waste  of  energy.  But  there  is  one  particular  place 
where  there  is  usually  much  greater  waste  than  anywhere 
else,  and  that  is  the  chamber  outside  the  wheel. 

423.  Just  when  the  water  leaves  the  wheel  a  large  portion 
of  its  energy  is  kinetic.  It  is  in  rapid  motion.  Now,  in  the 
large  discharge-pipe  there  may  be  as  little  kinetic  energy  as 
we  please.  Hence,  from  the  time  the  water  leaves  the  wheel 
till  it  enters  the  discharge-pipe  there  ought  to  be  great  care 
taken  in  allowing  the  kinetic  energy  to  become  converted  into 
pressure  energy. 

Professor  James  Thomson  discovered 
here  the  efficiency  of  a  whirlpool  cham- 
ber. When  we  let  water  escape  from  a 
wash-basin,  we  know  that  the  surface  of 
the  water  takes  a  shape  like  this  (Fig.  295). 
The  velocity  of  the  water  is  greater 
the  nearer  it  is  to  the  centre.  The  pres- 
sure  is  greater  the  farther  away  from  the 
centre.  The  spiral  motion  which  we  ob- 
serve in  this  case  is  the  only  steady  motion  of  water  which 
allows  a  constant  radial  discharge  without  the  water  making 


APPLIED   MECHANICS.  523 

objections,  setting  up  little  eddies  of  its  own,  and  thus  wasting 
energy  in  friction. 

Hence,  in  Thomson's  puuip,  after  the  water  is  discharged, 
it  circulates  in  this  cylindric  whirlpool  chamber,  which  he 
made  of  twice  the  diameter  of  his  wheel.  When  a  pound  of 
water  reaches  this  place,  in  consequence  of  its  radial  and 
circular  motions,  it  retains  more  nearly  the  whole  of  its  total 
energy  than  if  we  let  it  discharge  in  any  other  way.  It  has 
lost  much  of  its  velocity,  but  has  gained  in  pressure.  This 
whirlpool  chamber  of  Thomson's,  then,  did  for  the  water  what 
gradual  curves  do  for  the  water  in  a  pipe;  it  enables  the 
water  to  convert  its  kinetic  into  pressure  energy,  with  a  mini- 
mum of  waste  in  friction, 

It  has,  however,  to  be  remembered  that  even  here,  at 
the  outside  of  the  whirlpool  chamber,  the  water  retains  a 
very  considerable  amount  of  kinetic  energy,  even  when  the 
whirlpool  chamber  is  made  very  large,  and  much  of  this  is 
wasted  afterwards.  And,  although  no  one  believes  more 
firmly  in  the  reasoning  of  Thomson  than  I  do,  I  feel  that 
perhaps  the  whole  problem  admits  of  a  better  common- 
sense  solution.  Let  the  whirlpool  chamber  get  wider  or 
broader,  as  well  as  larger  in  diameter.  Let  the  wheel  have 
larger  orifices  on  its  outer  circumference  than  on  its  inner 
circumference.  The  water  will  lose  its  kinetic  energy  far 
more  rapidly  as  its  passage  widens  more  rapidly.  The  result 
of  this  will  be  that  although  in  this  rapid  change  there  is  more 
friction,  yet  when  the  water  is  only  a  short  distance  out,  it  is 
not  moving  much  faster  than  it  will  do  in  the  discharge-pipe, 
and  there  is  much  less  loss  in  entering  the  discharge-pipe. 
We  have  always  thought  that  since  Thomson's  chamber  is 
expensively  large,  we  ought  to  sub- 
mit to  a  modification  of  his  con- 
ditions even  from  the  place  where 
water  leaves  the  wheel.  But  when 
we  begin  to  consider  a  much  smaller 
chamber,  we  see  that  possibly  the 
vanes  ought  rather  to  slope  back- 
wards than  to  be  radial.  Let  the 
radial  velocity  at  M  N  be  vr.  The 
velocity  relatively  (see  Art.  36)  to  Fig.  296. 

the  vane  is  vr  -f-  sin.  0,  if  M  N  P  is  0. 

424.  Let  v  be  the  tangential  velocity  of  the  wheel  at  N. 


524  APPLIED   MECHANICS. 

Using  a  very  easily  understood  graphical  method  of  working,  let 
o  N  represent  the  radial  velocity  at  N  to  scale.  Let  M  N  and  N  p 
represent  the  direction  of  the  vane  and  rim  at  N.  Let  N  P 
represent  v,  the  tangential  velocity,  and  if  o  M  is  drawn  at 
right  angles  to  ON,  MN  represents  the 
velocity  of  the  water  relatively  to  the 
vane.  Hence,  as  the  resultant  of  M  N 
and  N  P  is  M  P,  M  P  is  the  total  velocity 
of  the  water  leaving  the  wheel.  It  has 
the  tangential  component  M  Q,  and  the 


radial  component  Q  P  which  it  had  in  the      u          Fi(r  297 
wheel.       Suppose  we  call  M  Q  by  the 

letter  v.  Now,  every  pound  of  water  leaves  with  the  tangential 
momentum  v  -r-  32  '2,  or  say  v/g.  Every  pound  per  second 
therefore  represents  a  tangential  retarding  force,  v/g,  acting  on 
the  rim  of  the  wheel,  and  this  multiplied  by  v,  or  v  v/g,  is  the 
work  done  usefully  per  pound  of  water.  If  we  take  it  that  in 
all  cases  there  is  a  loss  of  the  fraction  s  of  the  kinetic  energy 
due  to  tangential  motion,  only  s  v^/g  is  wasted  ;  v  v/g  is  total 
energy  ;  (v  v  —  s  v^/g  is  useful  energy,  or  height  to  which  the 
water  is  lifted.  We  may  say  roughly  that  the  efficiency  is 

11 

1  —  s  -  ;  s  depends  on  the  size  of  the  chamber.     What  the 

value  of  v,  and  therefore  the  angle  Q,  ought  to  be,  is  therefore 
a  question  of  minimum  total  waste  of  value,  in  interest  on. 
plant  and  waste  of  energy,  etc. 

425.  In  the  case  of  pumps  we  saw  that  each  pound  of  water 
gets  an  increased  store  of  energy,  which  may 
be  in  the  shape  of  pressure  energy,  or  kinetic 
energy,  or  both,  but  which  mainly  becomes 
potential. 

Now,  in  water-wheels,  turbines,  water- 
pressure  engines,  including  hoists  and  lifts, 
we  take  part  of  the  store  of  energy  from  each 
pound  of  water,  giving  it  to  machinery. 

As  a  simple  case  of  the  abstraction  of 
energy  from  water,  and  as  an  illustration  of 
the  acrobat  and  railway-train  principle,  con- 
sider the  vessel  (Fig.  298)  from  which  the 
water  is  flowing.  Water  leaves  this  vessel 
horizontally  from  an  orifice,  taking  away  with 
Fig.  29*  it  momentum.  The  quantity  of  momentum 


APPLIED   MECHANICS.  525 

it  takes  away  per  second  is  simply  the  force  acting  on  the 
vessel.  You  see  that  there  is  a  force  acting,  for  I  have 
arranged  the  vessel  as  the  bob  of  a  pendulum. 

426.  If  we  let  the  water  flow  from  an  orifice  through  which 
it  comes  in  parallel  streams,  it  is  easy  to  show  that  the  force 
acting  on  the  vessel  is  twice  the  total  pressure  which  would 
act  on  this  little  sluice  when  it  closes  the  orifice,  and  no  water 
is  flowing.  For  if  a  little  area  a  (square  feet)  of  the  orifice  is 
h  feet  below  still  water  level,  the  pressure  at  a  being  atmo- 
spheric, the  velocity  v  =  ^  2  g  h ;  the  volumetric  flow  is  a  v 
per  second ;  or  the  mass  per  second  is  w  a  v/g,  if  w  is  6 2  '3  Ibs. 

per  cubic  foot;    the  momentum  per  second  is  -     -  x  v,  or 

2  w  a  h.  When  the  orifice  is  closed  the  force  due  to  pressure 
upon  it  is  w  a  h.  There  have  been  no  very  sound  writers  on 
this  subject  except  Thomson,  but 
even  the  soundest  imagine  the 
force  to  be  less  when  the  vessel 
is  moving.  They  forget  in  their 
calculation  that  the  water  leaving 
the  vessel  had  at  the  beginning 

the  motion   of  the   vessel   itself.  

Fig.  299  shows  a  vessel  floating  on  Fig.  299. 

a   pond,  and,  moving  under  the 

action  of  its  jet ;  with  sufficiently  delicate  apparatus,  it  may  be 
shown  that  the  force  on  it  is  the  same  when  it  moves  as  when 
it  is  at  rest.  If  such  a  vessel  is  kept  supplied  with  water,  it  is 
easy  to  calculate  the  force  due  to  the  horizontal  velocity  of  the 
supply  water ;  in  fact,  we  must  consider  that  the  acrobats  enter 
the  train  (Art.  421)  as  well  as  leave  it.  Thus,  in  the  propul- 
sion of  a  ship,  a  large  centrifugal  pump  draws  water  from -be- 
neath the  ship,  and  propels  it  out  at  the  sides  and  stern  wards. 
Suppose  the  water  moves  through  the  nozzles  with  the 
velocity  of  30  feet  per  second,  and  that  the  ship  is  moving  the 
other  way  at  20  feet  per  second,  then  it  is  evident  that  the 
water  has  a  velocity  relatively  to  the  sea  of  10  feet  per  second. 
The  momentum,  therefore,  given  to  a  pound  of  water  is 
•sV  x  10,  and  this,  multiplied  by  the  velocity  of  the  ship, 
gives  6J  foot-pounds  of  energy,  which  each  pound  of  pumped 
water  imparts  to  the  ship.  Notice  that  all  the  kinetic  energy 
is  wasted,  or  J  ^  x  102,  or  1-56  foot-pounds  of  energy  per 
pound  of  water. 


626 


APPLIED  MECHANICS. 


It  is  easy  to  see,  if  we  had  no  friction  in  passages,  that  the 
greatest  efficiency  is  arrived  at  by  letting  the  water  take  with 
it  only  a  very  small  amount  of  kinetic  energy  as  it  mingles 
with  sea  water ;  that  is,  by  letting  the  backward  nozzle 
velocity  of  the  water  be  very  little  greater  than  the  forward 
velocity  of  the  ship.  But  of  course  this  is  not  at  all  a  prac- 
tical solution  of  the  problem  to  find  the  proper  speed  for 
maximum  good  result. 

427.  A  turbine,  water-wheel,  or  water-power  engine  takes 
energy  from  each  pound  of  water,  and  gives  it  to  machinery. 
Suppose,  for  example,  that  we  have  water  in  a  tank  or  dam, 
and  we  have  a  clear  fall  of  60  feet.  Now,  when  a  pound  of 
water  is  nearly  motionless  at  the  surface  of  the  dam,  it  has 
just  60  foot-pounds  more  energy  than  when  it  is  nearly 
motionless  in  the  tail  race  at  the  bottom.  A  water-power 
engine  of  any  kind  is  constructed  to  abstract  this  60  foot- 
pounds of  energy  with  as  little  waste  in  friction  as  possible. 
Instead  of  being  at  the  same  pressure  in  the  dam  and  tail 
race,  we  may  have  the  pressure  energy  much  greater  before- 
hand, as  well  as  the  potential  energy ;  but  in  every  case  we 
try  to  take  out  of  a  pound  of  water  the  total  difference  of 

energy.  Thus,  suppose  a 
pound  of  water  to  be  motion- 
less in  a  mill-dam  60  feet  high 
above  the  tail  race,  we  can- 
not take  more  from  it  than 
60  foot-pounds  of  energy. 
Suppose  a  pound  of  water  to 
be  motionless  60  feet  above 
the  tail  race,  but  that  it  is 
also  inside  an  accumulator, 
where  the  pressure  is  700  Ibs. 
to  the  square  inch;  we  can  take 
from  it  60  +  2-3  x  700,  or 
60  +  1,610,  cr  1,670  foot- 
pounds of  work. 

As  we  understand  the 
action  of  the  centrifugal 
pump,  we  have  no  difficulty 

in  understanding  the  action  of  the  turbine.  It  is  because 
we  have  studied  the  centrifugal  pump  that  we  dwell  upon 
the  Thomson  turbine.  Water  flows  from  a  pen -trough 


APPLIED    MECHANICS. 


527 


Fig.  301. 


through  cast-iron  pipes  to  A.  These  pipes  must  be*  bell- 
mouthed  ;  they  must  open  out  gradually  into  the  cistern : 
they  must  be  as  large  in  diameter  as  we  can  conveniently 
make  them.  In  that  case  the 
velocity  in  the  pipes  will  be  small, 
and,  therefore,  the  friction  will 
be  small.  Fig.  300  shows  a  plan 
of  the  chamber,  B,  into  which  the 
water  flows.  The  chamber  is 
so  large  that  the  velocity  there  is 
small,  and  the  water  finds  its  way 
equally  readily  into  the  central 
space,  whether  it  flows  between 
the  guide-blades  1  and  2,  or  2  and 
3,  or  3  and  4,  or  4  and  1.  We  are 
at  last  allowing  the  water  to  flow 
quickly,  for  the  guide-blade  cham- 
ber is  narrow.  When  the  water  is  just  leaving  the  guide-blades 
it  flows  rapidly ;  of  course  it  is  flowing  radially  as  well  as  tan- 
gentially  to  the  rotating  wheel,  P,  but  the  tangential  motion 
ought  to  be  equal  to  that  of  the  wheel. 

428.  Suppose  we  want  to  enter  a  moving  railway  train  or 
tram-car  without  shock,  we  try  to  get  a  velocity  equal  to  that  of 

the  train,  in  the  direction  of 
the  train's  motion,  before  we 
venture  to  enter  the  train  ; 
hence  the  tangential  velo- 
city of  the  water  must  be 
equal  to  that  of  the  end  of 
the  radial  vane  of  the  wheel, 
if  the  water  is  to  enter  it 
without  shock.  If  the  vane 
is  inclined  like  Fig.  302,  A, 
the  tangential  velocity  of 
^1  the  water  ought  to  be  less 
Jf  than  that  of  the  wheel  just 
/J  here.  If  the  vane  is  in- 
clined like  Fig.  302,  B,  the 
tangential  velocity  of  the 
water  is  made  greater  than 

that    of    the   vane.     In   fact,  the   relative  velocity  of  water 
and   vane  must  be  in  the  direction  of  the  vane,  if  there  is 


Pig.  302. 


528  APPLIED    MECHANICS. 

to  belio  shock.  Usually  the  vane  is  shaped  as  we  see  it  in 
Fig.  301,  which  is  an  enlarged  section  of  the  wheel,  F  ;  but 
we  sha-ll  suppose  it  to  be  radial  just  at  the  outside,  for 
simplicity  of  calculation.  We  must  remember,  then,  that 
somehow  or  other  we  must  try  to  get  a  tangential  velocity 
of  water  equal  to  the  velocity  of  vanes  there.  The  water 
now  flows  through  the  wheel,  which  lets  it  escape  at  the 
centre.  Here,  again,  we  must  remember  that  the  water  has 
to  escape  with  no  velocity  except  a  radial  one. 

If  we  wanted  to  let  a  stone  out  of  a  railway  carriage  so 
that  it  would  just  fall  to  the  ground  vertically,  so  that  it 
would  possess  no  forward  motion,  we  must  shy  it  backwards, 
with  respect  to  the  train ;  give  it  a  velocity  backwards  as 
much  as  it  has  forwards  already.  These  vanes,  then,  at  the 
centre,  let  the  water  out  backwards,  just  because  we  want  the 
water  to  have  no  forward  velocity  when  it  has  left  the  wheel. 
The  water  has,  of  course,  a  radial  velocity  everywhere,  which 
simply  depends  on  the  total  quantity  flowing  per  second, 
divided  by  the  tangential  areas  of  these  orifices. 

429.  We  want,  now,  to  know  how  much  store  of  energy  each 
pound  of  water  has  lost  in  passing  through  the  wheel,  and  we 
employ  the  rule  already  given.  Find  the  tangential  momentum 
of  the  water  at  p.  If  the  velocity  of  the  outside  of  the  wheel 
is  v,  then  v  -i-  32-2  is  the  forward  momentum  of  one  poun<l 
of  water.  This,  multiplied  by  v,  is  the  work  done  per  pound 
of  water,  or — 

v2  -v-  32-2  foot-pounds, 
because  it  enters  the  wheel.     Or  w  Ib.  of  water  per  second 

w 

means  a  momentum  of  ^K-S  v  Per  second,  and  this   is  force ; 
o'Ji'Z 

force  multiplied  by  velocity  v  gives  work  done  per  second,  and 
this  divided  by  w  gives  work  per  pound  of  water.  It  is  evident 
that  if  we  take  moment  of  momentum  lost  by  the  water  per 
second  in  passing  through  the  wheel,  and  multiply  by  the 
angular  velocity,  we  get  the  same  answer.  The  wheel  does 
no  work  on  the  water  as  it  leaves  at  K,  because  the  water 
leaves  with  no  forward  or  backward  momentum.  Hence  one 
pound  of  water,  from  the  time  it  enters  the  wheel  to  the  time 
it  leaves,  loses 

v*  -f-  32-2  foot-pounds, 

from  its  store  of  energy,  and  gives  this  store  to  the  wheel. 


MECHANICS.  529 

[f,  then,  it  loses  no  energy  by  friction  anywhere,  when  it 
enters  the  tail  race  it  has  just  this  much  less  energy  than 
when  it  left  the  pen-trough.  If  h  is  the  total  height  of  the 
fall,  evidently  one  pound  of  water  really  gives  out  h  foot- 
pounds of  energy,  so  that  h  is  twice  the  height  due  to  v.  We 
know  that,  in  practice,  what  the  water  gives  to  the  wheel  is 

less  than  h,  and  -^  -f-  h  is  called  the  hydraulic  efficiency  of 

OZi 

the  turbine.  It  is  the  ratio  of  the  energy  given  to  the  wheel 
to  the  total  energy  lost  by  the  water  in  falling  from  one  level 
to  the  other.  If,  then,  there  is  no  shock  to  the  water  in 
entering  or  leaving  the  wheel,  its  efficiency  is  twice  the  height 
due  to  the  velocity  of  the  rim  divided  by  the  real  total  fall  of 
the  water. 

Of  course  all  the  energy  given  to  the  wheel  is  not  utilised. 
There  is  friction  between  the  wheel-covers  and  the  wheel-case, 
friction  at  all  the  bearings,  etc.,  of  the  shafting  and  gearing 
which  transmit  the  power  of  the  wheel  to  a  mill.  We  are 
only  speaking  now  of  the  efficiency  of  the  passages,  which  is, 
however,  the  most  important  matter  in  connection  with 
turbines. 

Knowing  the  average  amount  of  water  passing  through  the 
wheel,  and  therefore  the  radial  velocity  at  K,  the  angle  of  the 
vanes  at  K  is  determined  if  we  know  the  average  speed  of  the 
wheel.  If  the  speed  and  quantity  of  water  were  exactly  pro- 
portional to  one  another ;  that  is,  if  the  speed  of  the  wheel 
were  exactly  proportional  to  the  horse-power,  the  inner  ends 
of  the  vanes  once  settled  would  remain  right  always.  But  if 
our  wheel  is  to  be  regulated  as  a  steam-engine,  so  that  quick- 
ening speed  causes  less  water  to  flow,  then  it  is  obvious  that 
the  inner  ends  of  the  vanes,  although  right  for  the  calculated 
flow,  are  not  properly  shaped  when  the  horse-power  diminishes 
or  increases.  The  loss  of  energy  here  is  not,  however,  likely 
to  be  great  in  any  case. 

It  is  different  at  the  entrance  to  the  wheel  P.  Unless  the 
guide-blades  are  directed  so  as  to  give  a  tangential  velocity  to 
the  water  equal  to  that  of  the  wheel,  there  is  a  considerable 
loss  by  friction  at  F.  (See  Appendix.) 

430.  Suppose  that  less  water  flows  through  the  turbine,  the 
inclination  of  the  guide-blades  ought  to  alter,  and  this  arrange- 
ment of  links,  which  we  see  in  the  drawing,  is  for  the  purpose 


530  APPLIED    MECHANICS. 

of  making  the  guide-blades  alter  their  inclinations  to  the 
wheel.  Each  guide-blade  is  pivoted  at  its  extremity,  K,  and 
when  one  is  shifted  they  are  all  shifted  in  position.  Unless 
there  is  a  great  variation  in  the  work  which  we  require  a 
turbine  of  this  kind  to  do,  it  is  not  necessary  to  apply  a 
governor  which  partially  stops  the  water  supply  when  the 
machinery  runs  a  little  too  quickly,  although  such  governors 
are  very  necessary  for  a  great  many  water-wheels  and  turbines. 

It  is  to  be  remembered  that  this  turbine  is  really  a  centri- 
fugal pump,  through  which  the  water  is  flowing  negatively. 
Increased  speed  tends  to  stop  the  flow.  If  the  wheel  were  at 
rest,  the  flow  would  be  very  much  greater  than  it  is.  Hence, 
increasing  the  speed  somewhat  stops  the  flow,  allows  less 
water  to  pass  through,  and  less  work  to  be  done.  This  action 
cannot  be  called  a  governor  action,  for  it  does  not  maintain  a 
constant  speed,  but  it  may  be  called  a  steadying1  action,  as  it 
prevents  any  great  change  of  speed,  even  for  a  considerable 
alteration  in  the  work  done. 

Except  at  the  speed  for  which  the  positions  of  the  guide- 
blades  are  fixed,  there  is  extra  loss  in  friction,  and  the  guide- 
blades  are  rearranged  should  any  considerable  change  be 
meditated  in  the  power  to  be  given  out. 

431.  In  arranging  a  turbine,  it  is  obvious  that  the  great 
point  to  settle  beforehand  is  this  : — What  ought  to  be  the  speed 
of  the  wheel  for  a  given  height  of  fall  1  If  there  were  no  loss  in 
friction,  we  could  say  at  once,  if  v  is  velocity  of  rim  of  wheel, 
-y2  -r  32,  the  total  loss  of  energy  by  one  pound  of  water,  ought 
to  be  equal  to  h  ;  that  is,  the  velocity  of  the  wheel  ought  to 
be  that  due  to  half  the  height  of  the  total  fall  of  the  water. 
Thus,  for  a  fall  of  60  feet  in  height,  half  of  this  is  30  feet; 
and  if  a  stone  fell  30  feet,  it  would  be  falling  with  a  velocity 
of  44  feet  per  second.  The  rim  of  the  wheel  ought  to  have  a 
velocity  of  44  feet,  then,  per  second,  and  it  is  easy  to  show 
that,  wherever  the  turbine  may  be  placed,  whether  it  has  a 
long  discharge  pipe,  or  is  submerged,  the  water  may  be  made 
to  flow  tangentially  into  the  wheel  with  the  same  velocity  as 
the  wheel  itself  has. 

But  we  have  usually  to  calculate  on  the  assumption  that  a 
certain  fraction  of  the  energy  of  the  water  is  wasted  in  the 
supply  and  discharge  pipes,  and  the  discharge  chamber,  and 
hence  the  velocity  of  the  wheel  is  less  than  that  due  to  half 
the  height  of  the  fall. 


APPLIED    MECHANICS.  531 

It  is  usual  to  assume  that  the  radial  velocity  of  the  water 
through  the  wheel  is  one-eighth  of  that  due  to  the  total  fall. 
Dividing  this  into  the  number  of  cubic  feet  of  water  flowing, 
we  know  the  total  tangential  area  of  the  space  between  the 
vanes  everywhere  in  the  wheel,  assuming  that  it  is  the  same 
everywhere  and  it  usually  is.  It  is  usual  to  take  the  inner 
radius  of  the  wheel,  E  M,  equal  to  the  depth,  M  G,  of  the  passages 
in  the  wheel,  so  that  both  these  dimensions  are  now  fixed. 
The  outer  radius  is  generally  twice  the  inner  one,  and  we  have 
already  calculated  the  tangential  velocity  of  the  outside,  so 
the  number  of  revolutions  per  minute  may  be  calculated.  The 
horse-power  given  out  is  usually  taken  to  be  less  than  three- 
fourths  of  the  true  horse-power  of  the  water.  Thus,  by  rules, 
partly  due  to  practical  experience  and  partly  due  to  imperfect 
theory,  we  are  able  to  fix  all  the  dimensions  of  a  turbine  of 
the  kind  we  have  been  describing.  (See  Appendix.) 

432.  We  think  that,  by  entering  thus  fully  into  the  theory 
and  construction  of  Thomson's  turbine,  we  can  dispense  with 
giving  a  catalogue  of  the  constructions  of  turbines  generally. 
This  turbine  is  said  to  be  one  of  "  inward  radial  flow."  We 
see  that,  for  a  given  quantity  of  water  flowing,  it  can  be  made 
hydraulically  perfect  \  that  is,  by  proper  construction  of  the 
guide-blades,  there  is  no  necessary  loss  of  energy,  any  more 
than  in  the  whirlpool  chamber  of  Thomson's  centrifugal  pump. 

In  the  same  manner,  we  could  discuss  the  action  of  water 
m  the  unsteady  "  outward  radial  flow  turbines,"  and,  again, 
in  the  axial-flow  turbines  of  Fourneyron  and  others.  The 
principle  of  our  stream  of  acrobats  jumping  on  and  off  a 
merry-go-round  will  in  every  case  tell  us  how  much  energy 
the  water  gives  to  the  wheel  of  a  turbine,  whatever  may  be  the 
nature  of  the  flow.  In  much  the  same  way,  also,  we  consider 
the  construction  of  the  floats  of  undershot  water-wheels,  and 
all  other  wheels  on  which  the  water  acts. impulsively. 

In  the  same  way  we  might  discuss  a  steam  turbine,  or  the 
action  of  air  in  motion  on  windmills.  Steam  turbines  are  dis 
cussed  in  my  book  on  the  Steam  Engine. 

When  the  available  fall  is  over  200  feet,  it  is  not  advisable 
to  use  a  turbine  water-wheel.  In  the  turbine,  as  we  saw, 
there  is  at  least  one  part  of  the  arrangement  in  which  about 
half  the  total  store  of  energy  is  in  the  shape  of  kinetic  energy ; 
and  when  the  energy  is  in  the  shape  of  kinetic  energy,  there 
is  a  great  waste  by  friction.  The  waste  is  proportional  to  the 


532  APPLIED    MECHANICS. 

kinetic  energy  —  that  is,  to  the  total  energy  —  and  hence  turbines 
are  at  least  not  more  economical  on  very  high  falls  than  on 
moderately  low  ones.  (See  Appendix.) 

433.  Steady  Motion  in  Fluids.—  The  mathematical  expression  of 
the  law  (Art.  410)  is  true  whether  the  motion  is  steady  or  not,  but  we 
give  to  the  mathematical  expression  the  energy  meaning  when  the 
motion  is  steady.  Motion  of  a  fluid  is  steady  when  every  particle 
at  any  point  moves  in  exactly  the  same 
way  as  all  its  predecessors  there.  The 
successive  positions  of  a  particle  mark 
out  a  stream  line,  and  a  bundle  of 
stream  lines  lie  in  a  stream  tube.  Let 
A  K  be  a  very  thin  stream  tube.  At  c 
let  the  pressure  be  p,  the  velocity  in  the 
direction  towards  u  be  v,  and  let  c  be  A 
feet  vertically  above  some  datum  level. 
Let  the  tangent  to  the  centre  line  c  D 
Pig.  303.  make  an  angle  a  with  the  vertical. 

Let  p  +  Sp,  v  +  $v,  h  +  8h  be  these 

quantities  at  D,  and  let  D  be  very  near  c.  Let  the  section  at  c  or 
at  n  be  a.  Consider  the  dynamic  condition  of  the  portion  of  fluid 
in  the  tube  between  c  and  D  in  its  motion  along  the  tube.  The 
force  urging  it  towards  B  is 

pa  -  (p  +  Sp}  a  -  w  .  c  D  .  a  .  cos.  o 

if  w  is  its  weight  per  unit  volume,  because  w  .  c  D  .  a  is  the  weight 
of  the  element,  and  we  resolve  this  weight  in  the  direction  of  the 
inclined  line.  We  must  state  that  this  force  is  equal  to  the  mass 
multiplied  by  its  acceleration.  The  mass  is  w  .  c  D  .  a  -r-  y.  If  8t 
is  the  time  which  a  particle  would  take  in  going  through  the 
distance  CD,  its  velocity  is  CD  -s-  8t,  and  its  acceleration  is  8v  ^  St. 
Hence  we  may  say 

pa  -  (p  +  Ip)  a  -  w  .  CD  .  a.  cos.  a  =  *°  '  CD  '  *  .  **. 

9  5t 

SV         CD, 

But  c  D  .  —  =  —  .  ov  =  v  .  Sv. 

01  01 

The  above  statements  are  true  only  when  c  D  is  thought  to  be 
smaller  and  smaller  without  limit.  Dividing  by  a,  and  recollect- 
ing that  c  D  .  cos.  a  =  5A,  we  find 

-  Sp  -  w  .  5h  =  -  v  .  5  v, 


Or,  as  we  may  write  it,  with  the  idea  that  the  values  of  ov,  etc. 
are  smaller  and  smaller  without  limit, 


APPLIED    MECHANICS.  533 

For  liquids,  w  is  constant,  and  hence,  integrating, 

JL  +  'L  +  h  —  constant  ....  (2). 
2ff      w 

If  w  is  not  constant,  we  cannot  integrate  until  we  know  how  w 
varies  as  a  function  of  p.     Until  we  know  thij,  we  can  only  write 


JL_  +  |  ^  +  h  =  constant  (3). 


We  note  in  (2)  that  v^g  is  the  kinetic  energy  of  a  pound  ol 
water,     h  is  its  potential  energy.     And,  he  it  because  of  Art.  410, 


or  merely  as  a  help  to  the  memory,  we  mean  to  call  I  dp/w,  or  p/w, 

the  pressure  energy  per  pound  of  fluid.  The  total  energy  of  a 
pound  of  fluid  remains  constant,  except  in  so  far  as  friction 
may  diminish  it. 

Example. — In  Fig.  306  we  see  some  stream  lines  of  a  fluid 
leaving  a  vessel  by  an  orifice.  We  assume  no  friction.  I  shall 
consider  the  various  orifices  shown  on  an  enlarged  scale  in  Fig.  286. 

We  wish  to  make  the  statement  that  (2)  is  the  same  at  A  as  at 
c.  Observe  that  at  A  the  pressure  is  atmospheric.  But  what  is 
the  pressure  at  c  ?  If  c  is  a  stream  line  touching  the  atmosphere, 
of  course  the  pressure  is  known;  but  inside  the  stream  at  c  the 
pressure  is  not  known  with  certainty.  We  assume  it  to  be 
atmospheric.  Now  a  pound  of  water  in  coming  from  A  to  c  has 
had  no  change  in  its  pressure  energy.  It  has  lost  potential  energy 
H  feet,  if  H  is  the  vertical  depth  of  c  below  A.  It  had  no  kinetic 
energy  at  A  ;  and  if  v  is  its  velocity  at  c,  its  kinetic  energy  there  is 
v2/2^.  Therefore  we  say  that  the  "loss  H  is  equal  to  the  gain  vz/2g, 
or  v  =  tj  2  g  H. 

It  will  be  noticed  that  we  choose  c  in  Fig.  286  at  places  where 
it  cannot  be  very  wrong  to  assume  atmospheric  pressure,  and 
where  the  stream  lines  are  all  presumably  normal  to  the  cross- 
section  a.  Hence  the  total  quantity  of  water  flowing  per  second  is 
Q  =  a  fJIg-R  .  .  .  .  (4)  cubic  feet.  Our  two  difficulties,  what  is 
a  ?  what  loss  is  there  by  friction  ?  are  solved  by  experiment.  It 
is  interesting  to  know  that  the  most  careful  measurements  of  a  and 
Q,  when  orifices  are  sharp-edged  and  water  is  flowing,  show  no 
perceptible  loss  by  friction.  This  is  one  ^reason  why  sharp-edged 
orifices  are  preferred  in  the  measurement  o*f  water  and  other  fluids. 
Another  reason  for  this  is  the  accuracy  with  which  we  can  verify 
the  shape  of  an  orifice.  Hence  in  square,  or  round,  or  triangular- 
shaped  orifices  we  have  taken  great  pains  to  find  such  a  place  as  c, 
and  to  find  a  there.  In  the  case  of  a  round  orifice  a  equals  the 
area  of  the  hole  multiplied  by  0'62  very  nearly.  The  part  c  is 
called  the  "  contracted  vein  "  often  referred  to  by  writers.  In  the 
case  of  Fig.  305,  a  is  half  the  area  A  of  the  small  hole  M  x.  This 
is  the  only  case  in  which  we  can  make  any  easy  attempt  to 
calculate  the  area  of  the  contracted  vein.  Note  that  with  this 
orifice,  whatever  be  the  shape  of  the  vessel,  the  total  pressure  of 


534  APPLIED    MECHANICS. 

the  fluid  upon  it  in  the  direction  opposite  to  the  arrow  at  c  is 
known  to  us,  being  A  w  H  if  H  is  the  depth  of  the  centre  of  A  below 
water-level,  because  the  velocities,  and  therefore  the  pressures, 
everywhere,  except  near  MN,  are  the  same  as  if  the  orifice  were 
closed;  and  near  MN  there  are  no  pressure-forces  parallel  to  the 
arrow  at  c.  Hence  A  w  H  is  the  momentum  leaving  the  vessel  per 
second.  It  is  only  when  the  orifice  is  small  that  we  can  be  sure 
that  the  average  velocity  at  c  is  ,J2,ffH,  and  the  volume  flowing 
per  second  is  a  ,J  1g  H.  The  mass  per  second  is  this  multiplied  by 
w/g,  and  the  momentum  per  second  is  this  multiplied  by  ^/  2  g  H  ; 
and  hence  A  w  H  =  2  aw  H,  or  A  =  2  a. 

434.  For  all  other  sharp-edged  orifices  than  that  shown  in 
Fig.  305,  we  rely  upon  experiment.     Q  the  quantity  in  cubic  feet 
per  second  flowing  from  a  sharp-edged  orifice  of  area  A  square 
feet,  the  centre  of  the  orifice  being  H  feet  below  still  water 
level,  Q  =  &A  \/2#H.      In  Fig.  305  k  is  ^,  as  we  have  seen. 
For  circular  orifices,  k  is   0*62   very  nearly,  even  when  the 
upper  edge  of  the  orifice  is  comparatively  near  the  upper  level, 
so   long  as  the  orifice  keeps  filled.      Again,  for  square  and 
rectangular  orifices  k  is  very  nearly  0'62.     We  shall  not  give 
the  great  table  of  numbers,  varying  from  0'61  to  0'63.  which 
have  been  experimentally  determined  for  various  sizes  and 
positions  of  rectangular  orifices,  because  we  do  not  think  it 
more  accurate  than  the  statement  that  0*62  is  very  nearly 
correct  for  all. 

435.  The  shape  of  the  stream  coming  from  a  rectangular 
orifice  is  very  interesting ;  and  a  student  must  meditate,  when 
looking  at  such  a  stream,  upon  the  way  in  which  its  component 
parts  collide  to  cause  the  curious  palpitating  change  of  shape 

M 


Fig.  304.  Fig.  305. 

and  section  which  is  going  on.     Fig.  306  gives  some  notion  of 
these  changes. 

436.  Experiment  has  shown  that  in  the  case  of  sharp-edged 
orifices  there  is  no  practical  difference  in  the  actual  flowing  of 


536  APPLIED   MECHANICS. 

water  from  what  the  flow  would  be  if  the  liquid  were  friction- 
less.  It  can  be  shown  that  when  liquid  is  frictionless  the 
stream  lines  from  similar  and  similarly  placed  orifices  in 
similar  vessels  with  the  same  kind  of  liquid  at  similar  heights 
are  similar,  the  corresponding  velocities  being  proportional  to 
the  square  roots  of  the  dimensions,  and  therefore  the  volumes 


A 


Pig.  307. 


flowing  being  proportional  to  the  two  and  a  half  powers  of  the 
dimensions.  If,  then,  water  flows  from  a  pond  over  a  sharp- 
edged  notch  shaped  like  a  right  angled  isosceles  triangle,  each 
of  the  edges  making  45°  with  the  horizontal,  as  in  Fig.  287, 
and  if  the  difference  of  level  from  B  to  A  is  H,  the  quantity  Q 
flowing  in  cubic  feet  per  second  is  proportional  to  ni  Prof. 
James  Thomson  gave  us  the  above  principle  and  this  method 
of  measuring  water.  By  careful  measurement,  he  found  that, 
H  being  in  feet,  Q  =  2'635H|.  .  .  .  (1).  The  rectangular 
notch  is  more  convenient.  Professor  Jas.  Thomson  showed 
that  the  empirical  formula  of  Mr.  Francis,  of  Lowell,  arrived 
at  with  great  care  and  at  great  expense,  is  a  rational  one. 

If  L  is  the  length  of  the  notch  in  feet,  H  being  vertical  height 
in  feet  from  sill  B  to  still- water  level,  for  a  given  H  there  is  a 
certain  value  of  L  beyond  which  increase  in  L  means  that  the 
increase  in  Q  is  proportional  to  the  increase  in  L.  In  fact,  we 
distinguish  the  flow  through  the  two  ends  of  length  m  H  at 
one  side  and  m  H  at  the  other,  and  the  flow  through  the  middle 
part  L  -  2  WH,  where  all  the  lines  of  flow  may  be  regarded  as  in 
vertical  planes.  We  have  good  reason  to  take  m  to  be  constant. 
Imagine  an  orifice  of  length  2  m  H.  The  flow  through  it  is 
ki  H  f ,  where  k\  is  some  constant.  The  flow  through  a  square 
orifice  of  height  H,  the  lines  of  flow  being  in  vertical  planes,  is 
&2  H  f ,  where  kz  is  some  constant,  and  therefore  the  middle  flow  is 
(L  -  2  m  H)/H  times  this,  or 

T       5        .    L  -  2  WH     6 

»  Q  =   &i  H  *   +   &2  H  ?. 

H 

This  will  be  found  to  reduce  to  Q,  =  b  (L  -  c  H)  H  i.  If  there  is 
only  one  end  contraction,  c  is  evidently  halved ;  and  if  there  are  no 
end  contractions,  c  is  Q. 

The  experiments  of  Mr.  Francis   give  us   the  values   of 
6  and  c,  so  that 

Q»  3-33(1-       H)H*....(2), 


APPLIED    MECHANICS.  537 

where  n  is  2  or  1  or  0,  according  as  we  have  all  the  edges 
sharp  or  we  have  the  edge  B  c  a  smooth  vertical  guiding  plane, 
or  both  B  c  and  D  C  smooth  v.ertical  guiding  planes. 

437.  In  a  gas,  we  have  w  a  p,   where  p  is  the  pressure  in 
pounds  per  square  foot,  if  the  temperature  could  be  kept  con- 

stant, or  we  have  the  rule  for  adiabatic  flow  w  <x  p  1  ,  where  y 
is  the  well-known  ratio  of  the  specific  heats.     In  either  of  these 

cases  it  is  easy  to  find  I  —  and  write  out  the  law.     This  law  is  of 

J*> 

universal  use  in  all  cases  where  viscosity  may  be  neglected,  and  is 
a  great  guide  to  the  hydraulic  engineer.     Thus  in  the  case  of 

adiabatic  flow,  w  =  cp  1  ,  the  integral  of  —  is 

1          -1-  1  i-i 


v2        1      y         1-- 
and  hence  A  +  —  +  ---  —^  p      i  =  constant  ....  (4). 

In  a  great  many  problems,  changes  of  level  are  insignificant, 

and  we  often  use  v2  H  --  -  -  *-a  pi—  constant  ....  (4)  for 

gases.  Thus,  if  p0  is  the  pressure  and  w0  the  weight  of  a  cubic 
foot  of  gas  inside  a  vessel  at  places  where  there  is  no  velocity,  and 
if  outside  an  orifice  the  pressure  is  p,  the  constant  in  (4)  is  evidently 

2<7       y        T— 

0  +   -?  —  '—-  jj    7     and  hence  outside  the  orifice 
c    y  -  1^° 

7-1  -1, 


and  as  c  is  w0  -f-  p0  ?,  it  is  easy  to  make  all  sorts  of  calculations  on 
the  quantity  of  gas  flowing  per  second.  Observe  that  if  p  is  very 
little  less  than  p0  and  if  we  use  the  approximation  (1  +  a)n  =  1+na, 

when  a  is  small,  we  find  v2  =  —  (p0  -  p}  ....  (6),  a  simple  rule 

which  it  is  well  to  remember  in  fan  and  windmill  problems.  In  a 
Thomson  water  turbine  the  velocity  of  the  rim  of  the  wheel  is  the 
velocity  due  to  half  the  total  available  pressure ;  so  in  an  air 
turbine,  when  there  is  no  great  difference  of  pressure,  the  velocity 
of  the  rim  of  the  wheel  is  the  velocity  due  to  half  the  pressure 
difference.  Thus,  if  p0  of  the  supply  is  7,000  Ibs.  per  square  foot, 
and  if  p  of  the  exhaust  is  6,800  Ibs.  per  square  foot,  and  if  we  take 
w0  =  0-28  Ib.  per  cubic  foot,  the  velocity  of  the  run  v  is,  since  the 
difference  of  pressure  is  200  Ibs.  per  square  foot, 

'  (100)  =  151  feet  per  second. 


538  ALLIED   MECHANICS. 

Returning  to  (6) :  Neglecting  friction,  if  there  is  an  orifice  of 
area  A  near  which  the  flow  is  guided  so  that  the  streams  of  air  are 
parallel,  Q,  the  volume  flowing  per  second,  is  a  =  v  A  ;  and  if  the 
pressure  is  p,  the  weight  of  stuff  flowing  per  second  is  w  =  v  AW 

or  v  A  cpi .     Using  v  from  (5),  and  letting  p\pQ  be  called  a,  we  have 
after  simplification  ' 

~7  *~- 

y  -  i  PO  \ 

Problem. — Find  p,  the  throat  pressure,  so  that  for  a  given 
inside  pressure  there  may  be  the  maximum  flow. 

It  is  obvious  that  as  p  is  diminished  more  and  more,  v,  the 
velocity,  increases  more  and  more,  and  so  does  Q.  But  a  large  Q 
does  not  necessarily  mean  a  large  quantity  of  gas.  We  want  w  to 
be  large.  When  is  w  a  maximum  ?  That  is,  what  value  of  a  in  all 
will  make 

2  7-U  2          1  +  1 

07(1-0    7    1  or  a  7  -  «        ?a  maximum  ? 

Differentiating  with  regard  to  o,  and  equating  to  0,  we  have 
2  l 


Dividing  by  a  1  ,  we  find  a 


In  the  case  of  air  7  =  T41,  and  we  find  p  = 

there  is  a  maximum  quantity  leaving  the  vessel  per  second  when 

the  outside  pressure  is  a  little  greater  than  half  the  inside  pressure. 

Problem. — Whenj?  is  indefinitely  diminished  what  is  v  ? 


Answer  :   v 


=   A  /  JL^L 
V         -   l 


This    is    greater    than    the    velocity   of    sound   in   the    ratio 
being  2'21  for  air.     That  is,  the  limiting  velocity  in 


V 


-  1 
the  case  of  air  is  2,413  feet  per  second  x   A  /  _L,  where  t  is  the 


A/1- 

V     273 


absolute  temperature  inside  the  vessel,  and  there  is  a  vacuum  outside. 
This  involves  the  idea  of  the  jet  creating  such  intense  cold  as  to  be 
at  the  absolute  zero  of  temperature.  (See  Appendix.) 

Returning  to  equations  (3)  and  (4),  we  assumed  h  to  be  of  little 
importance  in  many  gaseous  problems  of  the  mechanical  engineer. 
But  there  are  many  physical  problems  in  which  it  is  necessary  to 
take  account  of  changes  in  level.  For  example,  if  (3)  is  integrated 
on  the  assumption  of  corstant  temperature,  and  we  assume  v  to 


APPLIED    MECHANICS.  539 

keep  constant,  we  find  that  p  diminishes  as  h  increases,  according 
to  the  compound  interest  law.  Again,  under  the  same  condition 
as  to  v,  but  with  the  adiabatic  law  for  w,  we  find  that  p  diminishes 
with  h  according  to  a  law  which  may  be  changed  into  "  the  rate 
of  diminution  of  temperature  with  h  is  constant." 

A  great  number  of  interesting  examples  of  the  use  of  (2)  might 
be  given.  It  enables  us  to  understand  the  flow  of  fluid  from 
orifices,  the  action  of  jet-pumps,  the  attraction  of  light  bodies 
caused  by  vibrating  tuning-forks,  why  some  valves  are  actually 
sucked  up  more  against  their  seats  instead  of  being  forced  away  by 
the  issuing  stream  of  fluid,  and  many  other  phenomena  which  are 
thought  to  be  very  curious. 

438.  Example.  —  Particles  of  water  in  a  basin,   flowing  very 
slowly  towards  a  hole  in  the  centre,  move  in  nearly  circular  paths,  so 
that  the  velocity  v  is  inversely  proportional  to  the  distance  from  the 

centre.     Take"v  =  -,  where  a  is  some  constant  and  x  is  the  radius 
x 

a?         i) 
or  distance  from  the  axis.     Then  (3)  becomes  h  +  -  —  2  +  J—  =  Ct 

Now  at  the  surface  of  the  water  p  is  constant,  being  the  pressure 

eft 
of  the  atmosphere  ;  so  that  there  h  =  C  -  ~  —  ^  and  this  gives  us 

the  shape  of  the  curved  surface.  Assume  c  and  a  any  values,  and 
it  is  easy  to  calculate  h  for  any  value  of  x  and  so  plot  the  curve. 
This  curve  rotated  about  the  axis  gives  the  shape  of  the  surface, 
which  is  a  surface  of  revolution. 

A  student  who  depends  upon  a  text-book  to  give  him  complete 
information  is  not  learning  to  become  an  engineer.  Meditation 
when  looking  at  water  flowing  from  a  basin  ought  to  greatly  add 
to  what  he  will  obtain  from  such  a  book  as  this.  Perhaps  he  will 
begin  to  notice  that  it  is  centrifugal  force  due  to  whirling  motion 
which  maintains  pressure  at  a  place.  What  will  be  the  effect  of 
friction  at  the  solid  surface  of  the  basin  ?  It  will  diminish  velocity 
and  diminish  pressure.  Water  will  flow,  therefore,  down  the 
surface  of  the  basin  and  towards  the  hole.  If  this  is  well  under- 
stood, the  student  will  understand  how  it  is  that  at  a  bend  of  a 
river  the  earth  from  the  outer  bank  is  dragged  along  the  bottom 
and  deposited  on  the  inner  bank,  and  hence  that  a  river  through 
an  alluvial  plain  is  always  tending  to  get  more  crooked,  until  at 
length  it  cuts  off  a  bend  and  so  straightens  itself  in  a  new  channel. 
In  the  basin  problem  we  have  also  James  Thomson's  explanation 
of  the  phenomena  of  great  forest  fires,  and  also  of  the  prevailing 
wind  system  of  the  earth.  (See  Phil.  Trans.  A.,  Vol.  183.) 

439.  If  water  flowing  spirally  in  a  horizontal  plane  follows  the 
law  v  =  -,  where  r  is  distance  from  a  central  point;    noio  that 

W  =  c  -  \—  -y     The  ingenious  student  ought  to  study  how  p 


and   v  vary  at   right   angles   to   stream  lines.     He   has   only  to 
consider  the  equilibrium  of  an  elementary  portion  of   fluid,   cw 


540 


APPLIED    MECHANICS. 


(Fig.  308),  subjected  to  pressures,  centrifugal  force,  and  its  own 
weight  in  a  direction  normal  to  the  stream.  He  will  find  that  if 

-~  means  the  rate  at  which  p  varies  in  the  direction  of  the  radius 

dn 

of  curvature  away  from  the  centre  of  curvature,  and  if  a  is  the 

angle  D  c  E  (Fig.  303),  the  stream  being  in  the  plane  of  the  paper, 

which  is  vertical,  and  if  r  is  the  radius  of  curvature, 

f  =  ™V---w  sin.  «....(!). 
dn        g  r 

If  the  stream  lines  are  all  in  horizontal  planes, 
dp       w  v2  .  . 

dn  ~  g  r 
Stream  lines  all  circular  and  in  horizontal  planes  in  a  liquid,  so 

that  h  is  constant.     If  v  =  -,  where  b  is  a  constant, 

dp-_-w    IP 
dr  ~  g  '  r8' 

p  =  -  £  -  -2  +  constant  ....  (3). 

We  see,  therefore,  that  the  fall  of  pressure  as  we  go  outward  is 
exactly  the  same  as  in  the  exercise  at  the  end  of  last  article. 

Example. — Liquid  rotates  about  an  axis  as  if  it  were  a  rigid 
body,   so  that  v  =  br,  then    ~  =  —  fir,  p  =  -  -  £2r2  +  c.      This 

approximately  shows  the  law  of  increase  of  pressure  in  the  wheel 
of  a  centrifugal  pump  when  full,  but  when  delivering  no  water. 
It  is  the  answer  already  obtained  (Art.  175). 

EXERCISES. 

1.  The  pressure  at  the  inside  of  the  wheel  of  a  centrifugal  pump  is 
2,116  Ibs.  per  square  foot;  the  inside  radius  is  0'5  foot;  the  outside  radius 
1  foot.     The  angular  velocity  of  the  wheel  is  b  =  30  radians  per  second. 
Draw  a  curve  showing  the  law  of  p  and  r  from  inside  to  outside  when 
very  little  water  is  being  delivered.     If  the  water  leaves  the  wheel  by  a 
spiral  path,  the  velocity  everywhere  outside  being  inversely  proportional 
to  r,  draw  also  the  curve  showing  the  law  of  p  in  the  whirlpool  chamber 
outside. 

2.  The  expression    ^2        \ 


which  remains  constant  all  along  a  stream  line,  may  be  called  the  total 
store  of  energy  of  1  Ib.  of  water  in  the  stream  if  the  motion  is  steady. 
__  <ZE  il  dv  1  dp  dh  ,  ,.  .,. 

Now  -j—  =  -  v-^  H  —  4~  +  T-  becomes  from  equation  (1), 
dn      g    dn      w  dn      dn 

dv 


n  "being  in  a  normal  direction  away  from  the  centre  of  curvature  and  r 
the  radius    of    curvature.      This  expression  -=  (-  +  ^-  j  is  called  the 


APPLIED    MECHANICS.  541 

"average  angular  velocity  "  or  "  the  rotation "  of  the  liquid.     Hence 

dv      2v 

-r-  =  —  x  rotation  ....  (6). 

dn        g 

3.  Show  that  the  law  (1)  for  a  gas  under  adiabatic  conditions,  being 
4  of  Art.  437,  the  above  laws  (5)  and  (6)  hold  for  a  gas  as  well  as  a  liquid. 

4.  Circular  Stream  Lines. — What  is  v  as  a  function  of  r,  the  radius, 
if  the  flow  is  to  be  irrotational— that  is,  if  a  pound  of  water  or  of  gas  has 
the  same  energy  in  one  stream  line  as  in  another  ?     Here  v/r  +  dv/dr  =  0, 
or  dr/r  +  dvjv  =  0,  or  log.  r  +  log.  v  =  c,  a  constant,  or  v  a  1/r. 

5.  In  a  gas  flowing  irrotationally  in  circular  streams,  if  v  =  -, 
everywhere.     Inserting  v  =  b  in  (4)  of  Art.  437,  we  find 


6.  A  centrifugal  fan  makes  3,000  revolutions  per  minute.  It  is  2  feet 
in  inside  diameter,  4  feet  outside,  and  there  is  a  whirlpool  chamber  outside 
the  fan.  No  air  is  being  delivered  from  the  chamber.  The  pressure  just 
inside  the  wheel  is  1  atmosphere,  2,116  Ibs.  per  square  foot,  and  0°  C. 
Draw  curves  showing  the  pressure  at  any  distance  from  the  centre  of  the 
wheel,  both  in  the  wheel  and  the  whirlpool  chamber.  The  air  follows  the 
adiabatic  law.  If  the  speed  is  40  revolutions  per  minute,  and  the  fluid  is 
water,  draw  the  curves. 

440.  When  liquid  flows  by  gravity  from  a  small  orifice  in  a 
large  vessel,  where  at  a  distance  inside  the  orifice  the  liquid  may  be 
supposed  at  rest,  it  is  obvious  the  E  is  the  same  in  all  stream  lines, 

so  that  —  is  0,  and  there  is  no  "  rotation  "  anywhere.     It  can  be 

proved  that  if  a  portion  of  frictionless  fluid  possesses  "  rotation," 
this  can  never  be  destroyed.  Nor  can  rotation  be  created.  But 
the  student  must  refer  to  books  on  hydrodynamics  for  further 
information.  When  unstable  states  of  motion  set  in,  the  mathe- 
matics get  beyond  the  author's  powers,  even  in  straight  pipes. 
The  papers  of  Professor  Reynolds  may  be  consulted.  As  for  what 
occurs  in  viscous  fluid  at  a  bend  in  a  pipe,  nobody  has  done  more 
than  guess  as  yet.  At  small  velocities,  we  have  a  fairly  good 
knowledge  of  what  happens.  Suppose  very  viscous  liquid  is 
flowing  along  a  straight  pipe,  the  velocity  everywhere  is  such 
that  the  loss  of  energy  by  viscosity  is  a  minimum.  If  the  pipe 
has  a  curved  centre  line,  the  distribution  of  velocity  everywhere 
is  different  from  what  it  is  in  the  centre  line. 

When  fluid  flows  in  circular  cylindric  stream  lines,  a  stream  of 
internal  and  external  radii  r  and  r  +  8r,  of  unit  breadth,  is  acted 

on  by  tangential  force  on  its  internal  cylindric  surface  /*  (  .    —  J 

per  unit  area,  if  r  is  the  velocity.  These  forces  act  on  the  area 
2  ttr  at  a  distance  r  from  the  axis,  so  that  their  total  moment" 


512  APPLIED   MECHANICS. 

Let  the  moment  of  the  forces  on  the  outside  surface  be  M  -f  5  M. 
If  there  is  no  alteration  of  pressure  along  a  stream  line,  and  only 
balanced  forces  in  planes  at  right  angles  to  the  axis,  then  5  M  is  the 
only  cause  of  increase  of  moment  of  momentum  of  the  stream. 
The  state  into  which  the  system  settles  is  one  in  which  the  motion 

dM. 
is  steady.     In  fact,  --v-  is  0,  or 

d?v       1  dv       v  (t>. 

<P  +  rrfr-7*  =  0""(2)- 
The  general  solution  of  this  is 

*  =  Ar +  *....  (3). 

If  at  a  cylindric  surface  r  =  r0,  v  =  v0,  and  if  at  another  r  ~  "4, 
v  =  0,  then 


So  that  B  =  -  Arj2,  v0  —  A  (  r0  -  -1-),  A  =      *°V°  g,  and  hence  (3) 

\          ro  /  ro   ~  ri 

becomes  v  =     „  °'°  0  (  r  --  -  V     If,  then,  viscous  fluid  escapes 
r02  -  r-f  \         r  ) 

from  the  rim  of  the  wheel  of  a  centrifugal  pump  with  little  radial 
velocity  into  a  chamber  bounded  by  parallel  Motionless  faces,  if  r} 
is  the  outer  radius  of  the  chamber, 


And  if  TI  =  GO,  then  A  must  be  0  in  (3),  and  v0  =  —  ,  so  that  (3) 
becomes 


•  = 


The  loss  of  energy  per  second  is  -  fiv  (  ~  -  -  j  per  unit  volume, 
or  -  M  r°J°  (  -  2r°2°)  =  2p  T^.    The  volume  of  the  ring  between 

/*OO  j 

r  and  r  +  8r  is  2irr  .  Sr,  so  that  I     4  7r^r02^02  •    2  is  the  tota*  ^OS8>  or 

" 


4  ir/*r0t>02.  If  v  is  the  small  radial  velocity  of  fluid  leaving  a  wheel, 
the  weight  of  fluid  per  second  leaving  unit  breadth  of  the  wheel  is 
2  TrrQ  v  w.  The  kinetic  energy  of  it  due  to  its  tangential  velocity 

only    is    2*r0 — v02.      Its    kinetic    energy    per    pound    is      °. 

Total  kinetic  energy       v  w      .„     ,  .     , .  , 

si  — .     The  kinetic  energy  di 

•wasted  energy  2^u 


APPLIED    MECHANICS.  543 

velocity    only    is    2irr0.      If    the    law    were    v  =  m    (*!  —  r\ 

(_  j.  2  rz         \  /2r2v 

TV  -  1  -  ra  +  l  )  =  +  /**»»  \-^-)-    T*1*8  is  loss  of 

energy  per  second  per  unit  volume,  or  2rj2>nV  (*±  -  -\     Total 
loss  =  P  \  7rr> 


MISCELLANEOUS  EXERCISES. 

1.  In  a  force  pump  used  for  feeding  a  boiler  the  ram  has  a  diameter 
of  2  inches  and  a  stroke  of  24  inches.     How  many  gallons  of  water 
(neglecting  leakages)  would  be  forced  into  the  boiler  for  each  1,000 
double  strokes  (one  forward  and  one  backward)  of  the  pump  ? 

Ans.,  272  gallons. 

2.  How  many  gallons  of  water  will  be  delivered  per  hour  by  a  single- 
acting  pump  (diameter  of  plunger,  4  inches;  length  of  stroke,  12  inches) 
making  24  strokes  per  minute,  the  slip  being  15  per  cent.  ?    How  long 
would  it  take  this  pump  to  fill  a  tank  10  feet  by  6  feet  to  a  depth  of  3 
feet  ?  Am.,  665  gallons  ;  1  hour,  41  mins. 

3.  Certain  machinery  worked  from  an  accumulator  requires  20-horse 
power  for   a   quarter  of  an  hour   every  hour.      The   pressure   in   the 
accumulator  is  700  Ibs.  per  square  inch.     If  25  per  cent,  be  allowed  for 
frictional  losses,  find  the  size  of  a  single-acting  pump  which,  driven  for 
thirty-five  minutes  every  hour  by  a  donkey  engine,  making  50  strokes  a 
minute,  will  just  have  the  ram  at  the  top  of  its  stroke  at  the  beginning 
of  each  hour.     Assume  a  slip  in  the  pump  of  25  per  cent. 

Am.,  -094  cubic  foot. 

4.  In  a  double-acting  force-pump  the  diameter  of  piston  is  12  inches 
and  the  stroke  2  feet  6  inches.    The  distance  from  the  pump  to  the  well 
is  15  feet,  and  from  the  pump  to  the  place  of  delivery  is  35  feet.     Find 
the  horse-power  required  to  work  the  pump  if  30  per  cent,  is  wasted  in 
friction  and  the  number  of  strokes  be  40  per  minute.  Ans.,  21  16. 

5.  The  diameter  of  a  pump  bucket  being  6  inches,  and  the  vertical 
lift  from  the  well  to  the  point  of  delivery  being  40  feet,  find  the  load  on 
the  bucket.    "What  horse-power  will  be  necessary  if  the  stroke  be  15 
inches  and  there  are  20  double  strokes  per  minute  ?    Allow  30  per  cent. 
for  all  losses.  Ans.,  489  Ibs.  ;  -53. 

6.  The  discharge  from  a  pipe  is  12  gallons  a  second.     At  a  point  110 
feet  above  datum  level  the  diameter  is  5  inches,  and  the  pressure  2,050 
Ibs.  per  square  foot.     Find  the  total  energy  of  each  pound  of  water.     If 
at  a  point  where  the  pipe  is  2J  inches  diameter  and  10  feet  above  datum 
the  pressure  is  1,000  Ibs.  per  square  foot,  find  the  loss  of  energy  between 
the  two  points  mentioned.  Ans.,  146  ft.  Ibs.  ;  70'4  ft.  Ibs. 

7.  The  volume  of  water  passing  along  a  pipe  running  full  is  10  cubic 
feet  per  second.    At  one  section  the  area  is  2  square  feet  ;  at  another 


544  APPLIED    MECHANICS. 

place,  12  feet  below  the  level  of  the  former,  the  area  of  the  cross- 
section  is  1£  square  feet.  Find  the  difference  of  pressure  at  the  two 
sections,  friction  being  neglected.  Ans.,  710  Ibs.  per  sq.  i't. 

8.  At  a  certain  point,  15  feet  above  datum,  in  a  stream  flowing  from 
a  reservoir,  the  still  surface  of  which  is  150  feet  above  datum,  the  velocity 
is  20  feet  per  second.     What  is  the  pressure  at  this  point,  assuming  no 
loss  by  friction?  Ans.,  70 '4  Ibs.  per  sq.  in. 

9.  Find  the  power  of  a  waterfall  where  2,000  cubic  feet  of  water  pass 
per  minute,  the  height  of  the  fall  being  30  feet. 

In  a  waterfall,  20  tons  of  water  fall  from  a  height  of  36  feet  in  each 
minute,  and  are  employed  to  turn  a  turbine  which  transforms  six-tenths 
.of  the  energy  of  the  water  into  useful  work.  Find  the  horse-power  of 
the  turbine.  Ans.,  113-3  h.p. ;  29*32. 

10.  The  vertical  distance    from    still- water    level  to  the  lip  of    a 
rectangular  notch  was  observed  to  be  '3  feet  during  an  interval  of  six 
hours,  -65  feet  during  the  next  twelve  hours,  and  -4  feet  during  the  next 
six  hours,  the  width  of  the  notch  being  2£  feet.     Find  the  number  of 
gallons  passing  through  during  the  twenty-four  hours. 

Ans.,  1,567,000  gallons. 

11.  The  flow  of  water  in  a  certain  stream  is  measured  by  employing  a 
Thomson's  V-shaped  weir  gauge.     The  vertical  distance  from  still-water 
level  to  the  lowest  point  of  the  notch  is  observed  to  be  1-4  feet.     The 
water  which  passes  through  has  a  fall  of  20  feet,  and  is  employed  to  drive 
a  water-wheel  having  an  efficiency  of  60  per  cent.     Find  the  horse-power 
which  may  be  obtained  from  the  wheel.  Ans.,  8-33. 

12.  A  waterfall  is  to  be  utilised  for  electric  lighting.     The  engineer 
sent  to  inspect  the  place  finds  out  the  following  data  •  — The  water  at  one 
place  flows  in  a  straight  rectangular  channel  4|  feet  wide,  1\  feet  deep, 
with  an  average  velocity  of  3  feet  per  second.     The  available  fall  is  20 
feet,  and  the  water-wheels  to  be  used  have  an  efficiency  of  62  per  cent., 
the  dynamo  efficiency  being  80  per  cent.     Neglecting  all  other  losses  of 
energy,  find  approximately  how  many  60- watt  incandescent  lamps  may 
be  supplied.  Ans.,  471. 

13.  The  rim  of  the  wheel  of  a  centrifugal  pump  goes  at  30  feet  per 
second.     Water  flows  radially  at  5  feet  per  second.      The  vanes  are 
inclined  backwards    at   an   angle  of    35°  to   the  rim.      What  is  the 
absolute  velocity  of  the  water  ?    What  is  the  component  of  this  parallel 
to  the  rim?    If  120  cubic  feet  of  water  leave  the  rim  every  minute,  find 
the  tangential  retarding  force  at  the  rim.     What  is  the  work  done 
usefully  per  pound  of  water  ? 

Ans.,  23-4  ft.  per  sec.  at  12°'3  with  direction  of  rim ;  22-86  feet  per 
gee. ;  88-45  Ibs. ;  21-3  ft.  Ibs. 

14.  Suppose  water  to  flow  in  a  steady  stream  with  a  constant  total 
head  of  100  feet,  reckoned  from  the  datum  plane  and  from  zero  pressure. 
Determine  the  discharge  into  the  atmosphere  in  gallons  per  minute  from 
a  pipe  2  inches  diameter  at  a  point  5  feet  above  the  datum.     Ans.,  638. 

15.  An  orifice  1  square  inch  area  is  made  in  the  side  of  a  large  tank, 
at  a  depth  of  4  feet  below  the  surface  of  the  water,  and  the  issuing  jet  ia 
horizontal,     If  the  jet  falls  vertically  through  If  feet  in  a  horizontal 


APPLIED   MECHANICS.  545 

motion  of  5  feet,  and  the  discharge  be  16  gallons  per  minute,  find  the 
coefficients  of  velocity,  contraction,  and  discharge. 

Am.,  -97  ;  -64  ;  -62. 

16.  Find  the  discharge  in  gallons  per  minute  from  an  orifice  2  inches 
in  diameter  in  the  side  of  a  tank  under  a  constant  head  of  6   feet, 
measured  from  the  centre  of  the  orifice.     The  coefficient  of  discharge 
may  be  taken  at  '6.  Am.,  96. 

17.  In  an  inward  flow  wheel  the  velocity  of  flow  is  8  feet  per  second, 
the  internal  diameter  9  inches,  and  the  revolutions  10  per  second.     Find 
the  angle  of  the  vanes  at  exit,  so  that  the  water  may  leave  the  wheel 
radially.  Ans.,  18° -75. 

18.  Determine  the  velocity  with  which  the  water  enters  an  inward 
flow  turbine  under  a  head  of  36  feet,  the  speed  of  the  periphery  of  the 
wheel  being  32  feet  per  second.     The  vanes  of  the  wheel  are  radial  at 
entrance,  the  velocity  of  flow  is  constant,  and  the  water  leaves  the  whe&i 
with  no  tangential  velocity.  Ans.,  32  '56  ft.  per  sec. 

19.  What  horse-power  is  required  to  drive  a  radial- vaned  pump  of  15 
feet  diameter  at  50  revolutions  per  minute  when  delivering  15,000  gallons 
of  water  per  minute  ?     What  is  the  efficiency  if  the  lift  is  22  feet  ? 

Ans.,  219;  -46. 

20.  A  stream  of  water,  the  volume  of  flow  of  which  is  3,000  gallons 
per  minute,  has  a  velocity  of  20  feet  per  second.     It  impinges  on  a 
succession  of  curved  vanes  moving  with  a  velocity  of  8  feet  per  second  in 
a  direction  inclined  at  45°  to  the  direction  of  the  stream.     Determine  the 
direction  of  the  tangent  to  the  vane  at  entrance,  so  that  the  water  may 
impinge  without  shock.     If  the  vanes  are  circular  arcs  of  90°,  find  the 
resultant  pressure  on  the  vanes,  and  the  component  force  in  the  direction 
of  motion.  Ans.,  21°'5  with  direction  of  jet. 

21.  Find  the  horse-power  developed  in  a  Thomson  turbine  which  is 
supplied  with  15  tons  of  water  per  minute,  with  a  forward  tangential 
velocity  of  40  feet  per  second,  equal  to  the  speed  of  the  periphery  of  the 
wheel,  the  diameter  of  which  is  2  feet.     The  water  leaves  the  turbine  at 
a  radius  of  6  inches,  with  a  backward  tangential  velocity  of  10  feet  per 
second.  Ans.,  57'27. 

22.  The  diameters  of  the  inner  and  outer  circumferences  of  an  inward 
flow  turbine  are  2  feet  and  4  feet  respectively.     The  direction  of  the 
vanes  at  their  outer  ends  is  radial.     Determine  the  angle  at  which  the 
inner  ends  are  arranged,  supposing  that  velocity  of  flow  through  the 
turbine  is  one-eighth  the  velocity  due  to  the  total  head,  and  that  of  the 
outer  ends  is  that  due  to  half  the  head.  "  Ans.,  19° -6  with  rim. 

23.  A  turbine  with  radial  vanes  receives  50  gallons  per  minute  with 
an  effective  head  of  28  feet.     Find  what  should  be  the  total  area  of  the 
inlet  passages,  and  the  velocity  of  the  lips  of  the  vanes  for  maximum 
efficiency.  Ans.,  1'51  sq.  ft. ;  28  ft.  per  sec. 

24.  The  wheel  of  a   centrifugal  pump  is  '6  feet  in  diameter;  the 
turning  moment  on  the  spindle  is  12  pound-feet.      If    160  gallons  of 
water  are  raised  per   minute,  find   the   mean  velocity  with  which  the 
water  leaves  the  wheel,  assuming  that  on  entering  it  has  no  velocity  of 
whirl.  Ans.,  24-1  ft.  per  sec. 


546 


CHAPTER   XXV. 

PERIODIC     MOTION. 

441.  WHEN,  after  a  certain  interval  of  time,  a  body  is  found 
to  have  returned  to  an  old  position,  and  to  be  there  moving  in 
exactly  the  same  way  as  it  did  before,  the  motion  is  said  to  be 
periodic,  and  the  interval  of  time  that  has  elapsed  is  said  to 
be  the  periodic  time  of  the  motion.     Thus,  if  a  body  moves 
uniformly  round  in  a  circle,  the  time  which  it  takes  to  make 
one  complete  revolution  is  called  its  periodic  time. 

442.  When  a  body  moves   uniformly  in  a  circle,  as,  for 
instance,  the  bob  of   a   conical  pendulum,  if  we  look  at  it 
from  a   point  in  the  plane  of  its  circle,  it  seems  merely  to 
swing    backwards   and   forwards   in  a  straight   line.      Thus, 
it   is   known   that  Jupiter's   satellites   go    round   the   planet 
in   paths   which   are   nearly  circular,   but   a   person   on   our 
earth  sees    them   move   backwards   and    forwards   almost   in 
straight  lines.     Now,  if  we  were  a  very  great  distance  away 
from  the  bob  of  a  conical  pendulum  in  the  plane  of  its  motion, 
we  should  imagine  it  to  be  moving  in  a  straight  line,  and  the 

motion  which  it  would 

°'  appear  to  have — slow 

at  the  ends  of  its 
path,  quick  in  the 
middle — would  be  a 
simple  harmonic  mo- 
tion. To  get  an  exact 
idea  of  the  nature  of 
this  motion — in  fact, 
to  define  what  I  mean 
by  simple  harmonic 
motion — draw  a  circle, 
A.  o'  L  o"  (Fig.  308), 
and  divide  its  circum- 
ference into  any  even 
number  of  equal  parts. 
Draw  the  perpendicu- 
Fig.  308.  lars  B'B,  c'c,  etc.,  to 


APPLIED    MECHANICS. 


547 


any  diameter.  Now,  if  we  suppose  a  body  to  go  back- 
wards and  forwards  along  AOL,  and  if  it  takes  just  the 
same  time  to  go  from  A  to  B  as  from  B  to  C,  or  from 
any  point  to  the  next,  then  its  motion  is  said  to  be  a  simple 
harmonic  motion.  This  sort  of  motion  is  nearly  what  we 
observe  in  Jupiter's  satellites ;  it  is  almost  exactly  the  motion 
of  the  bob  of  any  long  pendulum  or  the  cross-head  of  a  steam- 
engine;  it  is  the  motion  of  a  point  in  a  tuning-fork,  or  a 
stretched  fiddle-string  when  it  is  plucked  aside  and  set  free ; 
of  the  weight  hung  from  a  spring  balance  when  it  is  vibrating ; 
of  the  up  and  down  motion  of  a  cork  floating  on  the  waves  in 
water ;  and  of  the  free  end  of  a  rod  of  metal  when  the  other 
end  is  fixed  in  a  vice  and  the  rod  is  set  in  vibration  ;  it  tells 
us  in  all  these  cases  the  nature  of  the  motion,  when  such 
motion  is  of  its  simplest  kind.  Thus,  for  example,  a  cork 
floating  on  water  may  really  have  a  very  complicated  motion, 
but  if  the  wave  in  the  water  is  of  its  simplest  kind,  the  cork 
goes  up  and  down  with  a  simple  harmonic  motion.  If  you 
study  the  figure  which  you  have  drawn,  and  then  watch  the 
vibration  of  a  very  long  pendulum,  you  will  learn  about  this 
kind  of  motion  what  cannot  be  learnt  by  reading. 

443.  Now  let  me  suppose  that  the  body  takes  one  second  to 
go  from  A  to  B,  or  from  B  to  c,  or  from  any  point  to  the 
next  in  Fig.  308.  Then  the  length  of  A  B  in  inches  represent* 
the  average  velocity  between  the  points  A  and  B,  and  in  the 
same  way  we  get  the  average  velocity  anywhere  else.  Thus, 
in  the  figure  from  which  the  woodcut  is  drawn  I  find 


A 

B 

0 

D 

E 

F 

0 

0 

H 

j 

K 

Velocity  from 

to 

to 

to 

to 

to 

to 

to 

to 

to 

to 

to 

to 

B 

C 

D 

E 

F 

0 

0 

H 

I 

j 

K 

L 

8  in  inches  per    "1 
second               J 

0-34 

1-00 

1-59 

2-07 

2-41 

2-59 

2-59 

2-41 

2-07 

1-59 

1-00 

0-34 

We  observe  that  the  velocity  increases  as  the  body  ap- 
proaches the  middle  of  the  path,  and  diminishes  again  as  it 
goes  away  from  the  middle.  Now  the  increase  in  the  velocity 
of  a  body  every  second  is  called  its  acceleration,  and  so  we  can 
observe  what  is  the  acceleration  at  every  place.  You  see 
that  the  velocity  changes  from  '34  to  I'OO  near  B  in  one 
second — that  is,  the  acceleration  near  B  is  -66  inch  per  second 


548 


APPLIED    MECHANICS. 


per  second.  Similarly  subtracting  1  -00  from  1  -59  we  find  the 
acceleration  at  c  to  be  0-59,  and  so  on.  Make  a  table  of  these 
values,  and  place  opposite  them  the  distances  of  the  points  B,  c, 
etc.,  from  the  centre.  In  this  way  we  find  from  the  figure  the 
following  Table  of  Values  : — 


Distance  from  o  to 

Acceleration  at 

Displacement  divided  by 
Acceleration. 

B  is  9-66 

B  is  0-66 

14-6 

c  is  8-66 

c  is  0-59 

14-7 

D  is  7'07 

D  is  0-48 

14-7 

E  is  5-00 

E  is  0-34 

14- 

p  is  2-59 

F  is  0-18 

14- 

o  is     o 

o  is     o 

G  is  2-59 

G  is  0-18 

14- 

H  is  5-00 

H  is  0-34 

14- 

I  is  7-07 

i  is  0-48 

14- 

j  is  8-66 

j  is  0-59 

14-7 

K  is  9-66 

K  is  0-66 

14-6 

From  this  it  is  evident  that  when  the  distance  of  a  point 
from  the  centre  is  divided  by  the  acceleration  at  the  point,  we 
get  about  14*6  in  every  case — that  is,  if  we  worked  more 
exactly  we  should  have  the  exact  law  that  the  acceleration  at 
a  place  is  proportional  to  the  distance  from  the  centre.  This 
curious  property  is  characteristic  of  the  kind  of  motion  which 
we  are  describing.  If,  again,  we  draw  a  number  of  figures, 
such  as  Fig.  308,  and  divide  the  circles  into  very  different 
numbers  of  equal  parts,  we  shall  find  that  in  every  case  the 
following  law  is  true : — The  periodic  time  of  a  simple  har- 
monic motion — that  is,  the  time  which  elapses  from  the 
moment  when  the  body  is  in  a  certain  condition  until  it 
gets  into  exactly  the  same  condition  again — is  equal  to  6-2832 
multiplied  by  the  square  root  of  the  ratio  of  displacement  to 
acceleration  given  in  the  third  column  of  the  above  Table. 
Thus,  in  the  Table  we  find  the  mean  value  of  the  ratio  (adding 
all  the  quotients  and  dividing  by  their  number  we  get  14-56) 
to  be,  let  us  say,  14'6.  Now  the  square  root  of  14 '6  is  3 '82, 
and  this  multiplied  by  6  -2832  is  24  seconds,  which  we  see  by 
inspection  is  the  periodic  time  in  Fig.  308. 

444.  We  see,  then,  that  if  the  force  acting  on  a  body  and 
causing  it  to  move  is  always  proportional  to  the  distance  of 


ALLIED   MECHANICS.  549 

the  tody  from  a  certain  point,  and  acts  towards  that  point, 
the  body  gets  a  simple  harmonic  motion,  and  we  have  a  rule 
for  finding  the  periodic  time. 

The  acceleration  is  always  towards  the  middle  point— that  IB, 
whilst  a  body  is  leaving  the  middle  its  velocity  is  being  lessened ; 
when  it  is  approaching  the  middle  its  velocity  is  being  increased. 
The  velocity  at  the  middle  is  equal  to  the  uniform  velocity  in  the 
circle  from  which  we  imagine  the  harmonic  motion  to  be  derived — 
that  is,  the  velocity  in  the  middle  is  equal  to  3-1416  times  the 
distance  A  L  divided  by  the  periodic  time. 

Suppose  the  body  to  be  at  Q,  Fig.  309,  moving  with  a  harmonic 
motion  in  the  path  AOL.     Describe  the 
circle,  draw  Q.  P  perpendicular  to  A  L,  then 
p  is  the  position   of  a  body  which  has 
corresponding  uniform  circular  motion. 

Let  the  time  t  seconds  have  elapsed 
since  the  point  o,  was  at  o.  The  corre- 
sponding point  P  was  then  at  c.  Let  the 
uniform  speed  of  p  be  v  feet  per  second ; 
then  c  P  =  vt.  Let  o  P  =  r ;  let  the  angle 
c  o  P  =  vt/r,  so  that  the  radius  o  p  moves 
with  the  angular  velocity  v/r.  The  time 
T  of  revolution  of  P  or  of  complete  oscilla- 
tion of  Q  is  2  TT  r  -j-  v,  or  2  TT  -T-  the  angular  velocity,  which  we 
shall  call  p.  Then  o  o,  =  o  P  .  sin.  o  P  Q  =  o  P  .  sin.  COP,  or,  if  we 
call  o  Q  =  x,  x  =  r  sin.  pt.  By  Art.  19, 

dx 

velocity  of  a  =  -,-  =  rp  cos.  pt, 

acceleration  of  Q  =  ^  =   -  rip1  sin.  pt. 
Hence  the  displacement  sc,  divided  by  the  value  of  the  acceleration, 

4  7T2 

if  we  neglect  the  -  sign,  is  1/p2  or  — ^  (since  T  =  2  v/p). 

Notice  that  the  acceleration  -  rp2  sin.  pt  is  the  resolved  part,  in  the 
direction  x  of  the  acceleration  of  P.  The  speed  of  P  is  v,  a  constant.  If, 
then,  it  has  an  a'cceleration,  it  cannot  be,  nor  any  part  of  it,  in  the  direc- 
tion of  its  path ;  it  is,  then,  in  the  direction  of  the  radius.  Let  the  acce- 
leration of  P  in  the  direction  o  P  be  a,  then1  the  resolved  part  in  the 
direction  o  L  is  o  cos.  P  o  L  =  acceleration  of  Q,  or  a  sin.pt  =  -  r#2  sin. 
pt,  or  a=  -rp2.  The  acceleration  of  P  is,  then,  centripetal  from  P 

towards  o,  and  this  amount  is  rp2.     Thus  p  =  -,  so  that  a  =  -  t>2/r, 

the  ordinary  formula  for  centrifugal  acceleration. 

Here  we  have  obtained  a  knowledge  of  the  centripetal  accelera- 
tion —  mathematically  from  knowing  what  is  meant  by  linear 

acceleration.  Conversely,  suppose  we  know  that  the  centripetal 
acceleration  of  P  in  the  direction  p  o  is  v2/o  p,  then  the  acceleration 
oi  Q  towards  the  centre  is  this  multiplied  by  cos,  POL;  that  is,  by 


550 


APPLIED   MECHANICS. 


o  d/o  p,  and  it  is  therefore  equal  to  —  .  —    or    o  Q  .  t^/o  r2,  90 

o  r     o  r  '       '  • 

that  it  is  proportional  to  o  Q.  Also,  o  Q  divided  by  the  acceleration 
(towards  the  centre)  is  o  p2/v2  or  4  7r2/x2  (since  2  ?r  .  o  F/V  =  T).  Wo 
ore  therefore  led  in  many  ways  to  the  rule 


Periodic  time  =  2  ,  A  /e^t 
'V    Acceleration 
(See  also  Art.  19.) 

445.  Example.  —  In  Fig.  310,  A  is  a  ball  of  lead  weighing  20 

Ibs.  carried  by  means  of  a  spiral  spring  whose 
own  weight  may  be  neglected,  let  us  suppose.* 
Find  by  experiment  how  much  the  spring 
lengthens  when  we  add  1  Ib.  to  the  weight  of 
A,  or  shortens  when  we  subtract  1  Ib.  from 
the  weight  of  A.  Let  it  lengthen  or  shorten 
0-01  foot.  Evidently,  if  ever  A  is  001  foot 
upwards  or  downwards  from  its  position  of 
rest,  it  is  being  acted  upon  by  a  force  of  1  Ib. 
tending  to  bring  it  to  its  position  of  rest.  We 
know  also  that  if  A  is  0-02  foot  or  0*03  foot 
above  or  below  its  place  of  rest,  there  is  a 
force  of  2  or  3  Ibs.  trying  to  bring  it  back. 
We  see,  then,  that  the  up  and  down  motion  of 
A  must  be  simple  harmonic.  When  the  dis- 
placement is,  say  0'02  foot,  the  force  acting 
on  A  is  2  Ibs.,  and  the  acceleration  of  A  is 
force  2  -=-  mass  of  A  ;  and  as  the  mass  of  A  is 
20  -f-  32-2,  or  0'621,  the  acceleration  of  A 
is  3  -2  2  feet  per  second  per  second  when  it  is 
displaced  0-02  foot  from  its  middle  position. 
Now,  employing  the  rule  given  above,  divide 
0-02  by  3-22  and  extract  the  square  root,  then  multiply  by 
6-2832,  and  we  get  0-495  second,  or  about  half  a  second  as  the 
periodic  time  of  the  swinging  ball.  When  we  make  experi- 
ments we  find  that,  unless  the  coils  of  the  spring  are  flat,  and 
the  rigid  support  of  A  exactly  in  the  axis,  the  ball  has  a 
tendency  to  turn  and  vibrate  laterally,  which  disturbs  observa- 
tions if  we  make  careful  measurements  of  the  length  of  swing. 

446.  Example.—  The   Simple  Pendulum.  —  A  simple  pen- 
dulum   consists   in   an   exceedingly   small    but    heavy    body 
suspended   by  means    of   a   long   inextensible   thread,    whose 

*  We  really  assume  that  one-third  of  the  mass  of  the  spring  is  added  to  A, 


Fig.  310 


APPLIED    MECHANICS. 


551 


weight  may  be  neglected,  capable  of  swinging  backwards  and 
forwards  in  short  arcs.  If  the  arcs  are  not  too  long,  the  time 
of  one  swing  is  always  the  same.  Thus,  in  Fig.  311,  s  is  the 
point  of  suspension,  s  P  a  silk  thread,  P  a  small  ball  of  lead. 
p  will  move  backward  and  forward  along  the  path  AOL  with  a 
motion  which  is  simple  harmonic,  provided  the  thread  is  so  long 
and  A  L  so  short  that  the  force  acting  on  the  ball  at  any  time  in 
the  direction  of  its  motion  is  proportional  to  the  distance  of 
the  ball  from  o.  To  show  that  this  is  so,  resolve  the  verti- 
cally acting  weight  of  the  ball  in  the  direction  of  its  motion 
along  A  o.  We  find  that  it  is  not 
quite  proportional  to  AO  unless  AO  is 
very  short,  but  if  this  slight  discrep- 
ancy is  neglected  the  force  urging  the 
ball  towards  o  is  the  weight  of  the  ball 
multiplied  by  o  A  and  divided  by  s  A, 
the  distance  from  the  point  of  support 
to  the  centre  of  gravity  of  the  ball. 
As  a  matter  of  fact,  the  nature  of 
the  vibration  does  not  depend  on  the 
weight  of  the  ball;  but,  to  fix  our 
ideas,  let  us  suppose  that  the  weight 
is  2  Ibs.,  then  the  mass  of  the  ball  is 
2  -f-  32 '2,  and  acceleration  along  A  o  is 


or 


the  force, 


the  force  -f-  mass, 

divided  by  —^  -the  mass,  so  that  the 

acceleration  is  32-2  x  A  o  -4-  s  A. 

Now,  our  rule  is  to  divide  A  o  by 
the  acceleration  at  A,  and  this  gives 

J^;  extract  the  square  root,  and  multiply  by  6  '2832  for  the 
periodic  time  of  oscillation  of  the  pendulum.  The  general 
rule  for  a  simple  pendulum  swinging  in  short  arcs  is  then  ; 

Time  of  a  complete  oscillation  =  6-2832  A/3JJJg*  **?*****- 
The  mass  or  inertia  is  2  -4-  32-2  at  all  places,  but  the  weight 
may  be  more  or  less  than  2  Ibs.  at  different  parts  of  the  earth 
in  the  proportion  of  g  to  32  -2,  and  hence  we  find 


as  our  general  rule. 

The  time  of  one  swing  is  half  this.     The  number  32-2  ex- 
presses the  effect  of  the  force  of  gravity  at  Lqndoji.     At  any 
s* 


552  APPLIED    MECHANICS. 

other  place  on  the  earth's  surface  it  would  be  different — that 
is,  at  different  places  on  the  earth  a  given  pendulum  has 
different  times  of  oscillation.  For  instance,  a  pendulum 
taking  2  seconds  for  a  complete  oscillation  at  Paris — that  is, 
taking  1  second  for  one  swing,  called  a  seconds  pendulum — if 
swung  at  Spitzbergen  would  gain  94  seconds  per  day ;  and  il 
swung  in  New  York  would  lose  30  seconds  per  day,  provided 
the  pendulum  did  not  alter  in  length  in  being  taken  from  one 
place  to  the  other.  Evidently  when  a  pendulum  gets  longei 
it  oscillates  more  slowly ;  hence  in  summer,  when  the  pen- 
dulum of  a  common  house-clock  expands  with  heat,  it  goe» 
more  slowly,  and  in  winter  it  goes  more  quickly,  unless  tha 
position  of  the  bob  is  adjusted.  A  pendulum  which  is  self- 
adjusting — that  is,  which  is  so  constructed  that  it  remains  oi 
the  same  length  whatever  be  the  temperature — is  called  a 
compensation  pendulum. 

447.  Example. — In  Fig.  312,  B  represents  a  strip  of  steel 
fixed  firmly  in  a  vice  at  c,  with  a  heavy  ball  A  fastened  at  its 
free  extremity.  Find  the  force  in  pounds  which  will  increase 


Pig.  312. 


the  deflection  of  A  by  O'Ol  foot ;  say  that  it  is  1  ib.  We 
know  that  a  force  of  2  or  3  Ibs.  will  cause  an  increased  de- 
flection of  twice  or  three  times  this  amount ;  and  as  the  force 
acting  on  the  ball  in  any  position  is  proportional  to  its  distance 
from  its  position  of  rest,  the  ball  will  swing  with  a  simple  har- 
monic motion.  If  we  can  neglect  the  weight  of  the  strip  of 
steel,  and  if  the  ball  is  small  in  comparison  with  the  length  of 
the  strip,  its  time  of  vibration  may  be  calculated  in  exactly 
the  same  way  as  that  of  the  ball  in  Fig.  311.  Experiments 
with  this  enable  Young-'s  modulus  E  to  be  determined. 

448.  Example. — Suppose  that  B  c  (Fig.  313)  is  a  bent  glass 
tube  of  uniform  section  containing  a  liquid  which  can  move 
without  friction  in  the  tube.  If  the  liquid  be  disturbed 
so  that  the  level  is  higher  in  B  than  in  c,  it  will  continue  to 


APPLIED   MECHANICS.  553 

swing  about  its  position  of  equilibrium — that  is,  the  position 
in  which  the  liquid  is  at  the  same  level  in  both  limbs  of  the 
tube.  Thus,  if  c  is  '01  foot  below  the  proper  level,  and  B  is 
•01  foot  above  this  level,  the  force  which  tends  to  cause  the 
liquid  to  return  to  its  proper  level  is  twice  the  weight  of  tha 
liquid  OB.  Suppose  the  weight  of  the  liquid  is  10  Ibs.  per 
foot  length  of  the  tube,  then  the  force  acting 
on  the  liquid  is  -02  x  10,  or  '2  Ibs.  If  the 
whole  length  of  tube  filled  with  liquid  is  6 
feet,  then  the  weight  of  liquid  which  has  to 
be  set  in  motion  is  60  Ibs.,  and  its  mass  is 
60  -f-  32-2,  or  1*863  ;  hence  the  acceleration 
is  -2  -^  1-863,  or  0-107  foot  per  second  per 
second.  The  displacement  is  '01,  and,  work- 
ing  by  our  old  rule,  displacement  divided  by 
acceleration  is  -0935.  The  square  root  of  pig.  sis 

this  is  -3058,  and  multiplying  by  6-2832 
we  get  1*92  second  as  the  periodic  time  of  the  oscillation. 
We  find  it  easy  to  prove  that  the  liquid  swings  in  the  same 
time  as  a  simple  pendulum  whose  length  is  half  the  total 
length  of  the  liquid  in  the  tube,  and  that  it  is  the  same  what- 
ever be  the  density  of  the  liquid — that  is,  whether  it  ia 
mercury  or  water. 

If  w  Ibs.  is  the  weight  of  liquid  per  foot  in  length  of  the  tuhe, 
if  x  is  the  displacement  OB  or  o  c,  the  force  causing  motion  is 
2  xiv.  (This  force  is  multiplied  by  g  -~  32*2  if  the  observation  is  not 
made  in  London. )  If  a  is  the  total  length  of  liquid  in  the  tube, 
the  weight  of  liquid  moved  is  aw,  and  its  mass  is  aw  -H  32-2. 

Hence  the  acceleration  is  2  xw  —  —  or  -  -  >  and  the  displacement 
divided  by  acceleration  is  x  -*-  — -^  or  ~  ;    so  that  the  periodic 

time  is  2  ir     /  JL     JL 
V     2   '   ff' 

449.  It  will  be  observed  that  in  all  these  cases  of  vibration 
of  bodies  there  is  a  continual  conversion  going  on  of  one 
kind  of  energy  into  another.  At  each  end  of  a  swing  the 
body  has  no  motion;  all  the  energy  is  therefore  potential, 
whether  it  is  the  potential  energy  of  a  lifted  weight  or  the 
potential  energy  of  strained  material.  In  the  middle  of  the 
swing  the  body  is  going  at  its  greatest  speed,  and  its  energy  is 
kinetic.  At  any  intermediate  place  the  energy  is  partly 
potential  and  partly  kinetic,  but  the  sum  of  the  two  remains 


554  APPLIED    MECHANICS. 

always  the  same,  excepting  in  so  far  as  friction  is  wasting  the 
total  store.  Now,  in  time-keepers  the  office  of  the  mainspring 
is  to  give  just  such  supplies  of  energy  to  the  balance  as  are 
necessary  to  replace  the  loss  by  friction  ;  and  we  have  to  ask 
the  question  —  At  what  part  of  the  swing  of  a  pendulum  or 
balance  can  we  give  to  it  an  impulse  which  shall  increase  its 
store  of  energy  without  disturbing  its  time  of  oscillation  1 
The  answer  is  this.  If  a  blow  is  given  to  the  bob  of  a  pendu- 
lum when  it  is  just  at  its  lowest  point,  energy  is  given  to  the 
pendulum  ;  we  give  it  power  to  make  a  greater  swing,  but  the 
time  which  it  will  take  to  make  this  greater  swing  is  just  the 
same  as  the  time  it  would  have  taken  for  a  smaller  swing. 
This  middle  point  is  the  only  point  at  which  we  can  give  an 
impulse  to  the  bob  without  altering  the  time  of  its  swing.  In 
the  lever  escapement,  and  in  other  detached  escapements  of 
watches,  the  impulse  is  always  given  just  at  the  middle  of  the 
swing. 

EXERCISES. 

1.  A  point  describes  a  S.H.  motion  of  1£  foot  amplitude  in  a  period 
of  -fth.  of  a  second.     Find  its  maximum  velocity  and  maximum  accelera- 
tion.     Ans.,  21,  v  feet  per  second;  294  ir2  feet  per  second. 

2.  A  piston  with  rod  and  crosshead  weigh  350  Ibs.     If  they  have  a 
S.H.  motion  with  amplitude  1*1  foot,  and  if  the  maximum  accelerating 
force  is  equal  to  that  produced  by  a  pressure  of  15  Ibs.  per  square  inch  on 
a  piston  14  inches  diameter,  what  is  the  periodic  time  ?        Ans.,  0'37  sec. 

OTHER    EXAMPLES    OF    PERIODIC   MOTION. 

450.  When  the  periodic  motion  of  a  body  is  not  simple  har- 
monic, we  find  that  by  imagining  the  body  to  have  two  or 
more  kinds  of  simple  harmonic  motion  at  the  same  time  we 
can  get  the  same  result.  Thus,  it  is  known  that  a  float,  em- 
ployed to  measure  the  rise  and  fall  of  the  tide  by  marking 
on  a  moving  sheet  of  paper  with  a  pencil,  has  a  motion  which 
is  periodic  and  not  simple  harmonic.  If  horizontal  distances 
represent  the  motion  of  the  paper  (unwound  from  a  barrel  by 

means   of   clockwork),  and 

'J^.  ,-\          therefore     represent     time, 

^  _  /  _  \  /      \^     and    if    vertical    distances 

mean   the    rise   or    fall   of 


Fig.  314.  water-level  in  feet,  we  get 

such  a  curve  as  is  shown  in 

Fig.  314.     Now  this  is  not  a  simple  harmonic  motion.     The 
difference  becomes  evident  if  you  plot  on  squared  paper  the 


APPLIED    MECHANICS. 


555 


Fig.  315. 


distances  o  A,  OB,  o  c,  o  D,  o  E,  OF,  etc.  (Fig.  308),  for 
equal  intervals  of  time,  for  you  will  get  a  curve  like  Fig.  315, 
which  is  easily  recognised, 
and  is  called  a  curve  of  sines 
or  cosines.  But  it  has  been 
found  that  if  we  take  certain 
curves  of  sines  whose  periodic 
times  are — 1,  the  semi-lunar 
day;  2,  the  semi-solar  day,  and 
some  others,  their  amplitudes 
and  epochs  being  properly  chosen-  and  draw  them  on  squared 
paper,  and  add  their  ordinates  together,  we  get  the  curve 
which  shows  the  real  rise  and  fall  of  the  tide.  In  the  very 
same  way  we  can  combine  simple  harmonic  motions  to  arrive 
at  any  periodic  motion.  A  good  way  of  combining  simple 
harmonic  motions  experimentally  is  to  let  a  body  hang  from  a 
string  which  passes  over  two 
or  more  movable,  and  the  same 
number  of  fixed,  pulleys.  These 
pulleys  are  pivoted  on  crank 
pins,  and  their  pivots  are  made 
to  revolve  at  any  desired  rela- 
tive speeds,  and  each  gives  to 
the  body  a  purely  simple  har- 
monic motion  by  its  action  on 
the  string.  The  body  gets  a 
motion  compounded  of  the  mo- 
tions of  the  pulleys,  and  if  it 
is  an  ink-bottle  or  pencil  press- 
ing on  the  paper  on  a  revolv- 
ing paper  roller,  we  get  a  time 
curve  of  the  periodic  motion. 
This  is  the  principle  of  the 
construction  of  Lord  Kelvin's 
Tide-Predicting  Machine. 

451.  When  a  body  can 
swing  east  and  west  under  the 
influence  of  forces  which  have  Fig.  310. 

no  tendency  to  move  it  except 

in  a  direction  due  east  and  west,  and  if  forces  acting  due  north 
and  south  can  make  it  swing  in  their  direction,  then  both  sets 
of  forces  acting  together  on  the  body  will  give  it  a  motion 


556 


APPLIED    MECHANICS. 


compounded  of  the  two  simpler  motions.  Thus,  a  ball  A 
(Fig.  316)  is  suspended  by  a  string,  PA,  which  is  knotted  at  p 
to  two  other  strings,  P  s  and  P  s',  equal  in  length,  and  fastened 
at  s  and  s'.  The  ball  may  swing  in  the  direction  E  o  w  as  if 
it  were  the  bob  of  a  pendulum  hung  directly  from  the  ceiling 
at  P',  but  it  may  also  swing  in  the  direction  N  o  s  at  right 
angles  to  E  o  w,  and  if  it  does  so  it  swings  as  if  the  point  p 
were  the  fixed  end  of  the  pendulum  A  p.  When  it  swings 
under  the  influence  of  the  two  sets  of  forces  tending  to  make 
it  move  both  ways  at  once,  the  motion  of  A  is  compounded  of 
the  other  two  simpler  motions.  If  P  A  is  one-quarter  of  the 
length  o  P',  then  the  east  and  west  swing  takes  twice  as  long 

as  the  north  and  south 
swing.  If  p  A  is  one-ninth 
of  o  P',  then  the  east  and 
west  swing  takes  three  times 
as  long  as  the  north  and 
south  swing.  The  motion  of 
A  is  sometimes  very  beauti- 
ful, and  the  experiment  is 
easily  arranged. 

452.  The  motion  is  quite 
easily  represented  on  paper. 
Thus,  in  Fig.  317,  A'  M  is  the 
north  and  south  direction, 
and  A  M,  at  right  angles  to 
it,  is  the  east  and  west  di- 
rection. Let  the  points  0, 
1,  2,  etc.,  in  each  of  these  lines  be  found  as  in  Fig.  308. 
Let  the  bob  be  supposed  to  go  from  0  to  1  in  A'M  in  the 
same  time  as  it  goes  from  0  to  1  in  A  M.  Notice  that  we 
have  twice  as  many  points  in  AM  as  in  A'  M,  showing  a 
slower  oscillation  in  the  direction  A  M.  We  can  begin  to 
number  our  points  anywhere,  remembering  that  when  the 
bob  completes  its  range  it  comes  back  again  in  the  opposite 
direction.  Now  put  marks  where  the  east  and  west  lines 
meet  the  north  and  south  ones,  drawn  through  corresponding 
points.  It  is  evident  that  the  curve  drawn  through  these 
successive  marks  is  the  real  path  traced  out  by  the  ball  when 
acted  upon  simultaneously  by  the  two  sets  of  forces  urging  it 
in  a  north  and  south,  and  an  east  and  west  direction. 

If  we  have  the  same  number  of  points  in  A'  M  as  in  A  M 


Fig.  317. 


APPLIED    MECHANICS.  557 

we  get  a  circle,  ellipse,  or  straight  line,  as  in  c,  B,  A,  Fig.  318. 
This  represents  the  motion  of  a  conical  pendulum  free  to 
swing  in  every  direction.  Again,  D,  E,  F,  and  many  other 
curves  that  might  be  drawn,  represent  the  case  which  we  took 
up  in  Fig.  317,  where  one  vibration  is  twice  as  quick  as  the 
other.  If  the  time  of  vibration  in  A  M  is  to  the  time  of 


c, 

-000< 


CJ^OOOO-COO  CCOOOOCOOT 


O 
O 


Fig.  318. 

vibration  in  A'  M  as  2  to  3,  we  get  curved  paths  like  G,  I,  J, 
and  so  on.  In  experimenting  with  the  pendulum,  Fig.  316. 
it  will  usually  be  found  that  slight  inaccuracies  in  the  lengths 
of  the  cords  will  cause  a  continual  change  to  go  on  in  the 
shape  of  the  path  traced  out  by  the  ball. 

We  can  produce  these  motions  by  spiral  springs,  and  in 
other  ways.     Thus,  for  example,  if  we  use  instead  of  the  strip 


558  APPLIED    MECHANICS. 

of  steel,  in  Fig.  312,  a  combination  of  two  strips,  B  and  B',  as 
in  Fig.  319,  so  that  the  heavy  bright  bead  A  is  capable  of 
vibrating  in  two  directions  at  the  same  time,  we  get  the  same 
combinations  of  simple  harmonic  motions,  depending  on  the 
point  at  which  B  is  held  in  the  vice  c. 

453.  When  a  body  has  a  periodic  rotational  motion  about 
an  axis  like  the  balance  of  a  watch  or  a  rigid  pendulum,  we  must 
no  longer  speak  of  the  force  causing  motion,  and  the  mass  of 
the  body,  and  the  distance  of  displacement ;  but  if  we  sub- 
stitute for  these  terms,  moments  of  forces,  moment  of  inertia 


Fig.  819. 


of  the  body  and  angle  of  displacement,  we  have  exactly  the 
same  rule  for  finding  the  periodic  time  of  oscillation.  The 
periodic  time  is  6'2832  times  the  square  root  of  the  angular 
displacement  of  the  body  at  any  instant,  divided  by  the  angular 
acceleration  at  that  instant.  And  we  know  that  angular 
acceleration  may  be  calculated  by  dividing  the  turning  moment 
acting  on  a  body  by  the  moment  of  inertia  of  the  body.  A 
point  in  the  balance  of  a  watch  swings  in  circular  arcs,  but  if 
we  only  take  account  of  the  distances  which  it  passes  through, 
and  suppose  it  moved  in  a  straight  line  instead  of  in  the  arc 
of  a  circle,  the  motion  is  very  nearly  simple  harmonic.  If 
there  were  no  friction  or  other  forces  acting  on  the  balance 
ex'cept  the  turning  moment  of  the  balance  spring  (see  Arts. 
521-2-3),  and  if  the  moment  of  the  spring  were  always  exactly 
proportional  to  the  angular  displacement  of  the  balance,  the 
motion  would  be  simple  harmonic. 

We  shall  see  in  Art.  522  that  the  turning  moment  due  to  the 

E  bfl 
spring  is  r~n~t  0>  if  E  is  the  modulus  of  elasticity  of  the  spring,  b  its 

breadth,  t  its  thickness,  and  I  its  length,  and  if  Q  is  the  angular 
displacement  in  radians.     Angular  acceleration  is  this  moment 

divided  by  moment  of  inertia  i  of  the  balance,  or  ^-.  — .      Hence 

12/i 
angular  displacement  9  divided  by  angular  acceleration  is         8 

so  that  the  periodic  time  of  the  balance  is  T  =  6-2832 


APPLIED    MECHANICS.  559 

Increasing  the  moment  of  inertia  of  the  balance  or  the 
length  of  the  spring  makes  the  vibration  slow.  Increasing  the 
breadth  and,  what  is  still  more  important,  increasing  the 
thickness  of  the  spring  makes  the  vibration  quick.  As  we  shall 
see  in  Arts.  521-3  that  our  calculation  of  the  turning  moment  of 
the  spring  is  not  quite  right,  that  the  dimensions  of  the 
balance  and  spring  alter  with  temperature,  and  that,  above  all, 
the  elasticity  of  the  steel  alters  with  temperature,  and  with  its 
own  state  of  fatigue,  the  rule  is  not  perfectly  true,  nor  can 
any  balance  be  regarded  as  taking  exactly  the  same  time  for 
its  oscillation  in  different  lengths  of  arc.  At  the  same  time  it 
is  of  great  help  to  the  watchmaker  to  know  that  with  con- 
siderable, although  not  with  perfect  accuracy,  the  time  of 
vibration  of  a  balance  is  proportional  to  the  square  root  of  the 
length  of  the  spring,  and  so  on.  For  example,  suppose  the 
spring  is  3  inches  long,  and  the  balance  makes  one  swing  in 
0'251  second,  now  if  he  wishes  it  to  make  a  swing  in  0-25 
second,  he  must  shorten  it  in  the  ratio  of  '251  x  '251  to 
•25  x  -25,  or  in  the  ratio  -063001  to  '0625,  so  that  the  length 
of  his  spring  ought  to  be  3  x  '0625  -f  -063001,  or  2-976 
inches — that  is,  it  ought  to  be  shortened  -024  inch.  In  the 
same  way  he  can  calculate  the  effect  of  adding  little  masses  at 
any  distances  from  the  centre  of  the  balance,  so  that  its 
moment  of  inertia  may  be  increased,  and  the  balance  made 
slower  in  its  swing.  The  same  law  tells  him  how  he  can 
compensate  the  balance,  so  that  when  in  summer  the  steel  of 
the  spring  loses  its  elasticity,  some  of  the  mass  of  the  balance 
will  come  nearer  the  centre,  in  order  that  the  moment  of 
inertia  may  diminish  in  the  same  proportion. 

454.  Compound  Pendulum. — The  simple  pendulum  described 
in  Art.  446  is  not  like  the  pendulums  used  in  practice.  In  these 
the  bob  is  not  so  small  that  we  can  consider  it  as  a  point ;  the 
long  part  is  not  a  thread  but  a  stiff  rod  of  metal  or  wood,  and 
there  is  usually  a  knife-edge  for  support,  about  which  it  can 
turn  with  little  friction.  In  common  clocks,  however,  the  top 
end  of  the  pendulum  is  a  thin  strip  of  steel  held  firmly  in  the 
chops,  but  the  easy  bending  of  this  strip  is  such  that  we  may 
imagine  an  equivalent  pendulum  to  move  freely  about  an  axis. 
Employing  our  general  rule  of  Art.  453,  we  find  how  to  calculate 
the  time  of  vibration.  This  compound  pendulum  vibrates  in 
the  same  time  as  a  certain  simple  pendulum,  called  the  equi- 
valent simple  pendulum,  whose  length  we  find  by  experiment. 


560 


APPLIED    MECHANICS. 


In  Fig.  320  let  S  be  the  axis  of  suspension,  G  the  centre 
of  gravity,  and  P  a  point  in  the  continuation  of  the  line 
s  O  such  that  s  P  is  the  length  of  the  equivalent  simple 
pendulum.  Then  p  is  called  the  centre  of 
oscillation,  and  it  is  also  known  to  be  the 
centre  of  percussion  of  the  pendulum  (see 
Art.  402).  It  can  be  proved  that  if  the 
pendulum  be  inverted  and  made  to  vibrate 
about  a  parallel  axis  through  p,  it  will 
vibrate  in  exactly  the  same  time  as  it  does 
about  8  ;  and  it  was  in  this  way,  by  in- 
verting a  pendulum  which  had  two  knife- 
edges,  and  adjusting  these  until  the  pendulum 
took  the  same  time  to  vibrate  about  one 
as  about  the  other,  and  then  measuring 
the  distance  between  them,  that  Captain 
Kater  found  the  length  of  the  simple  pen- 
dulum which  vibrates  in  a  given  time. 
This  method  is  still  employed  in  gravitation 
experiments  everywhere  to  find  the  value 
of  g,  which  is  32-2  feet  per  second  per  second  at  London. 
It  is  an  excellent  laboratory  exercise. 

If  s  is  the  axis  of  suspension,  o  the  centre  of  gravity,  w  the 
weight  of  the  pendulum,  then  the  moment  with  which  gravity 
urges  the  pendulum  to  return  to  its  position  of  rest  is  w  x  G  N  ; 
but  if  the  angle  o  s  o  be  measured  in  radians,  and  if  it  is  very 
small,  this  moment  is  almost  exactly  equal  to  w  x  so  x  angle  G  s  o. 
The  angular  acceleration  is  obtained  by  dividing  this  by  i,  the 
^ moment  of  inertia  of  the  pendulum  about  s,  and  our  rule  becomes 


Fig.  320. 


6-2832 


angle  G  s  N 
.  angle  G  s  N  +  I* 


6-2832 


....  (1). 


W  .  SG 

When  we  examine  this  formula  we  see  that  it  may  be  put  in 
another  form.  Find  a  point  K  such  that  if  all  the  mass  of  the 
pendulum  were  gathered  there  its  moment  of  inertia  about  s  would 

-ry 

be  the  same  as  at  present ;  in  fact,  such  that  ^oTo?  the  mass  of  the 

pendulum  x  s  K2,  would  be  equal  to  i.  The  distance  s  K.  is  called 
the  radius  of  gyration  of  the  pendulum  (see  Art.  112),  and  our  rule 
now  becomes 


=  6-2S32  A/ 


....  (2), 


g  .  s  o 


APPLIED    MECHANICS.  561 

where  g  is  32  '2.  In  the  simple  pendulum  SK  and  SG  are  equal, 
and  (2)  gives  the  same  rule  which-is  given  in  Art.  446.  However, 
in  an  ordinary  pendulum,  s  K  and  s  G  are  not  equal,  but  s  K2  -i-  s  G 
is  equal  to  some  length  such  as  s  p,  and  our  rule  becomes 


I     AP 

v  7-" 


T  =  6-2832^  f 

9 

Evidently  s  p  w  the  length  of  the  imaginary  simple  pendulum  which 
would  vibrate  in  the  same  time  as  our  real  pendulum.  The  imaginary 
point  P  has  been  called  the  centre  of  oscillation,  because  when  the 
pendulum  is  inverted  and  made  to  vibrate  about  an  axis  through  p 
it  vibrates  in  the  same  time  as  before. 

To  prove  this  it  is  necessary  to  return  to  equation  (1).  "We 
know  that  i  is  equal  to  the  moment  of  inertia  of  the  body  calculated 
as  if  all  its  mass  existed  at  G,  together  with  the  moment  of  inertia 
of  the  body  as  it  is  at  present,  but  calculated  about  an  axis  through 
G  parallel  to  the  present  axis  ;  that  is, 

I==^8a2  +  ^^, 

9  9 

where  k  is  some  length  unknown  to  us  just  now,  being  the  radius 
of  gyration  about  the  axis  through  the  centre  of  gravity.  Rule  (1) 
becomes 


I  =  6-2832 


V 

V 


w  .  SG 


or  T  =  6-2832      w      

9 

That  is,  the  length  of  the  simple  pendulum  which  will  vibrate  in 
the  same  time  ie  so  +  — ,  and  we  have  already  found  it  to  be  s  p 

in  equation  (3) ;  so  that  G  P  =  — ,  or  G  P  x  8  G  =  &2.     But  in  the 

very  same  way,  if  we  considered  the  pendulum  as  vibrating  about 
p,  we  should  find  the  length  of  the  equivalent  simple  pendulum  to 

be  greater  than  G  P  by  an  amount  equal  to  — ,  and  we  know  that 

s  G  is  equal  to  this  amount ;  so  that  s  P  would,  as  before,  be  the 
length  of  the  equivalent  simple  pendulum.  The  axes  of  oscillation 
and  suspension  are  therefore  interchangeable. 

455.  Examples.— The  bar  of  Fig.  321  with  two  adjust- 
able masses  may  be  fixed  to  one  end  of  a  wire,  the  other  end 
of  which  is  fixed  to  the  ceiling.  By  twisting  and  untwisting 
the  wire  the  bar  will  oscillate  with  a  motion  which  is  much 
more  nearly  simple  harmonic  than  that  of  the  balance  of  a 
watch,  Students  who  experiment  with  such  a  bar  can  adjust 


562  APPLIED   MECHANICS. 

the  weights  A  and  B  at  any  distance  from  the  axis  (there  ought 
to  be  an  engraved  scale  on  the  bar),  so  that  the  moment  of 
inertia  can  be  varied.  They  can  fasten  the  bar  at  the  end  of 
a  wire,  or  they  can  use  it  as  in  Fig.  321,  with  a  flat  spiral 


Pig.  821. 

spring,  or  as  in  Fig.  322,  with  a  cylindric  spiral  spring ;  and 
the  rate  of  its  vibration  gives  one  of  the  best  ways  of  in- 
vestigating the  twisting  moments  of  wires  and  such  springs 
when  strained  through  given  angles. 

In  the  case  Of  a  wire  the  twist  always  tends  to  bring  the 
bar  to  its  position  of  rest  with  a  moment  which  is  proportional 
to  the  angle  of  displacement  from  this  position — it  is  this 
property  which  causes  the  motion  to  be  simple  harmonic. 
This  moment  is  also  proportional  to  the  fourth  power  of  the 


APPLIED   MECHANICS.  563 

diameter  of  the  wire,  and  it  'becomes  less  as  the  length  of  the 
wire  is  increased.  By  means  of  a  circular  scale  and  a  pointer 
we  can  measure  the  extent  of  each  swing,  and  this  is  found  to 
decrease  gradually,  due  to  friction  with  the  air  and  the  in- 
ternal friction  or  viscosity  of  the  metal.  The  amount  of 


Fig.  322. 


diminution  of  swing  gives  us  a  means  of  determining  the 
viscosity,  and  the  apparatus  can  so  easily  be  fitted  up  that  no 
person  who  wishes  to  understand  the  properties  of  materials 
can  be  excused  from  making  these  experiments.  This  is  a 
common  method  of  finding  N,  the  modulus  of  rigidity  of  a 
material. 

If  the  length  of  the  wire  is  I  inches,  its  diameter  d,  and  if  x  is 
its  modulus  of  rigidity  (see  Table  XX.),  then  fro  i  Art.  295  we  see 
that  the  moment  with  which  the  wire  ants  on  the  bar.  when  its 


664  APPLIED    MECHANICS. 

angle  is  0  from  the  position  of  rest,  is  £-  N  -  d*.     If  the  moment  of 

'  >  —         I 

inertia  of  the  bar  is  i  (we  are  neglecting  the  fact  that  the  wire 
itself  has  some  mass  which  has  to  be  set  in  motion),  then  the 
moment,  divided  by  i,  is  the  angular  acceleration ;  and  using  thia 
quotient  as  denominator,  and  6  as  numerator,  extracting  the  square 
root,  and  multiplying  by  6-2832,  or  2  TT,  by  the  general  rule  of  Art. 
453,  we  find  the  square  of  the  period  of  a  complete  oscillation  to 

be  T9  =  -jp- .     If  we  are  in  doubt  as  to  our  calculated  i,  we  can 

find  it  experimentally ;  and  this  is  very  necessary  in  many  magnetic 
experiments.  We  add  a  known  moment  of  inertia  i  (say  two  equal 
small  masses  at  equal  distances  from  the  axis),  and  find  the  new  T 
(call  it  TJ).  Then  T^/x2  =  (i  +  »)/i,  and  i  may  be  found. 

When  motion  is  slow,  the  friction  in  fluids  is  proportional  to 
the  velocity,  and  any  friction  which  follows  this  law  is  called  fluid 
friction.  A  great  many  vibrating  bodies  tend  to  come  to  rest  by 
the  action  of  such  friction  as  this;  and  it  is  found  that  if  the 
friction  is  numerically  /  times  the  angular  velocity,  then  the 
logarithm  of  the  ratio  of  the  length  of  one  complete  swing  to  the 
next  is  nearly  equal  to  k  times  the  periodic  time.  Hence  this 
logarithmic  decrement,  as  it  is  called,  is  proportional  to  the  friction 
co-efficient.  If  we  observe  twenty-one  elongations  on  one  side  of 
the  middle  position,  then  one-twentieth  of  the  logarithm  of  the  first 
elongation  divided  by  the  last  is  k  times  the  periodic  time  of 
oscillation. 


EXERCISES. 

1.  Bifilar  Suspension. — In  many  measuring  instruments  a  body  is 
suspended  by  two  thin  wires  nearly  vertical.     If  the  vertical  length  of 
each  of  these  is  I,  the  distance  between  their  ends  at  the  top  a,  and  at  the 
bottom  b,  and  the  weight  of  the  body  10,  it  is  easy  to  show  that  for  a 
small  angular  displacement  0,  the  moment  tending  to  bring  the  body  to 
its  position  of  rest  is  very  nearly  (neglecting  torsion  of  the  wires  them- 

ab 
selves)  £y  w6.     Find  the  time  of  vibration  of  such  a  body  when  its 

moment  of  inertia  is  known.      In  truth,  the  constraining  moment  is 
proportional  to  sin.  9. 

2.  A  magnet,  turning  on  a  frictionless  pivot  at  its  centre  of  gravity,  is 
subjected  to  a  turning  moment  H  sin.  0,  or  very  nearly  H  0,  due  to  the 
earth's  magnetic  action,  if  it  makes  only  a  small  angle  0  with  its  position 
of  rest.     Find  the  time  of  a  vibration  if  the  moment  of  inertia  is  known, 
and  show  that  the  square  of  the  time  of  vibration  of  the  magnet  in 
different  places  is  inversely  proportional  to  H.     Find  what  is  the  effect  of 
adding  a  known  moment  of  inertia,  and  show  that  the  observations  on  the 


APPLIED    MECHANICS.  565 

two  times  of  vibration  enable  us  to  calculate  the  original  moment  of 
inertia  if  it  was  unknown. 

3.  Prove  that  the  time  of  complete  oscillation  of  a  ship  is  2irkj  \/gd, 
where  d  is  the  distance  from  the  centre  of  gravity  to  the  metacentre. 

456.  Stilling  of  Vibrations. — When  a  simple  harmonic 
motion  is  represented  on  paper  in  the  manner  described  in 
Art.  450,  we  have  a  curve  of  sines.  The  curve  may  be  ob- 
tained by  producing  the  lines  B  B',  c  c',  etc.,  of  Fig.  308, 
cutting  them  at  right  angles  by  equidistant  horizontal  lines, 
and  joining  the  successive  points  of  intersection  so  found.  It 
may  also  be  drawn  by  finding  from  a  book  of  tables  the  sines 
of  0°,  10°,  20°,  etc.,  and  plotting  0  and  sin.  0°,  1  and  sin.  10°, 
2  and  sin.  20°,  etc.,  on  a  sheet  of  squared  paper. 

A  curve  of  sines  expresses  the  fact  that,  if  d  represents  the 
displacement  of  a  vibrating  body  from  its  middle  position  after  an 
interval  of  t  seconds  since  it  was  at  the  middle  of  its  course,  then 
d  =  a  sin.  pt  where  a  is  the  greatest  displacement  of  the  body 
from  its  middle  position.  This  displacement  is  usually  called  the 
amplitude  of  the  vibration.  If  T  is  the  time  of  a  complete  vibration, 

it  is  easy  to  see  that  the  equation  is  d  =  a  sin.  —  t,  or  a  ein.  2irf .  t 
if/  is  the  frequency  or  number  of  complete  oscillations  per  second. 

If  we  make  the  bob  of  a  pendulum  terminate  below,  in  a 
tube  which  can  act  as  a  pencil-holder,  and  in  which  a  well- 
fitting  pencil  can  slide  freely,  and  if  we  move  a  sheet  of  paper 
at  a  uniform  rate  underneath  this  pencil  at  right  angles  to  the 
direction  of  motion  of  the  pencil,  a  curve  of  sines  will  be  traced 
out,  if  the  pendulum  swings  without  friction.  But  in  practice 
we  always  find  that,  what  with  the  friction  at  the  point  of 
support,  friction  with  the  atmosphere,  etc.,  a  pendulum's 
swings  get  smaller  and  smaller — that  is,  the  amplitude  of 
the  vibration  gets  less  and  less  as  time  .goes  on,  until  the 
pendulum  at  length  comes  to  rest. 

This  motion  is  not  a  simple  harmonic  motion,  but,  within 
certain  limits,  each  swing  may  be  regarded  as  very  nearly  a  simple 
harmonic  motion.  Practical  men  who  deal  with  oscillating  bodies, 
such  as  pendulums,  ships,  tuning  forks,  magnetic  needles,  and 
ouspended  coils  of  wire,  usually  assume  that  tho  motion  during 
each  swing  is  a  simple  harmonic  motion.  The  frictional  resistance 
to  motion  of  any  ordinary  vibrating  body  in  a  fluid  medium,  or  of 
a  magnetic  needle  vibrating  near  any  body  capable  of  conducting 
electricity,  is  almost  always  such  that  the  quicker  the  motion  the 


566  APPLIED    MECHANICS. 

greater  the  friction  (see  Art.  64) — that  is,  frictknal  resistance  is 
proportional  to  speed ;  and  in  this  case  it  is  not  difficult  to  show 

that,  instead  of  the  law  d  =  a  sin.     "*  t  .  .  .  .  (1),  we  have  the  law 

i  -  ae-at  sin.  —  *....  (2).    That  is,  if  the  strength  of  the  spring, 
TI 

or  other  governor  of  vibration,  and  the  character  of  the  vibrating 
body  are  such  that  without  friction  the  law  would  be  (1),  then, 
when  the  vibration  is  damped  by  frictional  resistance  of  the  above 
character,  the  law  of  the  motion  becomes  that  given  by  equation  (2). 
Here  a  is  a  constant  which  depends  on  the  character  of  the  friction. 
Thus  a  is  greater  when  a  pendulum  swings  in  water  than  when  it 
swings  in  air.  Also,  TI,  the  periodic  time  of  the  vibration,  is  no 
longer  the  same  T  as  it  was  for  undamped  vibrations,  and  the 

11  ° 

relation  between  T  and  TI  is  -^  =  — ^  +  7%  ....  (3) ;  or,  if  /  is 
the  undamped  frequency,  and  /j  the  damped  frequency,  then 

o 

fi  -f^  +    *     .  .  .  .  (4).      In  order  to   get   exact  ideas   on   this 

subject  of  the  damping  of  vibrations,  the  student  ought  to  plot  on 
squared  paper  a  curve  such  as  o  A'  B'  c'  D'  E'  F'  a'  H'  i,  Fig.  323, 
which  corresponds  with  equation  (2).  Thus,  let  us  suppose  that  a 
body  undamped  in  its  vibrations  gets  an  impulse  which  sends  it 
from  its  position  of  rest  in  such  a  way  that  its  amplitude  is  10 
inches,  and  let  the  time  of  a  complete  oscillation  be  T6  second. 

6*2832 
Then  the  law  of  its  motion  would  be  x  =  10   sin.      ,  .     t,  or 

I'D 

x  =  10  sin.  3-927  t  ....  (5)  where  x  is  in  inches,  t  in  seconds,  and 
the  angle  3-927<  in  radians.* 

If,  now,  the  friction  is  such  that  o  =  07,  we  find  from  (3)  that 
the  time  of  an  oscillation  is  practically  unchanged.  Find,  there- 
fore, the  original  curve  of  sines  by  calculating  the  second  column 
of  the  following  table.  The  numbers  of  the  first  two  columns 
plotted  on  squared  paper  would  represent  the  undamped  vibrations. 
But  for  damped  vibrations  the  numbers  of  the  second  column  have 
all  to  be  multiplied  by  €~^'>It ;  and  if  we  denote  this  multiplier  by 
the  letter  z,  we  see  that  z  being  e-0'7*,  or  log.  z  =  -  0-304 1.  We 
have  calculated  z  for  the  various  values  of  t,  and  placed  the  results 
in  the  third  column.  Multiplying,  therefore,  the  respective 
numbers  of  the  second  and  third  columns  together,  wo  get  the 
fourth  column  of  numbers ;  and  plotting  the  numbers  of  the  first 
and  fourth  columns  on  squared  paper,  we  find  the  curve  which 
shows  the  nature  of  the  damped  vibrations. 

*  We  may  write  (5;  in  the  form  x  =  10  sin.  225$.     Ii*  this  case  the  angle 
225f  is  expressed  in  degrees. 


568 


APPLIED    MECHANIC^. 


t  in  seconds. 

10  sin.  3-927  1,  or  10 
sin.  225  1,  if  angle  is 
taken  in  degrees. 

€-0-7* 

6  ~°'7*  10  sin.  3-927  f 

0 

0 

1 

0 

0-2 

7-07 

•869 

6-14 

0-4 

10 

•756 

7-56 

0-6 

7-07 

•657 

4-6& 

0-8 

0 

•571 

0 

1-0 

-  7-07 

•497 

-  3-51 

1-2 

-  10 

•432 

-  4-32 

1-4 

-  7-07 

•375 

-  2-65 

1-6 

0 

•326 

0 

2-0 

10 

•247 

2-47 

2-4 

0 

•186 

0 

2-8 

-  10 

•141 

1-41 

3-2 

0 

•106 

0 

3-6 

10 

•080 

•8 

4-0 

0 

•061 

0 

4-4 

-  10 

•046 

-  -46 

4-8 

0 

•035 

0 

5-2 

10 

•026 

•26 

5-6 

0 

•020 

0 

6-0 

-  10 

•015 

-  -15 

6-4 

0 

•Oil 

0 

457.  Some  students  may  find  it  as  instructive  to  first  draw  a 
curve  of  sines,  then  draw  the  logarithmic  curve,  correspond- 
ing to  column  three,  on  the  same  sheet  of  squared  paper,  and 
multiply  the  ordinate  of  one  curve  by  that  of  the  other  to  get 
the  ordinate  of  the  real  curve  which  exhibits  the  damped 
vibrational  motion.  This  is  what  has  been  done  in  Fig.  323 ; 
OABCDEFGHI  is  the  curve  of  sines,  L  p  Q  is  the  logarithmic 
curve,  showing  how  rapidly  the  amplitude  of  the  vibration 
diminishes,  and  o  A'  B'  c'  D'  E'  F'  G'  H'  I  is  the  curve  which  repre- 
sents the  actual  motion  of  the  vibrating  body.  In  this  figure 
the  logarithmic  curve  is  drawn  to  such  a  scale  as  seemed  con- 
venient for  showing  its  properties  distinctly.  It  would  be 
very  easy  to  dilate  on  the  nature  of  the  resulting  curve  o  A'  B', 
etc.,  but  this  book  is  written  to  help  students  who  are  earnest 
enough  to  calculate  the  above  numbers  and  plot  the  curve,  and 
when  they  perform  these  operations  they  will  have  very  clear 
notions  about  the  motion  we  have  been  investigating. 

Exercise. — A  heavy  disc,  suspended  by  a  wire,  vibrates  in  each 
of  a  number  of  fluid  media,  its  periodic  time  of  vibration  in  all 


APPLIED    MECHANICS.  569 

being  sensibly  the  same,  or  T5  second.  The  ratio  of  the  amplitudes 
of  two  successive  swings  in  one  direction  being  0'9  in  one  fluid — 
that  is,  the  second  swing  being  only  nine-tenths  of  the  first,  and 
the  third  being  only  nine-tenths  of  the  second,  and  so  on — and  0*8 
in  another  fluid,  and  0*7  in  another,  what  numbers  will  express  the 
relative  viscosities  of  these  fluids  ? 

Here  we  have,  taking  common  logarithms,  0'9  =  e-1'^a  for  the 
first  fluid,  so  that  -  log.  0'9  =  1-5  a  log.  e,  or  a  =  ~    °,g' ,  that 

is,  a  =  0'07.  In  the  same  way  a  =  0'15  and  o  =  0-24  for  the 
other  fluids,  and  hence  7,  15,  and  24  are  the  required  numbers 
expressing  the  relative  viscosities  as  measured  by  the  vibrating  disc 
method. 

A  very  slowly  swinging  disc  and  pointer  will  enable  us  to 
plot  the  complete  curve  from  actual  observations.  The  nature 
of  the  motion  when  the  friction  is  that  of  solids  rubbing  on 
solids  is  studied  in  my  book  on  the  Calculus. 


670 


CHAPTER    XXVI. 

MECHANISM. 

458.  IN  Art.  179  we  pointed  out  how  skeleton  drawings  and 
models  may  be  made  useful  incases  where  A'elocity  ratios  vary  greatly. 
We  shall  now  give  a  sketch  or  outline  of  a  general  theory  from 
which  students  may  find  benefit  if  they  fill  it  in  for  themselves. 
We  cannot  say  that  there  is  as  yet  any  satisfactory  treatise  on  this 
subject.     The  most  interesting  part  of  it   (to   us)  is  that   which 
concerns  the  mechanism  of  steam  and  other  engines,  and  this  will 
be  found  in  the  author's  book  on  the  Steam  Engine. 

459.  We  have  already  referred  to  spur  and  bevil  gearing  used  to 
drive  one  shaft  from  another  at  uniform  velocity  ratios.     Consider 
at  any  instant  two  teeth  in  contact ;  each  of  them  is  a  rigid  piece 
rotating  about  a  fixed  centre  and  acting  on  the  other  by  rubbing 
contact.     We  shall  now  briefly  refer  to  this  way  of  transmitting 

motion. 

Let  A  w  be  a  body 
moving  in  the  plane  of  the 
paper  about  the  axis  A,  let 
B  v  be  another  body  moving 
about  the  axis  B,  and  let 
these  two  bodies  keep  in 
contact,  as  we  sey  them  in 
contact  at  I .  Now,  this 
keeping  in  contact,  what 
does  it  mean?  It  is  that, 
whatever  may  be  the  rela- 
tive tangential  or  sliding 
motion  at  the  point  of  con- 
tact P,  they  have  the  same 
velocity  in  the  direction  of 
their  common  normal  there 
at  the  instant  when  we 
study  them.  Consider  their 
motions  during  an  exceed- 
ingly short  interval  of  time 

Fi     324  St.     There  are  two  points  P 

to  be  considered.    One  is  on 

the  body  A  w,  and  it  moves  to  Q,  where  P  Q  is  an  arc  of  a  circle  about 
A  as  centre ;  the  other  P  is  on  the  body  B  v,  and  it  moves  to  11, 
where  P  R  is  an  arc  of  a  circle  about  B  as  centre.  The  angular 
velocity  of  A  w  being  «,  the  angle  P  A  o.  is  a  .  St ;  the  angular 
velocity  of  B  v  being  b,  the  angle  P  B  R  is  b  .  St.  As  8t  is  con- 
sidered smaller  and  smaller,  P  o,  and  P  R  may  be  considered  more 
and  more  nearly  short  straight  lines.  P  o,  =  A  P  .  a  .  St  and 
P  R  =  B  P  .  b  .  St.  Now  observe  that  if  x  P  s  is  the  direction  of 


APPLIED    MECHANICS.  571 

the  common  normal  to  the  two  bodies  at  P,  the  normal  component 
p  s  of  the  motion  P  Q,  must  be  the  same  as  that  of  the  motion  P  R  ; 
in  fact,  the  straight  line  QSR  must  be  at  right  angles  to  the 
normal  T  p  s.  Now  draw  T  E  parallel  to  A  p,  and  observe  that  the 
two  triangles  T  E  p  and  Q  P  R  have  their  sides  Q  R  and  T  p,  R  p  and 
p  E,  P  Q,  and  E  T  respectively  at  right  angles  to  one  another,  and 
hence  they  are  similar ;  that  is, 

QR:RP:PQ  =  TP:PE:ET.  ...  (1), 
or  BJ!  =  ^. 

PQ,          ET 

Hence    2Jli*  =  1?  Or  -  =  AP-*B.- 
A  P  .  #      ET'       a       BP.ET 

But  because  T  E  is  a  line  drawn  parallel  to  A  P,  a  side  of  the 
triangle  A  p  B,  we  have 

AP          AB  PE          AT. 

E~T    =    B^andB~P   =   AV 

therefore     AP-FE=A_T 

ET . BP          BT 

Hence      *  =  ±1 (2). 

a        BT 

Hence  the  ratio  of  the  angular  velocities  is  inversely  as  the 
segments  into  which  the  common  normal  at  the  point  of  contact 
divides  the  line  of  centres.  Hence,  if  the  ratio  of  the  angular 
velocities  is  constant,  the  common  normal  at  the  point  of  contact 
p  passes  always  through  the  same  point  T  in  the  line  of  centres  A  B. 

Notice  also  that  5*?  =  — .     But  p  Q  ~  a  .  A  p  .  5*,  T  E  =  A  p  — , 

PQ          TB  A3 

Hence  it  will  DB  found  that 

Q  R  =  P  T  (a  +  b}  St. 
That  is,  the  slipping  speed  at  P  is  the  speed  at  the  end  of  a  radius 
T  P  when  the  radius  revolves  at  the  angular  velocity  a  +  b.   Hence 
there  is  always  slipping  at  p,  unless  P  is  on  the  line  of  centres  and 
the  speed  of  slipping  is  proportional  to  the  distance  P  T. 

The  student  will  find  that  when  there  is  friction,  if  DPT  = 
D'  P  T  =  angle  of  repose,  then  p  D  is  the  direction  of  the  mutual 
force  when  A  w  rotates  as  shown,  with  the  hands  of  a  watch ; 
whereas  P  D'  is  the  direction  of  the  mutual  force  when  A  w  rotates 
in  the  opposite  direction,  so  that  B  v  is  the  driver. 

If  we  have  given  the  shape  of  B  v  P,  and  we  desire  to  find  the 
shape  of  a  piece  A  w  P  which  will  gear  with  it  at  a  constant  angular 
velocity  ratio,  make  a  template  of  BVP,  and  arrange  that  when 
this  template  is  moved  about  the  fixed  centre  B  a  sheet  of  paper 
shall  move  about  A  through  the  proper  angular  distances.  If  for 
each  position  of  the  two  the  curved  shape  of  B'  p  v'  be  drawn  on 
the  paper,  the  pencil  marks  will  show  the  proper  shape  of  A  w  p 
if  it  is  to  touch  B  p  v.  In  fact,  A  w  p  will  be  the  envelope  of  the 
shapes  of  B'  p  v'.  When  B'  P  v'  is  circular  with  B  as  centre,  A  w  p 
will  also  be  circular.  True  rolling  will  be  possible,  for  p  will  be 
at  T,  and  as  T  will  be  constant  in  position  the  angular  velocity 


572  APPLIED    MECHANICS. 

ratio  will  be  constant.  When  B'  P  V  is  the  arc  of  an  ellipse  whose 
focus  is  at  B,  and  if  the  distance  B  A  is  equal  to  the  major  axis  of 
the  ellipse,  A  w  P  will  be  a  similar  ellipse ;  there  will  be  true 
rolling,  with  changing  angular  velocity  ratio.  The  other  foci  may 
be  connected  by  a  link.  If  B'  P  v'  is  shaped  like  an  equiangular 
spiral  whose  pole  is  at  B,  A  w  p  will  be  a  similar  spiral,  and  there 
will  be  true  rolling  between  them,  but  with  changing  angulai 
velocity  ratio ;  in  this  way  lobed  wheels  are  formed  to  gear  together. 
460  Discs  or  cylinders  touching  each  other,  their  axes  parallel, 
are  used  for  friction  gearing.  If  the  horse-power  H  is  to  be  trans- 
mitted, and  v  is  the  common  circumferential  speed  in  feet  per 
minute,  P  being  the  necessary  tangential  force,  P  =  33,000  H/V. 
Slipping  is  to  be  impossible,  and  therefore  P  -j-  /x  is  the  force 
necessary  to  press  the  cylinders  together.  If  this  force  acts 
through  the  bearings  of  the  two  shafts,  it  is  usually  found  in 
practice  that,  unless  at  very  high  speeds  and  with  small  power, 
there  is  so  much  practical  difficulty  that  the  gear  is  never  used. 
Compressed  paper  and  leather  have  been  used  to  work  with  iron. 
Sometimes  nest-gears  are  used  to  produce  the  necessary  pressure, 
but  when  the  necessary  pressure  has  been  produced  it  has  led  to 
disintegration  of  the  surface,  or  such  local  elastic  changes  of  shape 
as  produce  annoying  sound.  In  one  case,  where  the  driven  pulley 
is  very  heavy  and  the  pressure  is  produced  by  its  own  weight  and 
that  of  the  spindle  and  part  of  the  weight  of  the  rotating  armature 
of  a  little  dynamo  machine,  the  gear  has  been  used  satisfactorily. 
Wedge-shaped  grooves  and  projections  have  been  cut  in  the  rims 
of  the  pulleys,  and  sufficient  grip  has  been  produced  in  this  way, 
but  there  is  no  longer  true  rolling. 

461.  When  we  attempt  by  using  teeth  to  get  the  necessary  driv- 
ing forces  we  introduce  sliding  contact,  using  spur,  bevil,  and  skew 
bevil  wheels ;    the  names  pitch  circles,   pitch  cones,   and  pitch 
hyperboloids  being  used  for  the  friction  gear,  which  would  run 
with  the  same  velocity  ratios.     When  the  axes  are  not  in  one 
plane,  frusta  of  hyperboloids,  generated  by  the  rotation  of  the 
same  straight  line  round  both  axes,  will  gear  with  one  another, 
always  touching  along  a  straight  line.     But  there  will  not  be 
simple  rolling ;  there  is  sliding  along  the  line  of  contact. 

Every  student  ought  to  study  the  shapes  of  spur-wheel  teeth ; 
it  is  easy  to  apply  one's  knowledge  to  other  kinds  of  teeth  and 
rubbing  and  rolling  gear.  Nothing  illustrates  the  fact  that  we  do 
not  really  think,  so  well  as  this,  that  all  the  principles  for  the 
proper  construction  of  worm-wheel  teeth  and  chain  gearing  were 
to  be  found  in  books  many  years  before  there  existed  any  good 
worm-wheel  teeth  or  chain  gearing.  The  subject,  like  all  other 
parts  of  machine  design,  is  best  studied  when  one  draws  things  to 
scale,  and  there  are  now  many  books  to  assist  the  student  in 
machine  design.  Perhaps  it  will  be  well  to  neglect  almost  all  the 
mathematical  parts  of  such  books  on  strength. 

462.  Suppose  we  have  two  curved  rollers,  v  T  v  moving  about 
the  axis  B  and  w  T  w  about  the  axis  A,  and  suppose  that  these  are 
capable  of  rolling  on  one  another  as  they  rotate ;  they  touch  at 
T,  and  the  angular  velocities  about  A  and  B  are  as  B  T  to  A  T.    Now 


APPLIED   MECHANICS. 


573 


suppose  we  wish  wheels,  with  teeth  centred  at  B  and  A,  to  have 
exactly  the  same  angular  velocity  ratios  as  the  rollers ;  if  B'  p  V 
and  A'  P  w'  are  the  shapes  of  the  teeth  in  contact  at  P,  it  is 
necessary  that  during  the  motion  the  common  normal  at  p  should 
pass  through  the  pitch  point  T.  To  effect  this  object  it  is  necessary 
that  B'  P  v'  and  A'  P  w' 
should  be  two  troch- 
oidal curves,  generated 
by  the  rolling  of  a 
curve  inside  the  curve 
T  v  and  outside  the 
curve  T  w.  Thus  in 
Fig.  325  imagine  the 
curve  v  and  the  curve 
w  to  move  about  B  and 
A,  rolling  on  one  an- 
other and  keeping  in 
contact  at  T  in  the 
straight  line  B-A,  and 
imagine  the  curve  p  P'T 
to  roll  also,  keeping  in 
contact  with  both  v 
and  w  at  T.  It  is  really 
rolling  inside  v  and 
outside  w.  If  any  point 
of  P  P'  T,  such  as  P,  is 
always  on  the  contours 
of  the  two  teeth,  the 
straight  line  PT  is  al- 
ways normal  to  these 
teeth  at  P,  their  point 
of  contact,  and  we  have 
ensured  that  the  com- 
mon normal  to  both 
passes  through  the 

pitch  point  T.  w  and  v  may  be  ellipses,  but  they  are  generally 
circles,  A  and  B  being  their  centres.  If  p  P'  T  is  a  circle,  rolling 
inside  v  and  outside  w,  the  trochoidal  curves  are  called  hypo-  and 
epi-cycloids.  When  a  number  of  wheels  are  to  gear  any  one  with 
any  other,  we  usually  choose  one  rolling  circle  for  the  insides  and 
outsides  of  all  the  pitch  circles ;  it  is  taken  of  a  diameter  equal  to 
the  radius  of  the  smallest  pitch  circle.  When  it  rolls  inside  this 
smallest  pitch  circle  the  hypocycloid  is  a  radial  line.  A  rack  may 
be  regarded  as  part  of  a  wheel  of  infinite  diameter.  Sometimes  the 
trochoidal  curves  are  involutes  of  circles,  the  rolling  curve  being 
really  a  straight  line  or  an  infinite  circle.  Let  w  and  v  (Fig.  326)  be 
the  pitch  circles.  Draw  any  line  c  D  through  T,  the  pitch  point,  and 
describe  circles  with  B  and  A  as  centres  touching  this  line  at  D  and 
c  respectively.  Draw  through  T'  o  T  H,  the  involute  of  the  circle 
D  o,  and  F  T  E,  the  involute  of  the  circle  or.  If  G  H  and  E  F  be  the 
contours  of  teeth  rotating  about  B  and  A,  their  common  normal 
remains  always  c  D.  Thus  at  any  point  T'  in  c  D,  if  we  draw 


325. 


574  APPLIED    MECHANICS. 

involutes  to  the  two  circles,  they  must  touch  there,  because  a 
tangent  to  a  circle  cuts  all  the  involutes  at  right  angles.  It  is 
evident  that  a  pair  of  wheels  with  involute  teeth  may  have  less  or 
more  distance  between  their  centres  without  any  alteration  of  their 
velocity  ratio,  this  being  B  T  :  A  T  or  B  D  :  A  c.  If  there  is  no  friction, 
c  D  is  the  direction  of  the  driving  force.  It  is  usually  more  oblique 
to  the  line  of  centres  in  wheels  with  involute  teeth  than  in  cycloidal 
teeth.  For  the  most  accurate  work  the  actual  trochoidal  curves 
are  drawn  by  the  rolling  of  templates,  but  there  are  well-know  a 
drawing-office  rules  by  which  we  approximate  to  the  true  best 
curves  by  means  of  arcs  of  circles. 

463.  In  Art.  462  we  showed  how,  when  given  the  shape  of  a  tooth, 
to  find  the  shape  of  another  tooth  to  gear  with  it  with  constant 
velocity  ratio.  A  good  example  of  the  application  of  this  rule  is 
found  when  designing  the  shape  of  the  tooth  of  a  worm  wheel. 


The  shape  of  the  thread  of  the  worm  being  chosen,  any  section  of 
the  worm  thread  by  a  plane  at  right  angles  to  the  axis  of  the  worm 
wheel  may  be  drawn  by  elementary  practical  geometry.  Regard 
it  as  the  tooth  of  a  rack.  The  section  there  of  the  tooth  of  the 
worm  wheel  must  be  the  shape  of  a  tooth  to  gear  with  the  given 


APPLIED    MECHANICS.  575 

rack  tooth,  and  must  be  drawn  by  some  such  rule  as  we  have  given 
in  Art.  462. 

464.  Sometimes   pieces  centred  at  A  and  B,  acting   upon  one 
another,  do  not  act  directly  as  shown  in  Fig.  324,  but  through  the 


Fig.  827.  Fig.  328. 

agency  of  a  third  body.  Thus,  if  pins  on  v  and  w  are  connected  by  a 
link  w  v,  and  if  the  pins  are  Motionless,  we  know  that  the  direction 
of  the  force  at  any  instant  must  be  in  the  direction  of  the  centres  of 
the  pins  or  in  the  line  w  v.  It  is  easy  to  show  as  before  that  if  a 
is  the  angular  velocity  of  AW  and  b  the  angular  velocity  of  BV, 

then  -,  =  — ,  if  T  is  where  w  v  cuts  the  line  of  centres  A  B.     The 
0       TA 

complete  study  of  the  relative  motion  of  four  links  such  as  AW, 
w  v,  v  B,  and  B  A  is  of  course  a  very  complicated  business  if  we 
imagine  them  to  be  of  all  sorts  of  lengths.  We  may,  if  we  please, 
imagine  any  one  of  them  fixed  and  consider  the  motions  of  the 
others.  Thus  in  Fig.  328  consider  A  B  to  be  fixed.  Think  that 
A  and  B  are  merely  pins  in  some  fixed  object  of  any  shape  whatso- 
ever. Now  consider  the  other  pieces  to  be  of  any  curious  shapes  and 
lengths.  The  motion  is  taken  to  be  in  the  plane  of  the  paper,  w  v  is 
only  a  straight  line  joining  the  centres  of  the  two  pins  w  and  v  ; 
but  imagine  w  v  to  represent  any  curiously  skaped  body — we  might 
wish  to  know  the  motion  of  any  point  in  this  body.  The  student's 
great  aim  is  to  get  a  correct  mental  picture,  and  if  he  recollects 
that  AV  is  moving  about  A  as  an  instantaneous  centre  or  at  right 
angles  to  A  AV,  and  v  is  moving  about  B,  the  whole  body  w  v  must 
just  at  the  present  instant  be  moving  about  the  point  c  as  a  centre. 
We  have  produced  w  A  and  v  B  to  meet  at  c,  the  instantaneous 
centre  of  motion  of  the  whole  body  w  v.  One  has  then  a  mental 
picture  of  the  motion  of  any  point  whatsoever  in  the  body  w  v. 

465.  General  motion  of  a  body  parallel  to  a  plane. — Let  us 
simply  say  a  plane  figure  or  body  in  its  own  plane.     If  we  consider 
T 


576  APPLIED    MECHANICS. 

the  motion  of  any  two  points,  this  settles  the  motion  of  every  other 
point,  so  we  shall  speak  only  of  two  points.  It  is  easy  to  prove 
that  when  we  consider  two  positions  of  the  body  there  is  one  point 
of  the  body  whose  position  is  the  same.  Thus,  if  w  v  is  one 
position  and  w'  v'  is  another  position,  bisect  w  w'  by  a  line  at  right 
angles  to  it,  and  let  it  meet  the  rectangular  bisector  of  v  v'  in  c. 
Evidently  o  is  a  point  in  the  body  which  has  not  altered  its 
position  ;  it  is  the  instantaneous  centre  of  the  motion.  Notice 


Fig.  829. 


Pig.  330. 


that  if  w  w'  is  parallel  to  v  v',  c  is  at  an  infinite  distance ;  we  then 
say  that  the  motion  is  one  of  translation  merely.  In  Fig.  329  we 
have  a  piece,  w  v,  whose  ends  move  in  the  paths  w7  w  and  v'  v.  At 
any  instant,  if  we  draw  w  c  and  v  c,  normals  to  the  paths,  we  find 
c  the  instantaneous  centre. 

Consider  any  point  p  in  w  v,  and  join  PC.  P  traces  out  some 
path  during  the  motion.  Notice  that  P  c  is  the  normal  to  this 
path,  and  a  line  through  p  at  right  angles  to  p  c  is  the  tangent  to 
it.  If  we  want  to  consider  the  envelope  of  the  straight  line  w  v, 
notice  that  the  foot  of  the  perpendicular  from  c  upon  w  v  is  a  point 
in  that  envelope,  because  it  is  the  only  point  of  w  v  which  moves  in 
the  direction  of  the  length  of  w  v.  The  student  must  not  imagine 
that  because  at  the  instant  every  point  of  a  body  is  moving  about  c, 
therefore  c  is  the  centre  of  curvature  of  the  path  of  w,  or  of  v,  or  of 
p.  c  is  only  for  the  instant  the  centre  of  motion,  and  such  lines  as 
we,  v  c,  and  P  c  are  the  directions  of  the  normals  to  the  actual 
paths  of  these  three  points.  As  the  figure  moves  in  its  own  plane 
from  one  position  to  others  successively,  let  0103(33,  etc.,  be  the 
successive  points  of  the  figure  about  which  the  rotations  take  place, 
and  let  c\  c2c3,  etc.,  be  the  positions  of  these  points  on  the  fixed  plans 
when  each  is  the  instantaneous  centre  of  rotation.  Then  the  figure 
rotates  about  Ci  (or  c\,  which  coincides  with  it)  till  c2  coincides  with 
<?2 ;  then  about  cz  till  c3  coincides  with  c3)  and  so  on.  Hence,  if  we 
join  Ci  02  C3,  etc.,  in  the  plane  of  the  figure,  and  c\  c2c3,  etc.,  in  the 
fixed  plane,  the  motion  will  be  the  same  as  if  the  polygon  Ci  c2  03, 
etc.,  rolled  upon  the  fixed  polygon  c\czcy  etc.  By  supposing  the 
successive  displacements  smaller  and  smaller,  we  have  curves,  and 
hence  any  motion  whatever  of  a  plane  figure  in  its  own  plane  may 
be  imagined  as  produced  by  the  rolling  of  a  curve  fixed  to  the 
figure  upon  a  curve  fixed  to  the  plane.  It  immediately  follows 
that  any  displacement  of  a  rigid  solid  parallel  to  a  plane  may  be 
produced  by  the  rolling  of  a  cylinder  fixed  in  the  solid  on  another 


APPLIED    MECHANICS. 


577 


cylinder  fixed  in  space,  the  generating  lines  of  the  cylinders  being 
at  right  angles  to  the  plane. 

466.  What  we  have  said  about  motion  in  a  plane  is  true  about 
motion  of  figures  shaped  to  fit  a  spherical  surface ;  straight  lines  in 
the  one  case  corresponding  to  great  circles  of  the  sphere  in  tho 
other  case.  It  is  easy  to  have  a  mental  image  of  this.  Now 


Pig.  831. 


imagine  that  any  point  of  the  superficial  spherical  figure  is  joined 
to  the  centre  of  the  sphere,  and  we  see  that  (1)  if  a  rigid  body  has 
one  point  fixed  (imagine  this  the  centre  of  an  imaginary  spherical 
surface),  however  it  may  move,  in  any  two  positions  of  the  body 
there  is  one  line  of  the  body  which  is  common  to  the  two  positions. 
(2)  Any  motion  may  be  regarded  as  due  to  the  rolling  of  a  cone 
fixed  in  the  body  upon  a  cone  fixed  in  space.  What  we  have  said, 
therefore,  about  the  pieces  AW  and  B  v  of  Fig.  327  moving  about 
axes  at  right  angles  to  the  plane  of  motion  may  be  at  once  applied 
to  pieces  moving  in  a  spherical  surface  about  axes  meeting  in  the 
centre  of  the  sphere.  Or  the  cylindiic  pieces  A  w  and  B  v  may  be 
imagined  to  be  conical  pieces  moving  about  axes  A  and  B,  meeting 
at  the  vertex  of  thd  cones.  Hence  all  that  we  have  said  about  spur 
wheels  is  at  once  applicable  to  bevil  wheels,  which  are  suitable  for 
shafts  whose  centre  lines  or  axes  meet  at  a  point. 

467.  Fig.  329  is  worth  a  very  great  deal  of  consideration.  There 
is  a  link  w  v,  and  motion  of  a  point  w  in  it  is  known.  The  shape  of 
the  path  of  v  is  known,  to  find  the  motion  of  the  whole  link,  andjof 
any  point  in  it,  at  every  instant.  When  the  paths  of  w  and  v  are 
arcs  of  circles,  we  have  the  four-bar  kinematic  chain  of  Fig.  327. 
When  w  (a  crank  pin)  moves  in  the  arc  of  a  circle,  and  v  is  a  block 
moving  in  a  straight  slot,  one  particular  case  is  called  a  slider 
crank  chain  (see  Fig.  332).  When  both  w  and  v  are  blocks  moving 
in  straight  slots  at  right  angles  to  one  another,  we  have  the 
ordinary  trammels  for  describing  an  ellipse.  We  have  the  mathe- 
matical basis  of  the  elliptic  chuck,  for  we  may  imagine  any  part  of 
a  mechanism  to  be  at  rest,  and  tnen  the  relative  motions  of  the 
others  become  absolute  motions.  The  theory,  then,  of  Art.  464  ia 
the  same  for  a  very  great  many  mechanisms. 


578 


APPLIED    MECHANICS. 


468.  In  Fig.  327,  if  we  imagine  B  A  to  be  fixed,  B  in  the  position 
in  which  it  is  in  the  figure,  w  A  and  B  A  infinitely  long,  so  that  w's 
arc  of  a  circle  is  really  a  straight  line,  we  have  the  usual  crank  B  v 

and  connecting  rod  v  w, 
which  we  can  indicate 
in  its  simplest  form  by 
Fig.  332.  Now  take 
this  as  it  stands.  A 

Fig.  332.  piece  B  A  x  along  which 

w  slides,  and  the  con- 
necting-rod w  v  and  the  crank  v  B  ;  think  merely  of  the  relative 
motion  of  each  piece  to  the  rest.  You  may  imagine  any  one  of 
the  pieces  fixed,  and  you  may  consider  the  motion  of  any  point 
in  any  piece.  If  BX  is  fixed,  v  describes  a  circle,  w  a  straight 
line,  and  any  point  in  w  v  describes  a  curve  which  is  some- 
what like  an  ellipse,  only  blunter  at  one  end  than  the  other. 
Now  imagine  B  v  =  v  w :  v  will  not  make  a  rotation.  Imagine 
w  v  extended  beyond  v  as  far  again  :  that  point  will  describe  a 
straight  line,  and  we  have  a  parallel  motion.  A  student  must  try 
these  things  with  models.  Now  imagine  B  v  fixed,  and  let  v  w  turn 
uniformly.  [Draw  the  fixed  part  of  any  curious  shape,  the  frame 
of  a  machine,  v  and  B  here  being  two  pins.]  We  get  a  quick 
return  mechanism  for  shaping  and  other  machines.  Imagine  v  w 
fixed,  and  we  have  the  mechanism  of  oscillating  cylinder  engines 
and  of  other  machines  as  well.  Now  imagine  w  fixed,  and  guiding 
B  x,  so  that  it  shall  only  move  in  the  direction  of  its  length,  and 
we  have  a  well-known  pump  mechanism.  When  we  consider  that, 
even  from  Fig.  332,  with  the  motion  of  w  in  a  line  with  B,  we  have 
obtained  a  number  of  different-looking  mechanisms,  and  that  these 
can  be  varied  very  curiously  by  taking  various  lengths  of  the  parts, 
it  will  be  seen  that  there  is  a  nearly  endless  variety  of  forms 
derivable  from  the  mechanism  shown  in  Fig.  327.  Now  imagine  a 
pin  Y  in  the  piece  w  v  linked  to  z,  a  pin  in  an  arm  z  B,  and  we  have 
a  much  more  complicated  problem  to  study  in  the  relative  motion 
of  six  pieces.  Any  one  of  them  may  be  imagined  fixed,  and  they 
may  be  of  all  sorts  of  lengths. 

469.  The  following  example  is  of  geometrical  interest.      In 
the  Peaucellier  cell  (Fig.  333),  A  B  and  A  w  of  Fig.  327  are  made 


334. 


Pig.  333. 

equal  in  length,  and  so  are  B  v  and  v  w.  Two  pieces,  w  c  and  CB, 
equal  to  B  v  and  v  w,  are  added.  It  is  easy  to  see  that  A  c  v  is 
always  a  straight  line.  Also  the  positions  of  A,  c,  and  v  are  such 


APPLIED    MECHANICS. 


579 


that  A  c  .  A  v  remains  constant.  For  if  A  c  v  and  w  B  he  joined, 
and  they  meet  in  o,  the  centre  of  the  rhombus,  A  w9  =  A  o2  +  o  w2, 
and  v  w2  =  v  o2  +  o  w2.  Hence 

A  w2  -  v  w2  =  A  o2  -  v  o2  =  (A  o  +  v  o)  (A  o  -  .vo)  =  A  v  .  A  c. 

Hence,  if  A  is  fixed,  and  c  traces  out  any  curve,  v  will  trace  out 
the  reciprocal  curve.  If  c  is  constrained  by  a  link  o'  c  to  move  in 
a  circle,  and  if  O'A  =  o'c,  v  will  describe  the  reciprocal  curve/ 
which  in  this  case  is  a  straight  line.  If  o'  c  is  not  equal  to  o'  A,  v 
will  describe  the  arc  of  a  circle  of  much  greater  radius  than  o's. 
470.  When  the  four  links  of  Fig.  327  form  a  parallelogram,  as 


Fig.  335 


A  w  v  B  of  Fig.  334,  then,  if  p  is  a  point  anywhere  rigidly  part  of  the 
link  w  v,  but  in  the  line  w  v,  and  if  at  any  time  it  is  in  the  same 
straight  line  with  A,  and  a  point  Q,  which  is  rigidly  a  part  of  the 
link  BV,  and  in  the  line  BV,  then  p  and  Q  will  always  be  in  a 
straight  line  with  A.  We  can  imagine  now  that  either  Q  or  p  or 
A  is  a  fixed  point,  and  the  other  two  will  follow  paths  which  are 
similar  to  one  another. 

>  Thus,  for  example,  if  AWVB  (Fig.  335)  is  the  four-bar 
mechanism  of  Fig.  327  (A  B  need  not  be  drawn  because  we  imagine 
it  fixed),  and  if  A  w  D  is  one  piece,  v  p  and  p  D  being  links  equal  to 
w  D  and  w  v  respectively,  then  the  pin  P  will  travel  in  a  path 
similar  to  that  of  Q  if  Q  w  :  A  w  :  :  P  D  :  A  D.  If  Q  travels  very 
nearly  in  a  straight  line  (if  B  v  :  A  w  :  :  Q  w  :  Q  v,  we  have  a 
near  approach  to  straight-line  motion  of  Q,  so  long  as  B  v  and  A  w 
do  not  make  too  great  an  angle  with  one  another),  then  p  will  also 
travel  very  nearly  in  a  straight  line. 

471.  Again,  if  A  w  v  B  (Fig.  336)  is  a  parallelogram,  and  if  p  is  a 
point  rigidly  part  of  the  link  B  v,  and  if  Q,  is  a  point  rigidly  part  of  the 
link  w  v,  and  if  p  and  Q  are  so  placed  that  the  ratios  p  v  :  v  B  :  B  p  are 
the  same  as  the  ratios  v  Q  :  Q  w  :  w  v,  then  it  can  be  shown  that  A  Q 
and  if  p  and  Q,  are  so  placed  that  the  ratios  p  v  :  Y  B  :  B  p  are  the 
same  as  the  ratios  v  Q  :  o,  w  :  w  v,  then  it  can  be  shown  that  A  Q 
and  A  P  always  make  the  same  angles  with  one  another,  and  that 
the  ratio  of  the  lengths  AQ  and  AP  keeps  constant.  In  the 
triangles  p  v  B  and  v  Q  w,  the  angles  at  p  and  v,  at  v  and  Q,  and  at 
B  and  w  are  respectively  equal.  Hence  the  triangles  aw  A  and 
A  B  P  are  similar,  and  it  follows  that  the  angle  Q  A  p  keeps  constant 
in  any  motion  of  the  mechanism,  and  the  ratio  of  the  lengths  A  Q 


580 


APPLIED    MECHANICS. 

and  A  P  keeps  constant.  It  follows,  therefore,  that  if  A  is  fixed  and 
p  is  allowed  to  follow  any  curved  path,  Q  will  follow  a  similar  path. 
472.  Note  that  Fig.  334,  where  P  and  Q,  are  in  the  straight  lines 
connecting  w  v  and  B  v,  is  only  a  particular  case  of  Fig.  336.  Note 
also  that  if  A  w  and  w  v  are  a  pair  of  links  guided  at  A  and  v  along 
any  paths,  and  if  A  B  and  B  v  are  another  pair  of  links  whose  A  and 
v  follow  identical  paths  with  the  first  pair's  A  and  v,  then  for  each 
point  in  the  one  mechanism  there  will  he  a  corresponding  point  in 
the  other  which  follows  a  similar  path. 

It  is  interesting  to  prove  that  if  the  links  A  w,  w  v,  A  B,  B  v, 
forming  a  parallelogram  as  in  Fig.  336,  he  joined  up  as  in  Fig.  337, 

then  any  four  points  p  Q  R  s  in  a 
line  parallel  to  AV  remain  in  a 
line  parallel  to  AV,  however  the 
mechanism  may  alter  in  shape ; 
and  the  distances  P  Q,  Q  R,  R  s  are 
always  such  that  p  R  .  p  o.  is  con- 
Fig.  337.  stant.  Also  p  Q  =  R  s. 

473.  "We  study  mechanisms,  1st, 

Because  in  designing  new  machinery  we  ought  to  have  a  general 
fairly  exact  knowledge  of  the  sort  of  motion  which  each  piece  has 
in  existing  machines.  For  this  we  watch  existing  machinery,  and 
work  with  rude  models  whose  dimensions  may  be  varied.  2nd, 
Because  we  wish  to  perfect  an  existing  mechanism.  For  this  we 
need  a  more  exact  study  of  the  motion  of  every  part,  and  we  use 
skeleton  drawings,  algebraic  and  trigonometric  analysis,  or  graphical 
methods  such  as  I  have  described.  3rd,  Because  in  these  days  of 
increasing  speed  of  machinery,  the  forces  necessary  to  produce 
the  accelerations  of  all  the  parts  have  become  important,  both 
for  the  strength  and  stiffness  of  all  the  parts,  and  also  for  the 
effects  of  vibration. 

474.  In  Fig.  329  we  saw  that,  knowing  the  directions  of  motion  of 
w  and  v,  we  can  find  the  instantaneous  centre.  Now  let  w  v  get  any 
motion  whatsoever  in  the  plane  of  the  paper.  Let  the  velocities  of 
w  and  v  be  w  and  v  in  the  directions  marked.  If  we  draw  perpen- 
diculars to  w  and  v,  meeting  at  c,  the  angular  velocity  of  w  about 


f 


Fig.  338. 

c  must  be  the  same  as  that  of  v,  and  this  must  be  the  same  as  that 
of  any  other  point  K,  rigidly  attached  to  w,  about  c.  Hence,  join 
any  point  K  to  c,  the  velocity  of  K.  is  at  right  angles  to  CK,  and 
is  represented  in  amount  by  the  length  of  c  K.  Hence,  as  soon  as 
we  find  the  instantaneous  centre,  we  have  a  diagram  of  velocities 
of  all  points  in  the  body.  Notice  that  if  we  have  the  instantaneous 


Af*t>LtEl>   MECHANICS. 


581 


Fig.  839. 


Centre  c  and  the  velocity  of  any  one  point,  we  have  tne  whole 

diagram  to  scale.     If  c  D  is  a  perpendicular  to  v  w,  as  the  length 

of  every  line  now  in  the  diagram  represents  the  amount  of  a 

velocity  which  is  at  right  angles  to  its  direction,  c  D  is  the  com- 

ponent or  common  velocity  of  all  points  in  the  straight  line  w  v  in 

the  direction  w  v.     D  w  is  the  velocity  at  right  angles  to  the  direc- 

tion w  v  of  the  point  w,  and 

D  v  is  the  same  for  v,  hence 

w  v  represents  v's  velocity  in 

excess  of  w's  in  this  direc- 

tion.   Hence,  if  w  v,  v  x,  x  Y 

are  given  links,  and  if  we 

know  the  directions  of  the      "~ 

velocities  of  each  joint  at  the 

instant,  say  the  directions  of 

the  dotted  lines,  we  can  find  the  velocity  of  every  point  in  any 

hody  which  is  rigidly  a  part  of  any  of  the  links  if  we  know  the 

actual  speed  of  any  one  point. 

Choose  a  pole  o.  Draw  lines  from  o  at  rigty  angles  to  the 
velocities  of  all  the  joints—  o  w,  o  v,  etc.  Let  the  distance  along 
any  one  of  them  represent  the  velocity  of  that  joint  to  scale,  and 
now  draw  the  lines  wv,  vx,  xy  parallel  to  the  real  links.  Then  the 

lengths  o  w,  ov,  ox,  oy  re- 
present the  velocities  of  the 
joints.  To  prove  this,  drop  a 
perpendicular  o  D  from  o  upon 
any  of  the  link  directions  in 
our  diagram  ;  we  choose  vw. 
If  o  w  is  the  amount  to  scale 
and  is  at  right  angles  to  the 
velocity  of  w,  then  o  D  repre- 
sents the  velocity  of  w  in  the 
direction  of  the  link  w  v  ;  hut 
this  ought  to  be  the  same  for  v. 
We  have,  then,  the  amount 
OD  of  a  component  of  v's 
velocity,  and  we  have  the 
direction  of  its  whole  velocity,  and  our  construction  is  the  very 
one  which  we  should  adopt  to  obtain  v's  velocity.  If  w  and  Y 
are  fixed,  the  diagram  wvxy  is  a  closed  polygon,  in  the  present 
case  a  triangle. 

Notice  that  in  any  case,  if  from  o  we  drop  a  perpendicular  o  D 
on  a  side  of  the  diagram,  say  wv,  then  D  w  and  D  v  represent  the 
,  velocities  of  w  and  v  at  right  angles  to  the  length  of  D  w.  Hence 
the  difference  wv,  divided  by  the  actual  length  of  w  v,  represents 
the  angular  velocity  of  wv.  Thus,  in  Fig.  343,  take**~AWVB  as 
four  links,  A  and  B  having  no  velocities.  From  the  point  o,  which 
we  may  also  call  a  or  J,  draw  lines  parallel  to  A  w  and  B  v,  because 
we  know  that  these  are  at  right  angles  to  the  velocities  of  w  and  v, 
and  draw  vw  parallel  to  vw.  Then  bv  -«-  B  v,  aw  -^  AW,  and 
rw  -T-  v  w  represent  the  angular  velocities  of  the  three  links. 

470.  In  Fig.  339,  if  K  is  a  point  rigidly  attached  to  the  link  v  z 


Fig.  340. 


582 


APPLIED    MECHANICS. 


(in  the  diagram  find  k  so  that  vkx  is  a  triangle  similar  to  v  K  x),  then 
o  k  is  at  right  angles  to  and  represents  to  scale  the  velocity  oi  K.. 
Professor  K.  H.  Smith,  who  was,  I  believe,  the  inventor  of  this 


Fig.  842. 

method,  takes  his  radiating  lines  parallel  to,  instead  of  at  right 
angles  tov  the  actual  velocities  (Trans.  R.S.E.,  Jan.,  1885).  He 
gives  a  similar  construction  for  accelerations. 

If  a  point  B  is  fixed  and  a  body  rotates  in  the  plane  of  the 
paper  round  it,  the  accelerations  of  any  points  A  and  K  are  in  the 
directions  A  B  and  K  B,  and  they  are  proportional  to  the  distances 
A  B  and  K  B.  Let  A  K  B  be  a  rigid  body  with  a  motion  in  the  plane 
of  the  paper.  Let  the  actual  amounts  of  the  accelerations  of  A  and 
B  be  known  ;  represent  these  in  amount  and  direction  by  the  lines 
o  a  and  o  b.  Now  make  the  figure  akb 
exactly  similar  in  shape  to  the  real  body 
AKB.  The  true  acceleration  of  any  point 
K  is  represented  in  amount  and  direction 
by  ok. 

To  prove  this :  The  acceleration  of  A  is 
the  vector  sum,  acceleration  of  B  +  accelera- 
tion of  A  relatively  to  B.  Hence  la  repre- 
sents this  acceleration  of  A  relatively  to  B. 
But  the  motion  of  A  and  the  whole  body 
relatively  to  B  is  a  rotation,  and  hence  bk 
represents  the  acceleration  of  K  relatively 
to  B  to  the  same  scale.  The  vector  sum 
o  b  +  bk  —  o  k  represents  therefore  K'S  ac- 
celeration. 

Notice  that  the  true  acceleration  of  any 
point  A  is  the  vector  sum  of  two  accelera- 
tions— the  first  in  the  direction  of  motion, 
the  second  the  centripetal  acceleration  if 
the  path  of  A  is  curved. 
476.  Let  a  known  force  in  any  direction  be  applied  to  A,  a  point 
in  the  body  AKB  which  has  a  known  motion  in  the  plane  of  the 
paper. 

1st.  Imagine  the  body  divided  into  small  equal  masses.  The 
construction  of  Fig.  342  enables  us  to  find  the  forces  with  which 
these  masses  resist  their  accelerations.  They  may  be  combined 


Fig.  343. 


APPLIED    MECHANICS.  583 

with  the  weights  of  the  parts  to  get  what  may  be  called  the  load 
diagram  of  the  body. 

Let  it  be  known  that  there  is  another  force  of  unknown  amount, 
but  known  in  direction,  acting  at  A  (for  example,  the  guiding  force, 
with  or  without  friction,  at  the  crosshead  of  a  steam  engine).  It  is 
easy  to  see  that,  by  Art.  475,  we  can  find  the  amount  of  this 
guiding  force  at  A,  and  also  the  total  force  which  must  be  acting  at 
a  given  point  B. 

2nd.  In  Art.  475  we  have  a  construction  which  enables  us  to 
find  the  stress  at  any  point  of  any  section  of  the  structure  A  K  B,  if 
it  may  be  imagined  to  be  any  quasi-prismatic  structure  like  a  beam 
or  connecting-rod,  or  even  if  it  be  shaped  like  part  of  a  metal  arch. 
In  such  a  case  as  that  of  the  connecting-rod,  the  steam-engine 


- ._._»-. .o 

t^  0    A^  C 


maker  ought  to  study  the  motion  from  many  points  of  view,  o  v 
and  v  w  being  the  crank  and  connecting-rod  in  any  position. 
Produce  w  v  to  meet  o  B  in  B.  o  B  is  at  right  angles  to  the  line  of 
centres,  o  w.  Then  the  lines  o  v  and  o  B  are  at  right  angles  to  the 
velocities  of  v  and  w,  and  v  B  is  in  the  direction  of  the  link  w  v ; 
consequently,  o  v  B  is  (Art.  475)  a  diagram  of  velocities,  o  v 
representing  to  scale  the  constant  velocity  of  v,  OB  represents  to 
the  same  scale  the  velocity  of  w ;  also  the  distance  v  B  represents 
the  angular  velocity  of  the  connecting-rpd.  It  follows  from  this 
at  once  that  a  force  F  at  w  in  the  direction  w  o  would,  if  there 
were  no  friction  and  the  connecting-rod  were  massless,  produce  a 
turning  moment  F  x  o  B  on  the  crank  shaft. 

It  can  be  shown  that  if  we  draw  B  A  at  right  angles  to  w  B,  join 
T  A,  draw  B  c  parallel  to  v  A,  let  w  be  in  w  o  as  far  from  A  as  c  is,  but 
on  the  opposite  side  of  A,  and  join  w  v,  then  o  w  v  is  an  acceleration 
diagram  such  that  if  K  is  any  point  in  the  connecting-rod,  and  we 
let  the  points  rv,  k,  and  v  be  relatively  to  each  other  as  w,  K,  and  v, 
(k  is  evidently  in  the  same  horizontal  as  K),  then  ko  represents  the 
acceleration  of  K  in  direction  and  magnitude  to  the  same  scale  to 
which  vo  represents  the  centripetal  acceleration  of  v.  In  fact, 
o  w  lev  is  a  diagram  of  accelerations. 

477.  When  one  point  of  a  body  is  fixed,  and  we  may  neglect 
centripetal  accelerations,  we  may  take  up  the  problem  as  a  particular 
case  of  the  general  problem  already  considered,  or  we  may  attack 
it  analytically  as  follows  : — If  o  p  is  the  straight  centre  line  of  an 
arm,  o  being  fixed,  and  if  a  is  its  angular  acceleration ;  neglecting 
forces  parallel  to  o  r,  let  there  be  a  force  F  acting  at  p ;  the 
acceleration  at  any  point  Q  (neglecting  radial  acceleration)  is  xa,  it 
OQ,  is  x.  If  w  is  the  mass  per  unit  length,  the  load  due  to 
T* 


584  APPLIED    MECHANICS. 

acceleration  per  unit  length  is  max  ;    and  if  M  is  the  bending 
moment  at  the  cross-  section  (see  Art.  357), 


Assume  m  to  "be  some  function  of  x,  and  let  s  be  the  shearing 
force  at  a  section.  Let  \mx.dx  =  v,  I  v  .  dx  =  «,  and  let  the 
values  of  v  and  u  become  v\  and  ult  when  x  =  I.  Then 

•  =  ;£  =  «*  +  '•.  .-(2), 

M  =  au  +  ex  +  e  .  .  .  .  (3), 

P  =  oVi  +  c  .'.<?  =  F  -  avj, 

D  =  atii  +  (F  -  avi)  I  +  e  .'.  e  =  a  (lv\  -  u\)  -re. 

If  A  is  the  area  of  cross-section  and  p  is  the  mass  per  unit  volume, 
A  p  —  m.  If  we  wish  to  have  the  same  maximum  stress  /in  every 
cross-section,  and  if  z  is  the  strength  modulus  of  the  section,  then 


m  -  Ap 

Dividing,  therefore,  (3)  by  w,  we  know  the  value  of  Z/A  every- 
where, if  the  arm  is  to  be  of  uniform  strength.  Thus,  for  example, 
let  the  section  be  rectangular,  of  breadth  z  and  depth  (in  the  plane 
of  motion)  y.  Then  A  =  zy,  z  =  \  ty* ;  so  that  Z/A  ia  J  y.  Hence 

fy       M  6p  M 

from  (4)  we  have  -  =  -  or  y  =  -j  - ,  or 

y  =^n(au  +  ex  +  e) (5). 

Since  pzy  =  tn,  z  =  — ,  so  that  when  we  know  y  we  know  z ; 

Z  =  fp*  £  '     '  '  (6)' 
Example. — Let  m  =  a  -  bx, 
v  =  l>ax2  - 
u  = 
c  = 


478.  In  discussing  the  reciprocating  motion  of  a  point,  I  have 
found  in  practice  much  simplification  in  my  ideas  when  I  have 
.reduced  the  motion  to  some  such  shape  as 

x  =  a  sin.  (qt  +  4)  +  b  sin.  (2  qt  +  ez]  +  etc. 

where  x  is  the  displacement  of  the  point  from  some  fixed  point  in 
its  path,  and  q  is  2  TT  multiplied  by  the  frequency,  or  2  ir  divided  by 
the  periodic  time.  For  many  purposes  we  find  the  first  term 
sufficient.  In  most  valve  motions,  such  as  link  motions  and  radial 


APPLIED    MECHANICS.  585 

gear,  I  find  that  only  two  terms  are  ever  needed  even  for  a  very 
exact  definition  of  the  motion.  One  interest  attaching  to  this 
method  of  working  is  in  its  showing  how  the  b  term  becomes 
twice  as  important  in  the  velocity,  and  four 
times  as  important  in  the  acceleration  as  the 
fundamental  term.  Another  important  matter 
is  this :  in  modern  machinery  vibration  is 
becoming  very  important,  and  the  above  de- 
scription of  a  motion  is  the  one  that  lends 
itself  most  easily  to  a  discussion  of  the  vibra- 
tions which  want  of  balance  gives  rise  to.  (See 
the  author's  books  on  the  Calculus  and  on  the 
Steam  Engine.) 

479.  The     following    proposition    is    the 
foundation    of    most    calculations    on    motion 
communicated  through  links.    The  points  A  c  B 

are    in  a    straight   line.      They    move,    keen-  Fig-  345. 

ing  in  a  straight  line,   and  at  the  same  dis- 
tances from  one  another.      (In    engineer's  language  ACS'  is  a 
moving  link,  and  c  is  a  point  in  it.)     Prove  that  the  motion  of  o 

is  the  vector  sum  of  the  fraction  —  of  B'S  motion,  and  the  fraction 

A  B 

—  of  A'S  motion.    For  let  A  A',  B  B',  c  c'  be  any  displacements  of  A,  B, 

A  B 

and  c.     Join  A'  B.     Draw  c  c"  parallel  to  A  A',  and  join  c"  c'.    Now 
^  =  ^  =  ^,  so  that  c"  tf  is  paraUel  to  B  B'.    Hence,  as  c  c 

is  the  vector  sum  of  c  c"  and  c"  c',  we  have  proved  the  proposition. 
It  is  easy  to  extend  our  reasoning  to  motion  which  is  not 
parallel  to  one  plane.  The  motions  of  any  three  points  A  B  c  of  the 
body,  not  in  one  line,  define  the  whole  motion ;  and  when  we  are 
given  the  accelerations  of  A,  B,  and  c,  it  is  easy  for  anyone  who 
knows  descriptive  geometry  to  make  a  diagram  showing  the  accelera- 
tion of  any  point  of  the  body. 

480.  Newton's  great  one  law  of  motion  for  any,  however  complex, 
system  of    bodies  is  this :    Look  upon  the  rate  of    change  of 
momentum  of  any  small  portion  of  matter  of  a  system  as  a  force 
in  the  opposite  direction.    All  such  forces  are  in  equilibrium  with 
the  forces  which  act  on  the  system  from  the  outside.    In  most 
cases  it  is  this  most  general  way  of  stating  the  law  that  is  most 
useful  to  engineers.     Given  their  diagram  of  accelerations,  they 
really  have  a  force  or  load  diagram,  and  the  problems  to  be  dealt 
with  are  now  merely  worked  out  by  graphical  statics. 

481.  Most  men  are  led  to  their  study  of  this  subject  through  what 
is  called  D'Alembert's  principle.    This  is  a  principle  which  served 
a  very  useful  purpose.     At  a  time  when  English  mathematicians 
were    stagnating,  being    academically    learned    as    to    Newton's 
methods,  but  being  really  ignorant  of  them,  the  French  mathema- 
ticians   were    developing    kinetics    practically    independently    of 
Newton's  methods.     Nevton's  third  law  was  quite  misunderstood, 


586  APPLIED    MECHANICS. 

and  D'Alembert  discovered  a  principle  which  gave  the  power  of 
solving  dynamical  problems  with  certainty.  It  will  not  astonish 
anyone  who  knows  academic  methods  to  be  informed  that,  although 
D'Alembert  gave  what  is  now  seen  to  be  merely  a  rather  cumbrous 
explanation  (easily  misunderstood,  because  of  certain  technical 
terms  which  may  be  confounded  with  one  another)  of  Newton's 
third  law,  it  is  through  the  clumsy  explanation  that  the  subject  is 
nearly  always  approached.  We  believe  that  this  is  one  of  the 
greatest  reasons  why  many  engineers  are  so  disgusted  with  higher 
studies  in  dynamics.  There  seems  to  be  as  much  absence  of 
common  sense  now  in  academic  persons  as  there  was  in  the  time 
of  Erasmus.  He,  the  greatest  scholar  of  the  fifteenth  century, 
wrote :  "  They  are  a  proud,  susceptible  race.  They  will  smother 
me  under  six  hundred  dogmas.  They  will  call  me  heretic,  and 
bring  thunderbolts  out  of  their  arsenals,  where  they  keep  whole 
magazines  of  them  for  their  enemies.  Still,  they  are  Folly's 
servants,  though  they  disown  their  mistress.  They  live  in  the 
third  heaven,  adoring  their  own  persons,  and  disdaining  the  poor 
crawlers  upon  earth.  They  are  surrounded  with  a  bodyguard  of 
definitions,  conclusions,  corollaries,  propositions  explicit,  and 
propositions  implicit." 

482.  Let  the  engineer  take  Newton's  law  in  its  very  simplest 
form,  as  above  expressed,  and  he  will  have  no  difficulty  in  attacking 
the  most  complicated  problems,  for  the  dynamical  becomes  a  static 
problem  on  the  equilibrium  of  forces.     It  is  usual  to  express  part 
of  the  result  analytically  in  the  following  way  : — If  in  any  direc- 
tion, which  we  may  call  x,  the  small  mass  m  has  the  acceleration  #, 
then  the  resultant  force  acting  from  outside  the  system  in  that 
direction  is  equal  to  the  sum  of  all  such  terms  as  mae.     State  this 
as  being  true  in  any  three  directions,  and  of  course  it  is  true  in 
any  direction  whatsoever. 

Now  if  the  a^of  the  centre  of  gravity  of  the  whole  system  is  xt 
we  know  that  x"S,m  =  *$,mx.  Differentiate  with  regard  to  time 
once,  and  again,  and  we  see  that  the  whole  mass  2  m,  multiplied 
by  the  acceleration  of  the  centre  of  gravity,  is  equal  to  the  sum  of 
all  the  masses  multiplied  by  their  accelerations  in  the  direction  x. 
It  follows,  therefore,  that  the  motion  of  the  centre  of  gravity  of 
a  system  is  the  same  as  if  the  mass  of  the  system  were  collected 
there,  and  all  the  forces  acting  from  outside  on  the  system 
acted  there.  The  other  part  of  Newton's  law  is,  of  course: 
The  resultant  moment  of  all  the  outside  forces  about  any  axis 
is  equal  to  the  rate  of  change  of  moment  of  momentum  of  the 
whole  system  about  the  same  axis.  We  very  often  choose  as  our 
axis  an  axis  through  the  centre  of  gravity,  but  it  is  well  to  notice 
that  this  is  not  necessary. 

483.  Angular  Motion. — We  know  (Art.  92)  that  when  a  rigid 
body  can  only  rotate  about  an  axis,  if  the  sum  of  the  moments  of  the 
forces  acting  on  the  body  is  M,  when  the  body  moves  through  the 
angle  80  radians  the  work  done  is  M  .  §0.     If  any  little  portion  m  of 
the  mass  of  the  body  is  at  the  distance  r  from  the  axis,  and  the 


APPLIED    MECHANICS.  587 

body  is  rotating  with  the  angular  velocity  «,  so  that  ar  is  the 
velocity  at  the  place,  the  energy  stored  up  in  the  little  mass  is 
%ma?r2.  If  this  is  summed  up  for  the  whole  body,  we  see  that 
every  little  term  contains  %d2;  so  that  if  we  know  the  sum  of 
all  such  terms  as  mr2  (called  i,  the  moment  of  inertia  of  the  body 
about  the  axis),  we  can  say  that  the  kinetic  energy  is  £i#2.  When 
a  force  r  acts  on  a  body  through  the  distance  Ss,  the  work  done 
being  p  .  8s,  and  the  kinetic  energy  of  the  body  being  £  mvz,  by 
assuming  that  work  done  is  equal  to  gain  of  kinetic  energy,  we 
are  led  to  the  law  p  =  m  x  acceleration.  We  have  an  exactly 
analogous  set  of  terms  for  angular  motion.  The  work  M  .  50 
corresponds  with  F  .  5*;  the  energy  £ia2  corresponds  with  |mt?2. 
The  proof,  therefore,  is  exactly  as  in  Art.  497  —  namely,  if  a  moment 
M  acts,  through  the  angle  50,  on  a  body  moving  with  the  angular 
velocity  a,  and  whose  moment  of  inertia  is  i,  so  that  its  kinetic 
energy  is  \  la2,  and  if  a  +  Sa  is  its  new  angular  velocity,  then 

M  .  86  =  £i  (a  +  5«)2  -  Jia2  =  i  .  aSa  +  £i  .  (50)2, 


As  50,  and  therefore  8a,  become  smaller  and  smaller,  we  have 


or 

da 


if  a  is  used  for  —  ,  or  the  angular  acceleration.     This  is  exactly 

analogous  with  the  force  law  in  linear  motion,  i  a  is  called  the 
moment  of  momentum  of  the  body,  and  i  a  is  the  rate  of  change  of 
the  moment  of  momentum. 

484.  To  arrive  directly  at  the  moment  of  momentum  of  a  rigid 
body  about  any  axis,  consider  a  portion  of  mass  m  at  the  distance  r 
from  the  axis  ;  r  making  an  angle  6  with  a  fixed  plane  through  the 

axls>  x  =  r  cos.  e,y  =  r  sin.  0, 

dx  de  dy  ,    de  ... 


The  moment  of  the  momentum  m  -=-  about  the  axis  is  -  m  y  .  —  in 

dt  dt 

the  direction  of  increasing  0,  and  the  moment  of  momentum  of 

m  jf  is  mx  .  Jf,  and  the  sum  of  these  is 
at  at 


and  the  sum  of  all  such  terms  is  i-^,  or  i  a.     The  rate  of  change  of 
this  is  i  -,  or  10.     This  also  may  be  obtained  by  differentiating 


588  APPLIED   MECHANICS. 

(1),  and  so  finding  -^  and  3.     Taking  the  moments  of  the 

virtual  forces  -  m-^  and  -  #*-vf  about  the  axis,  we  arrive  at 

cPQ 
-  I-TTJ,  or   -  io,  which,  together  with  the  moments  due  to  the 

outside  forces,  produce  equilibrium;  or,  with  signs  changed,  are 
equal  to  them,  as  above.  Students  ought  to  think  of  other  ways 
of  reaching  these  results  concerning  rigid  bodies. 

485.  Kinetic  Energy  of  any  System.  —  Let  mbejthe  mass  of  a  por- 
tion of  the  system  in  the  position  x,  y,  z.  Let  x,  y,  z  at  this  instant 
be  the  position  of  the  centre  of  gravity.  The  whole  kinetic  energy 
is  the  sum  of  such  terms  as 


Let  x  —  x  +  xlt  y  =  ~y  +  yl,  z  =  z  +  z1  ;  so  that  x\  yl,  z1  show 
the  position  of  m  relatively  to  the  centre  of  gravity.    Note  that 

Sdx\*      (dx       dxl\*      Sdx\z         dx    dx1       /dxl\t 

(dt)  =  (di  +  Tt  ]  I  =  (dt)  +  2  dt  '  dt  +  (dt)  ' 


dt  '  dt  -    "dt~  "dT>          "dt  X1 
the  rate  of  change  of  2  m  x1  and  2  m  x1  =  0.     Hence 

/dxl\* 

2mUJ- 

The  first  of  these  terms  is  the  whole  mass  multiplied  by  the  square 
of  the  x  component  of  the  velocity  of  the  centre  of  gravity.  It 
follows,  therefore,  that  the  kinetic  energy  of  any  system  is 
equal  to  the  kinetic  energy  which  the  system  would  have  if  it  were 
all  moving  with  the  velocity  of  the  centre  of  gravity  +  the  kinetic 
energy  due  to  the  motions  of  all  the  parts  relatively  to  the  centre 
of  gravity. 

Now,  the  motion  of  a  rigid  body  relatively  to  any  point  of  it 
can  only  be  a  rotation.  Hence  for  rigid  bodies  we  have  the  rule  : 
If  the  velocity  of  the  centre  of  gravity  is  v,  and  if  there  is  a 
rotation  of  angular  velocity  9  about  some  axis,  and  if  i0  is  the 
moment  of  inertia  of  the  body  about  a  parallel  axis  through  the 
centre  of  gravity,  and  M  is  the  whole  mass,  the  kinetic  energy  E  is 

£  M  v2  +  1 10  (e)2.     If  io  =  M  A-2,  then  E  =  i  M  |  v2  +  k*  (0>  j  .    It 

is  only  in  the  case  of  a  rigid  body  (in  all  this  by  rigid  body  I 
mean' an  infinitely  rigid  body)  that  we  have  this  simple  rule.  In 
the  case  also  of  a  rigid  body  we  can  find  law  (Art.  482)  from  the 
law  of  energy,  as  we  do  in  Art.  495  ;  but  we  cannot  do  this  so 
easily  for  a  system  having  internal  relative  motion,  because  there 
is  internal  potential  energy,  which  may  alter.  The  virial  and 
other  laws,  which  may  easily  be  arrived  at,  do  not  concern 
engineering  applications  of  mechanics. 


APPLIED    MECHANICS. 


589 


486.  In  any  ordinary  elastic  body  the  internal  motions  are  vibra- 
tional.     The  momentum  and  moment  of  momentum  of  the  system 
are  therefore  practically  the  same  as  if  the  body  were  quite  rigid. 
This  is  not  the  same  in  regard  to  the  energy.     "We  say  that  it 
disappears  or  is  changed  to  heat  and  other  forms ;    this  means 
merely  that  it  has  become  molecular,  and  equations  which  regard 
the  body  as  if  it  were  rigid  are  quite  inapplicable.     Thus  it  is  that 
the  momentum  equations  are  what  we  rely  upon,  because  when  we 
study  the  motions  of  bodies  momentum  cannot  be  hidden  as  energy 
may  be. 

487.  When  we  work  exercises  on  rigid  dynamics  we  ought  to 
apply  Newton's  law  always  in  the  shape  given  in  Art.  480.     When 
we  work  a  new  exercise  let  it  be  regarded  as  a  new  illustration,  to  be 
worked  to  making  our  knowledge  of  the  fundamental  principle 
clearer.     Most  details  of  one's  theory  and  artifices  for  the  solution  of 
problems  will  disappear  from  view  in  one's  professional  work.     Let 
the  fundamental  notion  be  so  well  fixed  that  it  cannot  disappear. 

488.  Exercise. — In  the  compound  pendulum  of  Fig.  346  find  the  force  act- 
ing at  s,  the  point  of  suspension.    In  Fig.  347  let  o  be  the  centre  of  gravity. 


Fig.  346. 


Pig.  347. 


Let  x  and  Y,  as  shown,  be  the  horizontal  and  vertical  components  of  the 
force  at  s.  Let  G  s  y  be  0,  and  let  the  body  be  moving  so  that  0  increases. 
If  s  o  be  called  r,  the  velocity  of  a  in  its  path  is  rQ.  It  has  an  accelera- 
tion rfl  in  its  path,  and  r(0)2,  a  centripetal  acceleration  in  the  direction 
G  s.  I.  Regard  now  the  whole  system  of  forces,  which  are  supposed  to 
balance  if  they  were  to  act  at  one  point,  G.  Let  M  be  the  mass  or  w/^, 
we  have  -  M  r9  in  the  direction  G  A,  -  M  r(0)2  centripetal  in  the  direction 
o  s,  w  downwards  (the  weight  of  the  body),  x  horizontally,  and  Y  down- 
wards.  Resolve  these  in  any  direction  whatsoever,  they  are  to  balance. 
Thus,  resolve  them  horizontally  and  vertically, 

x  -  M/-0  cos.0  +  Mr(0)2  sin.0  =  0  ....  (1), 

y  +  w  -f  M  rQ  sin.0  +  M  r(0)2  cos.0  =  0  ....  (2). 

Observe  that  the  forces  in  G  A  and  G  s  are  virtual  forces,  or  forces 
equal  and  opposite  to  mass  multiplied  by  acceleration,  II.  Now  take 
moments  about  any  axis.  The  most  convenient  seems  to  be  s.  w  is  the 
only  externally  applied  force  that  has  any  moment  about  8,  and  so 

W  r  sin.0  4-10  =  0....  (3), 


590 


APPLIED    MECHANICS. 


if  i  is  the  moment  of  inertia  about  s.     From  these  equations  x  and  Y,  in 
terms  of  9  and  0,  may  be  calculated.     We  first  express  9  in  terms  of  6 

from  (3).     If  i  =  ^  £2,  (3)  gives 


=       r  cos.9 


Y  = 


-  w  -  -  r  sin.0  (  -   p  sin.0)  -    -  r  (0)2  cos.0, 


will  depend  on  the  limit  of  the  swing,  and  may  have  any  value. 

A 


489.  Vibration  Indicator. — Fig.  348  shows  an  instrument  which 
has  been  used  for  indicating  quick  vertical  vibration  of  the  ground. 
The  mass  c  P  <j  is  supported  at  p  by  a  knife-edge,  or  by  friction- 
wheels.  The  centre  of  gravity  o  is  in  a  horizontal  line  with  p  and 
Q.  Let  P  o  =  0,  GQ  =  £,  po,  =  a  +  b  =  l.  The  vertical  spring 
A  R  and  thread  R  o,  support  the  body  at  Q.  As  a  matter  of  fact,  A  R 
is  an  Ayrton -Perry  spring,  which  shows  by  the  rotation  of  the 
pointer  R  the  relative  motion  of  A  and  Q.  Let  us  neglect  its 
inertia  now,  and  consider  that  the  pointer  faithfully  records 
relative  motion  of  A  and  Q.  It  would  shorten  the  work  to  only 
consider  the  forces  at  p  and  Q  in  excess  of  what  they  are  when  in 
equilibrium ;  but,  for  clearness,  we  shall  take  the  total  forces. 

When  a  body  gets  motion  in  any  direction  parallel  to  the  plane 
of  the  paper,  we  get  one  equation  by  stating  that  the  resultant 
force  is  equal  (numerically)  to  the  mass  multiplied  by  the  linear 
acceleration  of  the  centre  of  gravity  in  the  direction  of  the 
resultant  force.  We  get  another  equation  by  stating  that  the 
resultant  moment  of  force  about  an  axis  at  right  angles  to  the 
paper  through  the  centre  of  gravity  is  equal  to  the  angular 


APPLIED    MECHANICS.  591 

acceleration,  multiplied  by  moment  of  inertia  about  this  axis 
through  the  centre  of  gravity.     I  shall  use  x,  x,  and  x  to  mean 

displacement,  velocity,  and  acceleration,  or  x,  ~y  and  ~. 

Let  P  and  A  have  a  displacement  xl  downward.  Let  Q  be 
displaced  x  downward.  Let  the  pull  in  the  spring  be  a  =  Q0  +  c 
(x  -  Xi)  where  c  is  a,  known  constant  (c  is  the  reciprocal  of  the  h 
used  in  Art.  514).  Let  w  be  the  weight  of  the  body.  Then  if  PO 
and  QO  be  the  upward  forces  at  the  Doints  marked  P  and  a,  in  the 
position  «f  equilibrium, 

00  (a  +  b)  =  w  a  and  PO  +  %  =  w. 
Hence 


Q  =  Q0  +  C  (X  -  Xi). 

Now  o  is  displaced  downwards  -  ^  xl  +  -  1  x,  so  that 
o>  +  o  a,  -f-  o 


The  body  has  an  angular  displacement  0  clockwise  about  its  centre 
of  mass,  of  th 
inertia  about  o, 


of  mass,  of  the  amount  -  ~.     So  that  if  i  is  its  moment  of 

a.  +  o 


Hence  (2)  and  (3)  give  us,  if  M  stands  for  —  ,  and  if  i  =  M/fc2  where 

a 


is  the  radius  of  gyration  about  o, 


If  k±   is  the   radius  of  gyration  about  P,  we  find  that   (4) 
simplifies  to 

*  +  n*x  =  ^xi  +  »2*,  ----  (6) 

if  n  stands  f  or  j-     /  '—  =  2  »  x  natural  frequency  and  &  stands 
KI  v     M 

al 
for  1  -  |-g.     Call  x  -  xi  by  the  letter  y,  because  it  is  really  y  that 


592  APPLIED   MECHANICS 

an  observer  will  note,  if  the  framework  and  room  and  observer 
have  the  motion  x\.  Then,  as  y  =  x  -  x\>  or  x  =  y  +  x\, 

y  +  Xi  +  n2  (y  +  x.)  =  e*xi  +  ri*x\. 
So  that 

if  +  n-y  =  (*2  -  1)  x,  ....  (6), 
or 

y  +  n*y  +  —  *,  =0 (7). 

Thus  let  xv  =  A  sin.  qt. 

We  are  neglecting  friction  lor  ease  in  understanding  our 
results,  and  yet  we  are  assuming  that  there  is  enough  friction  to 
destroy  the  natural  vibration  of  the  body.  We  find  that  if  we 
assume  y  —  a  sin.  qt,  then 

al       a* 
a  =  j-*  -~ — 5  A. 

k?  n*  -  q* 

That  is,  the  apparent  motion  y  (and  this  is  what  the  pointer  of  an 
Ayrton-Perry  spring  will  show ;  or  a  light  mirror  may  be  used  to 

throw  a  spot  of  light  upon  a  screen)  is  ^  -^~ — %  times  the  actual 

motion  of  the  framework  and  room  and  observer.  If  q  is  large 
compared  with  n — for  example,  if  q  is  always  more  than  five  times 

n — we  may  take  it  that  the  apparent  motion  is  ^-2  times  the  reaJ 

motion,  and  is  independent  of  frequency.  Hence  any  periodic 
motion  whatever  (whose  periodic  time  is  less  than  £th  of  the 
periodic  time  of  the  apparatus)  will  be  faithfully  indicated. 

Note  that  if  al  =  kf,  so  that  o.  is  what  is  called  the  point  of 
percussion,  Q  is  a  motionless  or  "  steady  "  point.  But  in  practice 
the  instrument  is  very  much  like  what  is  shown  in  the  figure,  and 
Q  is  by  no  means  a  steady  point.  Apparatus  of  the  same  kind  may 
be  used  for  east  and  west,  and  also  for  north  and  south  motions. 

490.  We  did  not  think  it  necessary  to  interrupt  our  account  of 
simple  harmonic  motion  to  speak  about  the  analogies  between 
linear  and  angular  motion,  although  we  began  to  use  them  in  Art. 
453.  We  see  (Art.  482)  that  any  motion  whatsoever  of  a  rigid  body 
may  be  most  simply  studied  as  a  motion  of  translation  and  a 
motion  of  rotation,  and  we  shall  find  it  best  to  take  the  translational 
motion  of  the  whole  body  as  the  actual  motion  of  its  centre  of 
gravity,  and  the  relation  as  one  about  an  axis  passing  through  the 
centre  of  gravity.  When  we  do  this,  we  shall  find  the  following 
angular  formulse  valuable,  not  merely  for  the  motion  of  a  body 
whose  axis  is  fixed,  but  for  any  motion  whatsoever.  The  only 
point  in  which  the  analogy  between  linear  and  angular  motion 
fails  is  this :  The  mass  or  inertia  of  a  body  is  independent  of  the 
direction  of  motion ;  the  moment  of  inertia  of  a  body  is  usually 
different  about  different  axes  of  rotation. 


APPLIED   MECHANICS. 


593 


COMPARISON   OF  LINEAR  AND  ANGULAR  MOTIONS. 

0  =  angular  space. 


s  =  space,  t  —  time. 
v  =  velocity  =    -,  or  s. 


a  =  angular  velocity  =  ^r,  or  6. 


a  =  acceleration  =  -^5,  or  V,  or  v.         a  =  angular  acceleration  =  -^A,  01 


m  =  mass  or  inertia  =  w\g. 

F  =  force. 

mv  =  momentum. 


If  a  body  has  two  displacements 
(or  velocities,  or  accelerations,  or 
momenta)  given  to  it  simultane- 
ously, represented  in  magnitude 
and  direction  and  sense  by  o  p  and 
OQ,  they  are  equivalent  to  the 
displacement  (or  velocity,  or  ac- 
celeration, or  momentum)  o  K  if  o  p 
and  o  a  are  the  sides,  and  o  R  the 
diagonal  of  a  parallelogram,  the 
arrows  being  all  difluent  or  con- 
fluent. 


Forces  represented  by  OP  and 
o  Q  have  a  resultant  o  R. 


Impulse  =  change  of  momentum 


change  of  mo- 


Average  force 
mentum  -7-  time. 


Work  done  =  force  x  distance. 

Space  average  of  force  =  work 
done  -~  distance. 

Kinetic  energy,  £  mv2. 

Under  uniform  acceleration 
from  rest  at  time  0,  v  =  at,  t  = 


0,  or  a. 

i  =  moment  of  inertia. 
M  =  moment  of  forces  or  torque. 
la  =  angular  momentum,   usually 

called  moment  of  momentum. 

If  a  body  has  two  angular  dis- 
placements (or  velocities,  or  accele- 
rations, or  momenta  [moments  of 
momentum])  about  the  axes  o  p  and 
o  Q,  and  if  the  lengths  of  the  lines 
o  P  and  o  Q  represent  these  to  scale, 
and  if  the  arrows  indicate  positive 
direction,  as  that  in  which  a  right- 
handed  screw  moves,  they  are 
equivalent  to  the  angular  displace- 
ment (or  velocity,  or  acceleration, 
or  moment  of  momentum)  repre- 
sented in  axis,  amount  and  sense, 
by  OR. 

Moments  of  force  or  torque 
represented  by  OP  and  OQ  (the 
direction  of  the  line  being  the  axis 
about  which  the  moments  of  forces 
are  taken,  and  the  sense  of  the 
arrow-heads,  as  in  the  last  case) 
have  a  resultant  o  R 

a  =  M  -5-  I. 

Angular    impulse  =  change    of 


moment  of  momentum  = 


Average    torque  =  change 
angular  momentum  -r-  time. 

Work  done  =  torque  x  angle. 

Angular  average    of  torque 
work  done  ^  angle. 

Kinetic  energy,  £  i  «2. 

a  =z  at,  6  =  i  of2,  a2  =  2  a*. 


of 


594 


APPLIED    MECHANICS. 


Simple  harmonic  motion  if  T  is  the  periodic  time,  A  is  the  amplitude, 
e  is  the  lead 


=  A  sin. 


2-jr  /2ir  \ 

9  or  a  =  A  —  cos.  (  —  t  +  eY 


*  or  0 


A  body  moves  backwards  and 
forwards  tinder  the  action  of  a 
variable  force,  which  is  always 
proportional/ to  the  distance  of  the 
body  from  its  middle  position,  and 
which  always  acts  towards  this 
position;  and  if  the  force  at  a 
distance  of  1  foot  is  5  Ibs.,  then  the 
time  of  vibration  is  2  TT  times  the 
square  root  of  the  quotient  of  the 
mass  of  the  body  divided  by  5. 


If  m  is  the  mass ;  if  the  force  of 
friction  is  b  times  the  velocity;  if 
the  constraining  force  is  n  times 
distance  from  centre, 


If  a  body  vibrates  about  a  fixed 
axis  under  the  action  of  the  torque 
(say  from  a  spiral  spring  or  twisted 
wire),  so  that  it  is  always  propor- 
tional to  9,  the  angular  displace- 
ment of  the  body  from  its  mean 
position,  and  if  the  torque  i«t  5 
pound-feet  when  the  body  is  1 
radian  from  the  mean  position, 
then  the  time  of  a  vibration  is  2* 
times  the  square  root  of  the  quotient 
of  the  moment  of  inertia  divided 
by  5. 

If  i  is  moment  of  inertia,  fric- 
tional  torque  is  b  times  the  angular 
velocity.  If  the  constraining  torque 
is  n  times  the  angular  distance 
from  the  mean  position, 


*  or  6  =  A  «-•#  sin.  (a*  +  «)' 


where/  =  £  b/m,  or  £  J/i,  and 


A  body  of  mass  m  moves  with 
constant  speed  v.  At  any  instant 
its  momentum  is  in  the  direction 
o  K,  and  is  represented  to  scale  by 
the  amount  OB.  If  a  centripetal 
force  F  of  constant  amount  is  always 
acting  at  right  angles  to  the  direc- 
tion o  E,  the  effect  is  to  make  the 
point  E  travel  with  linear  velocity 
equal  to  p.  That  is,  the  direction 
of  motion  alters  with  an  angular 
velocity  f/mv. 


A  body  whose  moment  of  mo- 
mentum is  represented  to  scale, 
and  the  direction  of  its  axis  by  o  E, 
is  acted  on  by  a  constant  torque  M, 
which  is  always  about  an  axis  at 
right  angles  to  o  E.  The  effect  is  to 
make  the  point  E  travel  with  linear 
velocity  equal  to  M  ;  that  is,  the 
axis  OE  has  an  angular  velocity 
(towards  the  axis  of  the  torque) 
whose  amount  is  M  -5-  moment  of 
momentum  10. 


APPLIED    MECHANICS. 


595 


EXERCISES. 

1.  A  top  (Fig.  351)  rotates  at  r  radians  per  second  about  its  axis  o  c. 
The  axis  rotates  (or  precesses)  at  n  radians  per  second  in  a  conical  path 
round  the  vertical  o  z.  Represent  this  motion  by  one  cone  rolling  on 
another. 


ForM 
O 
Big.  360. 

Answer, — If  o  c  is  the  axis  of  the  rolling  cone  of  angular  velocity  r 
of  which  o  i  is  the  slant  side,  if  o  z  is  the  axis  of  a  fixed  cone  of  which  o  i 
is  the  slant  side,  at  any  instant  o  i  is  a  line  in  the  rolling  cone,  which  is  at 
rest,  and  about  which  it  rotates  with  angular  velocity  u.  Let  z  o  c  =  a, 
i  o  c  =  j8.  Draw  j  K  and  j  L  at  right  angles  to  o  z  and  o  i.  We  see 
that  j  would  go  into  the  paper  with  the  velocity  n  .  K  j,  because  j  is 
moving  in  a  horizontal  circle  whose  fixed  centre  is  K  ;  also  J  goes  into 
the  paper  with  the  velocity  o> .  L  j,  because  L  is  fixed ;  and  these  are  equal ; 
so  that  n  .  K  j  =  co  .  L  j,  or,  since  K  j  =  o  j  .  sin.  a,  L  j  =  o  J  -  sin.  /3, 

n  sin.  a  =  co  sin.  ft  .  .  .  .  (1). 
Again,  co  about  o  i  is  the  total  spin  of  the  top.     It  has  components 

co  cos.  )8  =  r  about  the  axis  o  c  }  .„» 

u>  sin.  £  or  r  tan.  £  about  the  axis  o  A  J  *  "  *  *  *  '* 

if  o  A  is  at  right  angles  to  o  c ;  and  hence  (1)  may  be  written 
fl  sin.  a  =  r  tan.  0  .  .  .  .  (1), 

jo  that  j8  is  now  known.     All  the  lines  drawn  are  in  the  plane  of  the 
paper  at  tb^  instant. 

2.  If  the  moment  of  inertia  of  the  top  about  a  c  is  c  and  about  o  A  is  A, 
find  the  resultant  moment  of  momentum. 

Answer. — c  .  r  is  the  moment  of  momentum  about  o  c ;  A  .  r  tan.  /3  is 
the  moment  of  momentum  about  o  A.  The  resultant  has  an  axis  o  H,  its 
amount  being  c  r  .  sec.  7.  If  7  is  H  o  c, 

tan.  7  =  A  r  tan.  /3  -5-  o  r  =  A  tan.  j8  -j-  c  .  .  .  .  (3). 

3.  If  the  distance  o  H  represents  the  moment  of  momentum  of  a  top 
about  the  axis  o  H,  and  a  torque  w  h  sin.  a  about  an  axis  at  right  angles 
to  the  paper  tends  to  send  o  H  away  from  o  z,  the  effect  is  to  make  the 
horizontal  velocity  of  the  point  H  at  right  angles  to  the  paper  be  w  A  sin.  a. 
What  is  the  angular  velocity  of  o  H  about  o  z  ? 


596  APPLIED   MECHANICS. 

Answer. — The  point  H  moves  in  a  circle  of  radius  o  H  .  sin.  z  o  H,  and 
the  linear  velocity  of  H,  or  w  h  sin.  a  divided  by  radius  gives  angular 
velocity,  or 

w  h  sin.  a  -r  o  H  sin.  z  o  H, 

_  _         w  h  sin.  a  w  h  .  sin,  a 

o  H  .  sin  (0  —  7)  or.  sec.  7  .  sin.  (a  -  yY 

Hence  Her  (sin.  a  -  cos.  a  .  tan.  7)  =  w  h  sin.  a. 

But  tan.  7  we  found  to  be  A  tan.  /8  •*•  c,  and  hence 

XI  c  r  (sin.  a  -  cos.  a  .  -  tan.  0\  =  w  h  sin  a. 

But  (1)  gives  us  tan.  /3  =  fl  sin.  a/r,  and  hence 
tier  -  fiz  .  cos.  a  .  A  =  w  A,     Q?  -  tl  .  —  sec.  a  .  +  —  sec.  a  =  0 

A  A 

From  this  we  may  calculate  fl,  the  rate  of  precession  of  the  top.  The 
student  will  find  a  value  of  r  for  which  any  processional  motion  is 
impossible,  and  the  top  must  fall.  He  may  also  consider  the  meaning  of 
two  different  real  values  for  fl.  We  have  given  this  exercise  because  it 
gives  a  good  drilling  in  the  use  of  the  above  rules.  Note  that  a  real  top 
has  not  an  infinitely  sharp  peg.  It  rolls  and  slides  on  the  table,  and  we 
shall  not  consider  this  much  more  complicated  problem. 


597 


CHAPTER   XXVII. 

CENTRIFUGAL     FORCE. 

491.  Centrifugal  Force. — If  a  body  is  compelled  to  move 
in  a  curved  path,  it  exerts  a  force  directed  outwards  from  the 
centre,  and  its  amount  in  pounds  is  found  by  multiplying  the 
mass  of  the  body  by  the  square  of  the  velocity  in  feet  per 
second,  and  dividing  by  the  radius  of  the  curved  path.     We 
evidently  get  the  same  answer  if  we  multiply  the  mass  by  the 
radius  and  by  the  square  of  the  angular  velocity  in  radians 
per  second.     Thus,  a  weight  placed  at  the  end  of  an  arm  like 
the  arm  of  a  wheel  exerts  a  pull  in  the  arm.     If  a  body  moves 
round  an  axis  20  times  per  minute  in  a  circle  whose  radius  is 
3  feet,  you  can  determine  the  centrifugal  force  by  first  finding 
the  velocity  of  the  body  and  using  the  above  rule,  or  you  may 
proceed  as  follows : — The  weight  of  the  body  multiplied  by  3 
multiplied  by  the  square  of  20  divided  by  2,937  is  the  centri- 
fugal force. 

Suppose  a  wheel,  whose  total  weight  is  20  tons,  or  44,800 
Ibs.,  has  its  centre  of  gravity  0'4  foot  away  from  the  axis — 
that  is,  suppose  the  wheel  to  be  eccentric — then  if  the  wheel 
makes  50  revolutions  per  minute,  the  centrifugal  force  is 
44,800  x  0-4  x  2,500  -r-  2,937,  or  15,253  Ibs.— that  is,  6-81 
tons.  This  force  acts  on  the  bearings  of  the  shaft,  always  in 
the  direction  of  the  centre  of  gravity  of  the  wheel. 

492.  Anyone  who  wants  to  get  clear  ideas  about  centri- 
fugal force  ought  to  make  experiments  of  his  own.     Unfor- 
tunately, although  there  are  many  toys  made  to  illustrate  the 
effects  of   centrifugal   force,  we  know  of   only  one   piece  of 
apparatus  which  enables  the  laws  to  be  systematically  experi- 
mented upon.     Fig.  352  is  a  drawing  of  it.     A  represents  a 
little,  flat,  cast-iron  box,  like  a  narrow  drum ;   one  drum-head, 
as  it  were,  being  replaced  by  a  thin  steel  plate,  so  as  to  be 
strong  and  flexible.     B  is  a  glass  tube  which  enters  the  box. 
Mercury  fills  the  box  and  the  tube  to  the  level  6,  and  when  c, 
the  centre  of  the  steel  plate,  is  pulled  or  pushed,  although  wo 
cannot  see  much  yielding  in  c,  we  can  observe  the  mercury 
rise  and  fall  in  the  tube.      There  is  a  screw,  D,  entering  the 
box  at  the  back  ;   by  means  of  this  screw  we  can  adjust  the 


APPLIED   MECHANICS.  599 

height  of  the  mercury  in  the  tube.  The  box  is  in  the  centre 
of  a  circular  table,  E,  which  can  be  whirled  round  at  any  speed 
we  please,  and  the  tube  is  exactly  in  the  axis  of  rotation,  so 
that  the  height  of  the  mercury  can  be  measured  whatever  be 
the  speed.  Fastened  to  the  centre  of  the  corrugated  plate  at 
C  is  a  long  brass  rod,  p,  which  is  supported  at  J  on  the  end  of 
a  little  rocker,  so  that  it  can  move  backward  and  forward 
with  less  friction  than  if  it  were  made  to  slide  on  a  bearing. 
At  any  place  along  F  we  can  clamp  a  weight,  H,  which  we  may 
alter  as  we  please  from  0*5  to  8  Ibs.  We  can  clamp  it  near 
the  axis  or  one  foot  away,  the  radius  of  the  circle  described  by 
its  centre  of  gravity  being  measured  by  the  scale  in  tenths 
and  hundredths  of  feet  marked  on  the  rod.  The  experimenter, 
turning  the  handle  which  drives  the  pulley  g,  keeps  his  eye 
upon  the  mercury  level,  and  is  able  to  maintain  a  very  great 
constancy  of  speed.  He  counts  the  revolutions  of  his  hand, 
and  one  of  his  companions  takes  the  time,  so  that  no  speed- 
counter  or  indicator  is  necessary.  Now,  the  centrifugal  force 
due  to  the  rod  and  sliding  weight  causes  c  to  be  pulled  out 
very  slightly,  and  this  causes  the  mercury  to  fall  in  the  tube ; 
and  it  is  easy  with  a  fixed  vertical  scale  alongside  the  tube  to 
measure  this  rise  and  fall.  We  usually  get  a  spring  balance, 
or  a  cord,  pulley,  and  weights,  and  before  our  experiments 
begin  we  pull  the  end  of  the  rod  p,  noting  the  height  of  the 
mercury  for  a  pull  of  1  lb.,  2  Ibs.,  etc.,  and  in  this  way  we  can 
afterwards  tell  the  value  of  our  scale  measurements.  We  also 
make  a  number  of  experiments  when  the  sliding  weight  is 
removed  from  the  rod  F,  to  tell  us  the  centrifugal  force  of 
everything  else  at  different  speeds,  and  this  we  subtract  from 
our  subsequent  observations.  We  see,  then,  that  we  cans 
measure  the  centrifugal  force,  in  pounds,  of  various  masses, 
from  0*5  to  8  Ibs.,  moving  at  any  speed  in  a  circle  whose 
radius  may  be  as  little  as  1  *5  inch  and  as  much  as  1 1  '5  inches. 
With  this  instrument  it  is  easy  to  test*  the  law  which  is 
usually  given,  and  without  working  with  some  such  instru- 
ment we  question  if  students  are  likely  to  have  any  but  vague 
notions  about  centrifugal  force. 

493.  There  is  another  method  of  experimenting  which 
suggests  itself,  with  apparatus  which  anyone  may  fix  up  for 
himself,  but  it  does  not  give  such  a  thorough  understanding 
of  the  law  to  the  person  who  experiments.  In  Fig.  353,  A  is 
ft  leaden  ball  at  the  end  of  a  silk  thread,  p  A,  fastened  at  P. 


600  APPLIED    MECHANIC^. 

A  is  kept  out  from  its  natural  position  in  the  vertical  by 
means  of  a  horizontal  thread  in  the  direction  A  B.  Now,  if 
we  pass  the  horizontal  thread  AB  over  a  pulley  and  hang  a 
weight  at  its  end,  we  find  that  the  force 
acting  in  A  B  is  to  the  weight  of  A  as  the 
distance  K  A  is  to  the  distance  KP.  The 
body  A  is  acted  upon  by  three  forces  :  its 
weight  downwards  in  the  direction  A  a,  the 
horizontal  force  A  B,  and  the  pull  in  the  string 
A  P.  The  triangle  of  forces  tells  us  that  as 
A  K  P  is  a  triangle  whose  sides  A  K  and  K  p 
and  P  A  are  parallel  to  the  three  forces,  then 
the  horizontal  force,  acting  in  A  B,  is  to  the 
vertical  force  which  is  the  weight  of  A  as 
the  distance  K  A  is  to  the  distance  p  K. 
Suppose,  for  example,  the  weight  of  A  to 
be  4  Ibs.,  the  height  p  K  to  be  8  feet,  and 
the  distance  A  K  1  foot,  then  A  K  is  one-eighth  of  K  p,  and  we 
are  sure  that  the  horizontal  force  needed  just  to  keep  A  in 
this  position  is  0'5  lb.,  for  it  must  be  one  eighth  of  the 
weight  of  A. 

Now,  let  such  a  ball  as  A,  hung  by  a  thread  p  A,  go  round 
and  round  in  a  circle.  Measure,  as  accurately  as  you  can,  K  A, 
the  radius  of  the  circle,  and  K  P,  the  vertical  height  from  tho 
ball  to  the  point  of  suspension.  Also  count  how  many  revo- 
lutions the  ball  makes  per  minute.  The  centrifugal  force  is 
now  doing  what  the  horizontal  string  did  before,  and  we 
know  how  much  it  is.  In  fact,  the  centrifugal  force  is 
obtained  by  multiplying  the  weight  of  the  ball  by  KA,  the 
radius  of  its  circle,  and  dividing  by  the  vertical  height  K  p. 
You  can  test  if  the  centrifugal  force  law  is  true,  therefore,  by 
means  of  your  measurements. 

Conversely,  if  the  weight  of  the  ball  is  w,  if  A  K  is  r,  if  p  A 

W  v*2 
is  Z,  and  PK  is  ht  the  centrifugal  force  is  -       ,  and  we  see  by 

T 

the  triangle  of  forces  that  this  must  be  equal  to  j  w.     Now,  if 
the  time  of  one  revolution  is  T,  then  v  =  2  TT  r  -h  T,  and  hence 

^  =  ^  becomes  T  =  2*-  A  /T.  ...  (1) 
y  r          h  V     g 

A   ball    going    round   in    this    way   is    called    a    conical 


APPLIED    MECHANICS.  601 

pendulum,  and  we  have  in  (1)  found  its  periodic  time.  Its 
periodic  time  is  that  of  a  simple  pendulum  (Art.  446)  whose 
length  is  the  same  as  the  h  of  the  conical  pendulum. 

EXERCISES. 

494.  1.  A  pendulum  bob  weighing  20  Ibs.  moves  with  the  velocity  of  5 
feet  per  second  at  the  middle  of  its  path.  What  force  is  tending  to  make  the 
pendulum  longer,  in  addition  to  the  weight  of  the  bob  ?  The  length  of  the 

9ft  9*i 

pendulum  is  15  feet.  Ans.,    ™    x  _  =  1-035  Ibs. 

32'2        15 

2.  A  locomotive  weighing  35  tons  travels  at  50  miles  an  hour  round  a 
curve  of  2,000  feet  radius.     What  is  the  centrifugal  force? 

Ans.,  2-92  tons. 

3.  The  centre  of  a  ball  of  10  Ibs.  is  at  4  feet  from  an  axis,  and  revolves 
at   500  revolutions  per  minute.     What  is   its  centrifugal  force  ?    Its 
centrifugal  force  is  to  be  balanced  by  those  of  two  balls  attached  to  the 
sh  aft,  each  of  20  Ibs.,  revolving  in  circles  whose  planes  are  4  feet  and 
1  foot  from  the  .plane  of  the  first  circle.    At  what  distance  is  each  of 
them  from  the  axis  ?  Ans.,  3,400  ;  0-4  foot ;  2  feet  nearly. 

4.  The  rim  of  a  pulley  has  a  mean  radius  of  20  inches  ;  its  section  is 
6  inches  broad,  and  average  thickness  £  inch.     If  it  revolves  at  200 
revolutions  per  minute,  what  is  the  centrifugal  force  per  inch  length  of  rim  ? 

Ans.,  18-6  Ibs. 

5.  A  vehicle  describes  a  horizontal  circle  of  600  feet  radius  with  a 
velocity  of  40  miles  per  hour ;  find  the  direction  of  the  resultant  due  to 
gravity  and  centrifugal  force.     If  it  is  a  railway  of  gauge  1  metre,  how 
much  higher  ought  the  outer  rail  to  be  than  the  inner  one  ? 

Ans.,  10°-1  with  vertical;  6-9  inches. 

6.  A  skater  describes  a  circle  of  100  feet  radius  with  a  velocity  of 
18  feet  per  second  ;  what  is  his  inclination  to  the  ice?  Ans.,  84°. 

495.  7.  The  law  of  energy  states  that  a  body  of  weight   w   and 
mass  M  falling  from  rest  without  friction  along  any  path,  at  any  instant 
being  at  the  height  h  and  having  the  velocity  v,  has  the  sum  of  its 
potential  and  kinetic  energies  constant,  or 

w  h  +  TJ  M  v2  =  constant  ....  (1). 

Show  from  this  that  force  is  equal  to  mass  multiplied  by  acceleration. 
If  *  is  distance  along  the  path  measured  in  the  direction  of  motion, 

differentiating,  we  have  w  —  .  +  M  v  .  ~  =  0.      But    -  dhlds  is  sin.  a 

as  as 

if  a  is  the  inclination  of  the  path  to  the  horizontal,  and    v  =  ds/dtt 

dv 
and  hence  w  sin.  a  =  uv  -^  .  .  .  .  (2), 

or.since  ,*  =  „**  =  * 

ds          dt     ds       dt 

dv 
w  sin.  a  =  M  j~t  .  .  .  .  (3). 

We  see,  therefore,  that  the  resultant  force  (w  sin.  o)  acting  on  the  body 


602  APPLIED    MECHANICS. 

is  equal  to  the  rate  of  change  per  second  of  the  momentum.    If  a  is  90°, 
so  that  the  body  falls  vertically,  then  -|  is  called  ff,  and  (3)  is  merely 
M  =  w/ff  .  .  .  .  (4). 

Students  ought  to  familiarise  themselves  with  this  and  other  ways  of  show- 
ing that  the  law  of  energy  leads  to  the  law  :  force  =  mass  x  acceleration. 

496.  We  think  that  Newton's  method,  the  Thomson  and  Tait  method, 
is  very  much  to  be  preferred  to  any  other  method  of  starting  in  the  study 
of  mechanics.  Every  engineer  ought  to  have  his  T  and  T'  (the  elementary 
treatise),  a  well-thumbed  book.    Nevertheless,  it  is  good  to  look  at  our 
fundamental  notions  from  the  energy  point  of  view.     The  older  proofs 
of  statical  principles  were  based  on  the  idea  of  work,  as  in  our  Chapter 
XI.    Assuming  that  mechanical  energy  is  not  lost  by  conversion  into 
heat  or  other  forms  and  that  a  total  store  remains  constant,  we  get  good 
working  views  of  our  subject.     Underlying  our  notions  are  our  specula- 
tions as  to  how  matter  performs  attractions;  but  leaving  out  ideas  of 
ethereal  stress  and  strain  energy  and  confining  our  attention  to  the  idea 
that  the  sum  of  the  potential  and  kinetic  energies  of  a  system  remains 
constant,  we  have  a  working  idea  of  great  usefulness.     Unfortunately 
many  men  forget  that  mechanical  energy  may  be  converted  into  heat, 
and  so  they  make  such  mistakes  as  to  calculate  the  force  acting  on  a  pile 
being  driven  as  the  space  rate  of  conversion  of  energy. 

497.  Let  us  take  another  example.      If  a  constant  force  F  acts  in  the 
direction  of  motion  of  a  body  for  a  time  5*  through  a  space  8s,  increasing 
Its  velocity  from  v  to  v  +  Sv,  the  work  done  by  the  force  is  F  .  Ss,  and 
this  is  spent  in  increasing  the  kinetic  energy  of  the  body  from  £  mv2  to 
|  m  (v  +  £v)2.     Hence 

F  .  5s  =  tnv  .  Sv  +  -J  vn  ,  (ov)  , 

$v    .    i      ^y  *  n\ 

or     P  =  mv  j-  +  £  m  —  Sv  . .  . .  (1). 

As  we  imagine  is,  and  therefore  $v,  to  be  smaller  and  smaller  without 
limit,  the  last  term  in  (1)  gets  to  be  nearer  and  nearer  0,  and  v  gets  to  be 
better  and  better  represented  by  SsjSt.  Hence 


or  force  is  equal  to  mass  multiplied  by  acceleration,  or  force  is  the  rate  of 
change  of  momentum. 

498.  8.  So  long  as  we  deal  with  force  in  the  direction  of  the  motion  of 
a  body,  there  is  no  difficulty  in  showing  that  the  law  of  work  or  the 
conservation  of  mechanical  energy  leads  to  the  rule,  "  Force  is  rate  of 
change  of  momentum."  When  a  point  is  moving  in  a  curved  path,  we 
can  say  that  the  component  of  the  force  in  the  direction  of  motion  is 
equal  to  the  rate  of  change  of  momentum  in  the  direction  of  motion.  But 
what  of  the  other  component  at  right  angles  to  this  ?  Here  it  is  necessary 
to  observe  that  mere  change  of  direction  of  motion,  and  not  merely  of 
speed,  indicates  that  force  is  acting  on  a  body.  It  is  a  matter  of 
common  observation  that  a  centripetal  force  is  necessary  to  keep  a  body 


APPLIED    MECHANICS. 


603 


moving  in  a  circle,  and  that  the  body  exercises  a  centrifugal  force  against 
those  constraints  which  compel  it  to  move  circularly.  When  a  curved 
surface  gradually  changes  the  direction  of  a  moving  body,  these  two 
forces  act  at  the  point  of  contact.*  To  get  clear  ideas,  consider  the 
conical  pendulum  of  Fig.  354.  The  body  p  revolves  in 
a  circle  at  constant  speed.  To  keep  it  at  the  constant 
distance  a  P  =  r  from  the  centre  a,  we  know  that,  if  it 
were  at  rest,  a  force  i1  must  act.  But  it  is  in  motion, 
and  yet  keeps  out  at  the  distance  r.  There  is  a  centri- 
petal force  whose  amount  is  known  to  us,  w  tan.  o,  and 
it  is  evidently  balanced  by  what  we  can  only  call 
a  centrifugal  force  of  this  amount  which  is  due  to  the 
motion.  Observe  that  we  here  have  a  case  of  centri-  - 
petal  force  acting  on  a  body,  creating  no  increase  of 
kinetic  energy,  and  creating  no  change  of  any  kind  of 
energy,  for  when  P  comes  round  to  the  same  position 
again  everything  is  just  as  before,  although  the  force  has 
been  acting  for  a  whole  round.  Mathematically  there  is  no  great 
difficulty.  We  assume  that  we  have  proved  that  forces  in  directions  at 
right  angles  to  one  another  may  be  studied  independently  and  may  be 
combined  as  vectors.  Consider  circular  motion  in  the  plane  of  the  paper. 
Constant  speed  v  of  P  means  a  velocity  v  cos.  9  in  the  direction  o,  A  and 
-  v  sin.  6  in  the  direction  Q  c.  The  two  accelerations  are  -  v  sin.  9  .  a 


de 


,  the  angular  velocity,  or  -. 


If  m  is  the 


and   -  v  cos.  9  .  a,  a  being 

mass  of  the  body,  we  know  that  there  are  two  forces  acting  upon  the 

body  :  -  mv  sin.  9  .  a  in  the  direction 
Q  A,  -  mv  sin.  9  .  a  in  the  direction 
QC.  Forces  always  compound  accord- 
ing to  the  vector  law,  as  may  be 
proved  from  the  law  of  work  (see 
Art.  495),  and  hence  we  see  that  the 
force  acting  upon  P  in  the  direction 
P  Q  is  a  centripetal  force  mvw  or  m 

— ,  and  the  acceleration  of  P  is  a  cen- 
tripetal acceleration  vz/r.  We  have 
seen  (Art.  493)  another  way  of  arriving 
at  the  amount  of  the  acceleration. 
In  the  very  same  way  it  is  easy 
to  show  that  .when  a  small  body 
Fig.  355.  moves,  not  merely  in  a  circle,  but  in 

*  People  may  have  notions  of  force  that  seem  to  be  quite  different  from 
a  metaphysical  point  of  view,  but  which  are  really  the  same  mathematically. 
Thus  one  man  puts  it  that  there  is  no  such  thing  as  force ;  we  have  only  mass 
X  acceleration.  Another  says — "Yes,  we  have  a  centripetal  force  ;  say,  acting 
on  a  body  which  is  whirled  round  at  the  end  of  a  string,  but  it  is  not  right  to 
speak  of  the  equal  and  opposite  force, sometimes  called  centrifugal  force. "  Now, 
if  we  only  think  of  the  body,  it  may  be  enough  to  speak  only  of  centripetal 
force  and  centripetal  acceleration  ;  but  a  string  in  tension  is  really  acted  on  by 
equal  and  opposite  forces.  If  one  of  these  is  exerted  towards  the  centre,  the 
other  is  exerted  outwards  by  the  body,  and  we  call  it  a  centrifugal  force. 


604  APPLIED    MECHANICS. 

any  curved  path,  the  force  acting  upon  it  has  two  components — one 
in  the  direction  of  the  path,  equal  to  its  mass,  multiplied  by  the  rate 
of  change  of  its  mere  speed ;  the  other  at  right  angles  to  the  path, 
equal  to  its  mass,  multiplied  by  its  linear  velocity,  multiplied  by  its 
angular  velocity. 

499.  It  is  only  when  we  have  looked  at  the  subject  from  the  energy 
point  of  view,  and  in  many  others  in  which  it  will  strike  a  thought- 
ful student  to  make  experiments,  that  the  beauty  of  Newton's 
generalisation  comes  home  to  us  and  we  see  how  all  the  results  of 
observation,  experiment,  and  speculation  are  given  to  us  in  his 
statements.  In  fact,  we  gradually  get  to  know  that,  whether 
acceleration  of  a  small  body  is  along  the  path  or  at  right  angles  to 
it,  force  is  equal  to  the  vector  rate  of  change  of  momentum.  Now, 
this  is  Newton's  definition  of  force,  and  we  may  begin  our  study 
•of  mechanics  from  this  point  of  view.  But  speculation  of  the 
above  nature  is  very  far  from  being  useless. 

Taking  up  our  subject  from  the  easiest  point  of  view,  accelera- 
tion is  the  time  rate  of  change  of  velocity,  and  force  is  the  time 
rate  of  change  of  momentum.  We  first  consider  acceleration  in 
the  direction  of  motion,  and  then,  if  a  body  changes  the  direction 
of  its  velocity,  the  lateral  rate  of  change  of  velocity  or  lateral 
acceleration  must  be  considered.  A  velocity  is  a  vector  quantity, 
and  velocities  are  added  as  all  vectors  are  added.  A  velocity  of 
5  feet  per  second  eastward,  added  to  a  velocity  of  5  feet  per  second 
southward,  are  equal  to  a  velocity  of  7'070  feet  per  second  south- 
eastward. 

600.  When  I  speak  of  the  motion  of  a  body,  I  usually  mean  the 
motion  of  its  centre  of  mass.  If  a  body  is  at  p  moving  along  a 
curved  path  with  the  constant  speed  of  v  feet  per  second,  when  it  is  at  Q 


Fig.  857. 

its  velocity  is  in  a  direction  making,  let  us  say,  an  angle  $6  with 
its  old  direction.  In  Fig.  357  let  o  B  represent  to  scale  the  velocity 
at  P  ;  and  let  o  c,  equal  in  length,  but  making  an  angle  c  o  B  =  50, 
with  o  B,  represent  the  velocity  at  Q.  Now,  in  vector  addition, 
OB  +  BC  =  OC;so  that  if  St  is  the  time  taken  by  the  body  to 
move  from  p  to  Q,  in  that  time  there  has  been  the  lateral  change  of 
velocity  BC.  B  c  zi:  o  B  .  £0  more  and  more  nearly  as  59  is  made 
smaller  and  smaller,  or  B  c  =  v  .  £9.  Hence  the  lateral  accelera- 
tion, which  is  lateral  change  of  velocity  divided  by  the  time  of  the 

The  academic  person  may  be  quite  right  to  stick  to  his  conventions  as  to 
force,  because  he  never  has  to  think  of  the  medium  (string  or  other)  through 
which  forces  are  exerted  when  he  is  working  his  problems.  The  engineer  is 
compelled  to  deal  with  the  larger  question  ;  he  very  wisely  converts  all  his 
dynamical  problems  into  statical  problems,  and  all  his  forces  are  always 
balanced. 


APPLIED    MECHANICS.  605 

«0  S9     ds         _50 

change,   is  v  .  —  or  v  —  .  -  ==  y2  —  =  v*  x  curvature,  because 
8r,  os      ot  os 

our  definition  of  curvature  is  j.     In  a  circle  the  curvature  d9/da 

happens  to  be  the  reciprocal  of  the  radius.  Hence  we  speak 
of  this  acceleration  towards  the  concave  side  of  a  body's  path  as  if 
it  were  moving  in  a  circle  (curiously  enough,  the  fact  seems  always 
to  be  forgotten  that  the  path  is  usually  not  a  circle,  nor  even 
the  very  smallest  arc  of  a  circle)  of  radius  r  with  a  centripetal 
acceleration  v2/r.*  The  centripetal  force  causing  the  change  of 
motion  is  the  mass  multiplied  by  this.  The  engineer  is  usually 
concerned  with  the  equal  and  opposite  force  which  the  body  exerts 
upon  the  constraints,  and  calls  it  centrifugal  force.  We  see  that 
it  is  mvz/r  or  mcPr  or  t0rw2/2937,  where  m  is  the  mass  of  the  body 
or  w  is  its  weight  in  pounds  at  London,  r  the  radius  of  curvature 
of  the  path  in  feet,  v  the  velocity  in  feet  per  second,  a  the  angular 
velocity  in  radians  per  second,  n  the  number  of  revolutions  per 
minute.  Observe  that  if  masses  mi  and  mz  are  attached  to  the 
same  shaft  at  distances  r\  and  r2  from  the  axis,  their  centrifugal 
forces  are  in  the  proportion  of  m^  r\  to  wa  rz ;  if  we  have  half  the 
mass  at  twice  the  distance,  we  have  the  same  centrifugal  force. 

If  a  body  changes  in  its  speed  v,  and  so  has  an  acceleration  in 
the  direction  of  its  path,  its  total  acceleration  is  the  vector  sum  of 
the  two  accelerations,  and  the  resultant  force  acting  on  the  body  at 

any  instant  is  the  resultant  of  m  —  along  the  path  and  m  — 

at  right  angles  to  and  in  the  plane  of  the  path. 

Example. — A  body  of  w  Ibs.  is  moving  along  a  curve  with  a 
velocity  v  and  an  acceleration  a  in  the  direction  of  motion;  the 
radius  of  curvature  of  the  curve  is  r.  What  is  the  total  accelera- 
tion and  the  force  causing  it  ?  Here  we  have  —  and  a  two  accelera- 
tions at  right  angles  to  one  another.  The  answer  is,  an  acceleration 

V  ~2  +  o2  in  the  direction  making  an  angle  9  with  the  direction 

gi 
of  motion,  where  tan.  6  =  —  •   Multiply  by  the  mass  w  ~  32-2,  and 

we  have  the  total  force. 

601.  Centrifugal  Force  in  Belts  or  Ropes.  —In  Fig.  356  let  p  Q  be 
a  small  portion  of  a  flexible  body  of  w  .  8s  Ibs. ;  its  centrifugal  force 

*  To  keep  in  our  minds  the  fact  that  all  motion  is  relative,  we  ought  to 
remember  that,  relatively  to  the  body,  other  bodies  have  a  centrifugal 
acceleration.  The  words  "centripetal"  and  "centrifugal"  are  technical 
terms  now ;  their  origins  seem  to  have  had  something  to  do  with  the  notion 
that  a  curve  has  millions  of  centres.  It  is  all  very  well  for  the  mathe- 
maticians to  speak  of  a  small  part  of  a  curve  as  being  an  arc  of  a  circle,  but  the 
engineer  knows  that  it  is  only  in  this  matter  of  curvature  or  rate  of  change 
of  e  with  *  that  it  is  like  an  arc  of  a  circle,  for  he  knows  that  complete 
information  about  the  very,  very  smallest  portion  of  any  curve  implies  ft 
complete  knowledge  of  the  whole  curve. 


APPLIED    MECHANICS. 

is  *'  Ss  .  !?,  or  -  5s  .  »«  .  ??,  or  ™  v2  .  80.     If  the  tension  is  T  Iba., 
g         r         g  5s        9 

we  know  from  Fig.  357  that  T  .  80  =  -  v2  .  50,  so  that  T  -  -  vz. 

9  9 

If  a  is  the  section  in  square  feet,  and  /  is  the  tensile  stress  per 
square  foot,  and  w0  is  the  weight  of  1  cubic  foot  of  the  material, 

w  =  aw0,  T  =  af\  so  that/=  —  vz  being  independent  of  the  radius 
of  the  path. 

Example.  —  What  velocity  will  produce  a  tensile  stress  of 
3,000  Ibs.  per  square  inch  in  the  thin  rim  of  a  cast-iron  pulley  ? 
Here/  =  3,000  x  144,  and  w0  =  -26  x  1,728.  Hence 


. 

v  =  175  feet  per  second  ; 

or,  in  engineers'  language,  a  velocity  of  10,500  feet  per  minute 
will  produce  what  is  usually  taken  as  the  working  tensile  stress  in 
cast  iron. 

602.  If  the  plane  of  the  path  alters,  if  the  plane  rotates  about  the 
tangent  to  the  path  through  the  angle  5<£  in  the  distance  5s,  then 
dQlds  is  called  the  tortuosity  of  the  path.  When  a  body  moves 
in  a  tortuous  curve  it  has  acceleration  dvjdt  along  the  path  and 
t?2/r,  a  centripetal  acceleration  in  the  plane  of  the  path,  or  the  plane 
of  curvature,  as  it  is  called.  And  if  a  model  made  of  three  very 
short  pieces  of  wire,  o  s,  o  R,  and  o  N,  be  made,  the  angles  between 
os,  OR,  and  o  N  being  right  angles,  and  if  we  conceive  o  s  to  keep 
parallel  to  the  path  Ss,  if  o  R  keeps  pointing  to  the  centre  of  curva- 
ture, then  the  angles  turned  through  about  the  axes  o  N  and  o  R  per 
unit  length  of  the  path  represent  the  curvature  and  the  tortuosity. 
A  student  ought  to  make  a  model  of  a  curve  with  wire  and  let  a 
little  frame  like  this  slide  along  it,  and  study  the  matter  for 
himself.  A  spiral  path  in  which  the  curvature  and  tortuosity  are 
constant  is  particularly  interesting.  If  we  refer  the  position  of  a 
particle  to  three  axes  of  reference,  its  total  acceleration  at  any 
instant  is  compounded  of  the  three  <Px/dP,  d^y/di2,  dfyjdt2.  The 
three  components  of  the  resultant  of  all  the  forces  which  are  acting 
are  m  times  these,  if  m  is  the  mass  or  inertia  of  the  particle. 


THE  BALANCING  OF  MACHINES. 

503.  If  a  wheel  is  fixed  eccentrically  on  its  shaft,  or  if  to  a 
shaft  there  is  attached  any  object  whose  centre  of  gravity  is 
not  exactly  in  the  axis  when  the  shaft  rotates,  centrifugal 
force  causes  pressures  on  the  bearings  of  the  shaft  which 
are  always  in  the  direction  of  the  centre  of  gravity  of  the 
rotating  mass.  In  this  case  there  is  said  to  be  a  want  of 
balance.  If  you  wish  to  observe  the  effect  produced  by  such 
want  of  balance,  mount  an  axle  to  which  a  wheel  is  keyed  on 


APPLIED    MECHANICS.  607 

any  support  which  is  not  very  firm  ;  fix  a  small  weight 
on  one  of  the  arms  of  the  wheel,  and  rotate  it  rapidly. 
You  will  find  that,  even  if  the  weight  is  small,  surprising 
effects  are  produced,  and  show  themselves  in  a  shaking  of 
the  supports  ;  and  the  evil  effects  are  four  times  as  great 
at  200  revolutions  per  minute  as  at  100  revolutions  per 
minute.  Centrifugal  force  is  proportional  to  the  mass  of 
a  rotating  part  multiplied  by  the  distance  of  its  centre  of 
gravity  from  the  axis  of  rotation,  multiplied  by  the  square  of 
the  number  of  revolutions  per  minute. 

504.  If  a  number  of  bodies  are  attached  to  a  shaft  and  are 
whirling  round  with  it,  each  of  them  at  any  instant  exerts  a 
force  on  the  shaft  which  can  be  calculated,  and  the  resultant 
effect  on  the  two  bearings  may  easily  be  determined,  just  as 
easily   as  in  the  static  problem  of  Art.  99.     If  the  axis  of 
rotation    passes    through   the   centre   of  gravity    of   all    the 
rotating   parts,    the   pressure   on   one   bearing   is   equal   and 
opposite   to   the    pressure   on   the   other;    and   by   properly 
placing   the  masses,  the  pressure   on   either  bearing  may  be 
reduced  to  nothing.     Thus  it  is  evident  that  when  two  masses 
are   directly  opposite   to  one  another  on  a  shaft,  their  cen- 
trifugal forces  may  be  made  to  balance  one  another.     When 
not  opposite  they  cannot  be  made  to  balance,  but  two  masses 
may  balance  one  which  is  directly  opposed  to  the  resultant 
force  of  the  two.     When  there  is  no  pressure  on  either  of 
the  bearings,  so  that  there  is  no  tendency  to  change  the  direc- 
tion of  the  axis,  it  is  said  to  be  the  permanent  axis  of  the 
rotating  masses.     All  axes  of  rotation  in  machines  ought 
to  be  permanent  axes.     When  this  is  the  case  in  a  rotating 
machine,  and  it  is  suspended  by  ropes  and  made  to  work,  there 
are  no  visible  oscillations. 

505.  The  balancing  of  a  machine  consists  in  adding  masses 
in    such    positions,     or    re-arranging    the    positions    of    the 
existing  masses   so  that  the  centrifugal  forces  due   to  their 
rotation   are   just  able  to  balance  the   otherwise  unbalanced 
forces  which  act  on  the  various  shafts.      The  student  will  find 
that  the  study  of  one  problem  in  balancing  will  make  him 
familiar  enough  with  the  method  of  calculation  for  its  applica- 
tion  to   almost   any  other   case  which  is   likely  to   occur  in 
practice.     The  most  usual  case  for  the  student  to  take  up  is 
that  of  the  locomotive  engine,  because  want  of  balance  in  the 
locomotive  is  capable  of  producing  very  serious  effects  indeed. 

u 


608 


APPLIED    MECHANICS. 


Fig.  358  shows  an  electromotor  driving  a  shaft  on  which  a 
number  of  discs  are  keyed.  Weights  may  be  fastened  on 
these  discs ;  the  want  of  balance  is  evident  when  the  shaft 


Fig.  358. 

rotates,  and  students  will  find  it  easy  to  illustrate  how  the 
shaft  may  be  balanced  by  other  weights  properly  placed. 
They  will  see  that  when  motions  are  merely  rotatory  we  can 
always  have  a  perfect  balancing  of  machinery. 


APPLIED   MECHANICS.  609 

506.  Example. — It  has  been  shown  by  experiment  that  the  appli- 
cation of  suitable  balance  weights  is  attended  by  a  sensible  reduction 
of  resistance  on  railways  at  high  speeds.  Locomotive  engines 
unbalanced  cannot  attain  as  high  speeds  as  when  balanced,  with 
the  same  consumption  of  fuel.  There  are  two  separate  sets  of 
unbalanced  forces  acting  on  the  crank  shaft  of  a  locomotive.  (1) 
The  centrifugal  force  of  the  crank,  crank-pin,  and  as  much  of  the 
connecting-rod  as  may  be  supposed,  roughly,  to  follow  the  path  of 
the  crank-pin  (say  one-half  of  it).  The  mass  or  weight  of  each  of 
these  multiplied  by  the  distance  of  its  centre  of  gravity  from  the 
axis,  divided  by  the  length  of  the  crank,  gives  the  mass  which,  on 
the  crank-pin,  would  produce  the  same  centrifugal  force.  Let  this 
weight  be  called  w  Ibs.  In  designing  engines,  we  consider  half  the 
connecting-rod  to  act  as  if  collected  at  the  crank-pin,  the  other  half 
to  be  moving  with  the  piston.  At  the  end  of  the  stroke,  when  the 
horizontal  component  of  the  centrifugal  force  is  greatest  and  the 
vertical  component  vanishes,  the  horizontal  pressure  on  the  axle 
caused  by  the  centrifugal  force  is 

W         Vz  W     /2lTR7l\2    I 

3*2  '  R  or  32^2  I  -60V  i  °r  "*"  *  2937' 
R  being  the  length  of  the  crank  in  feet,  and  n  the  number  of 
revolutions  per  minute.  (2)  We  have  the  force  due  to  the  mo- 
mentum of  the  reciprocating  mass,  including  piston,  piston-rod, 
slide,  and  the  second  half  of  the  connecting-rod.  The  loss  of 
momentum  is  most  rapid  just  at  the  end  of  the  stroke ;  and  as  loss 
of  momentum  per  second  is  what  we  call  force,  the  force  acting 
on  the  axle  at  the  end  of  the  stroke  due  to  this  cause  is  easily 
found,  and  proves  to  be 

wnna  *  2,937 

where  w  is  the  weight  of  the  total  reciprocating  mass. 

Now  a  weight  w\,  or  weights  whose  sum  is  t0i,  may  be  placed 
on  the  driving-wheel  or  wheels  at  a  distance  r  from  the  axis,  such 
that  the  centrifugal  force  of  w\  may  be  equal  to  the  sum  of  the 
above  forces.  This  leads  to 

w\r  =  WB  +  M>R; 

and  if  we  assume  any  distance,  r,  we  can  calculate  the  balance 
weight  or  weights,  w\. 

607.  Now,  for  the  axis  to  be  permanent  in  inside-cylinder  engines, 
w\  must  be  divided  into  two  parts,  one  for  each  wheel,  inversely 
proportional  to  the  distances  of  the  wheels  from  the  crank.  For 
outside-cylinder  engines  we  get  balance  weights  for  the  two  wheels 
whose  difference  is  w\,  and  they  are,  as  before,  inversely  proportional 
to  the  distances  from  the  wheels  to  the  crank  in  question.  Hence, 
a  consideration  of  each  cylinder  gives  two  balance  weights,  one 
usually  much  smaller  than  the  other.  As  the  cranks  are  at  right 
angles,  the  balance  weights  ought  to  be  90°  apart  on  each  wheel. 
Instead  of  using  these  two,  we  can  use  one  weight  placed  between 
their  positions,  so  that  its  centrifugal  force  is  the  resultant  of 
theirs.  Thus,  if  we  found  20  Ibs.  and  6  Ibs.  for  the  two  placed  at 
the  same  distance  from  the  axis  but  90°  apart,  make  o  A  equal  20, 


610  APPLIED   MECHANICS. 

and  o  K,  at  right  angles  to  o  A,  equal  to  6  according  to  any  scale ; 
complete  the  parallelogram,  and  o  c  represents  on  the  same  scale 
the  weight  which  will  replace  them.  It  ought  to  be  placed  at  just 
the  same  distance  from  the  axis  as  they  were  supposed  to  be 
placed ;  and  in  position  it  makes  the  angle  A  o  c  with  the  larger 
weight.  In  this  case  it  wiU  be  found  that  20-88  Ibs.  placed  18-3° 
from  the  position  which  the  weight  of  20  Ibs.  might  have  occupied 
will  be  required  to  replace  the  two. 

508.  It  often  happens  in  outside-cylinder  engines  that  the 
distance  from  one  wheel,  or  rather  from  the  centre  of  gravity  of  a 
balance  weight,  to  the  crank,  is  so  little  that  the  corresponding 
weight  for  the  other  wheel  is  very  small,  and  may  even  be  neglected. 
In  inside-cylinder  engines  it  will  be  found  that,  whereas  the  cranks 
are  at  right  angles  to  one  another,  the  balance  weights  on  the  two 
wheels  on  the  opposite  side  of  the  axis  to  the  cranks  are  often  only 
50°  apart.      In  inside-cylinder  engines  with  coupled  wheels  the 
outside  coupling  rods  and  cranks  are  usually  made  to  balance  the  inside 
moving  parts.    These  engines  work  very  smoothly  indeed.    Outside- 
cylinder  engines  with  coupled  wheels  are  very  unstable,  from  the 
use  of  small  wheels  requiring  very  rapid  revolution  of  the  crank 
axle;  from  the  cylinders  being  farther  apart  than  usual,  so  that 
the    coupling-rods    may  have    room,   and  from  the   number  of 
reciprocating  parts  being  increased.     The  conditions  seem  to  admit 
of  no  remedy  for  these  defects.     The  balance  weight  ought  to  be 
distributed  over  two  or  three  of  the  spaces  of  the  wheel,  that  the 
tire  may  not  be  unduly  strained. 

509.  We  have,  then,  the  following  easy,  approximately  correct 
rules  for  locomotives  : — If  R  is  length  of  crank,  r  the  distance  of 
centres  of  gravity  of  every  balance  weight  from  centres  of  wheels, 
e  the  distance  apart  of  the  centre  lines  of  cylinders,  d  the  distance 
apart  of  the  wheels  or  centres  of  gravity  of  the  balance  weights, 
w  the  total  weight  of  crank  (referred  to  the  pin),  pin,  connecting- 
rod,  piston,  slide,  and  piston-rod,  A  the  angle  which  the  position  of 
centre  of  gravity  of  balance  weight  makes  with  near  crank : 

(1)  Inside-cylinder  engines  with  uncoupled  wheels.     Each 

balance  weight  =  ^-5-  >J  2  d?  +  2e2,  tan.  A  =  -^ ;   so  that  A  is 

easily  obtained  from  a  book  of  tables. 

(2)  Outside-cylinder  single  engines  with  uncoupled  wheels. 

Each  balance  weight  =  — ,  A  =  180° ;   so  that  in  this  case  the 

balance  weight  is  placed  exactly  opposite  to  the  crank. 

(3)  Inside-cylinder  engines  with  wheels  coupled.     Find  by 
rule  (1)  if  the  weight  of  the  coupling-rods,  etc.,  is  too  great.     If  so, 
let  counter  weights  equal  to  the  difference  be  placed  opposite  the 
outside  cranks.     If  too  small,  the  difference  must  be  made  up  with 
balance  weights,  as  in  rule   (1).     The  positions  of  the  outside 
cranks  are  found  by  rule  (1). 

(4)  Outside-cylinder  coupled  engines.     Find  revolving  weight 
of  coupling-rods,  etc.,  for  each  wheel.    Also  find  Sum  of  the  weight 


APPLIED    MECHANICS.  611 

of  the  piston,  rod,  slide,  and  half  connecting-rod.  Divide  this 
latter  among  the  wheels,  adding  the  given  revolving  weight 
already  on  them.  Let  this  be  used  on  each  wheel  according  to 
rule  (2). 

610.  We  have  dwelt  upon  these  practical  rules  for  balancing  in 
locomotives,  because  they  give  good  illustrations  of  centrifugal 
force.  But  the  student  ought  clearly  to  see  that  it  is  only  when  a 
body  rotates  about  an  axis  that  we  can  exactly  balance  the  forces. 
A  body  with  reciprocating  motion  can  only  be  balanced  by 
another  body  with  a  reciprocating  motion ;  and  hence  it  is  that, 
after  much  expense  and  quarrelling  with  persons  complaining  about 
the  vibration  of  their  houses,  many  electric  lighting  companies  have 
discarded  reciprocating  steam-engines,  replacing  them  with  steam 
turbines.  In  Arts.  427  and  429  we  give  the  principles  on  which  the 
subject  may  be  studied.  I  give  practical  examples  of  their  use  in 
my  book  on  steam,  gas,  and  oil  engines.  In  that  book  I  give  the 
more  exact  constructions,  which  are  so  commonly  taught  now  to 
advanced  students,  to  find  the  forces  at  the  ends  of  a  connecting- 
rod  for  any  position.  It  seems  never  to  strike  a  student  or  his 
teacher  that  such  elaborate  calculations  may  possibly  not  give  a 
very  different  result  from  what  they  may  obtain  by  the  simple 
assumption  like  what  I  have  stated  above.  One  of  my  students 
has  made  the  comparison.  His  three  weeks'  constructions  to  find 
the  forces  acting  on  the  frame  of  a  Willans  engine,  the  turning 
moment  on  the  crank  shaft,  etc.,  nowhere  differ  appreciably  in 
their  results  from  those  obtained  by  assuming  that  half  the  mass 
of  the  connecting-rod  has  the  motion  of  the  cross-head,  and  the 
other  half  that  of  the  crank  pin.  The  very  much  more  important 
matter,  the  effect  of  the  motion  of  the  rod  upon  its  strength  as  a 
laterally  loaded  strut,  seems  never  to  engage  mnch  attention. 

EXERCISES. 

1.  A  crank  pin  4  inches  in  diameter  and  6  inches  long  has  to  be 
balanced.     If  the  length  of  crank  be  9  inches,  and  the  balance  weights 
are  placed  directly  opposite  each  crank  arm,  find  the  weights,  the  centre 
of  gravity  of  each  being  6  inches  from  the  centre  of  the  crank  shaft. 

Am.,  16-24  Ibs. 

2.  A  shaft  is  in  balance  under  the  action  of  three  weights,  one  of  300 
Ibs.,  at  a  distance  of  12  inches  from  the  axis;  another  of  100  Ibs.,  at  a 
distance  of  20  inches  from  the  axis,  on  the  opposite  side,  and  30  inches  to 
the  right  of  the  first.  How  much  must  the  third  weight  be,  and  where  must 
be  its  position  along  the  shaft,  if  its  distance  from  the  axis  is  28  inches  ? 

A ns.,  57 '1  Ibs.;  37.5  inches  to  left  of  first  weight. 

3.  In  a  locomotive  the  distance  between  the  centre  lines  of  the 
cylinders  is   27£  inches.      Balance  weights  are  fixed  at  a  horizontal 
distance  apart  of  59  inches,  the  centre  of  gravity  of  each  describing  a 
circle  of  55  inches  diameter.     If  the  weight  of  the  reciprocating  masses 
for  each  cylinder  be  400  Ibs.,  and  the  stroke  be  25   inches,  find  the 
position  and  magnitude  of  balance  weights  to  counteract  the  horizontal 
and  alternating  force  and  couple. 

An*.t  At  160°  with  near  crank;  142  Ibs.  each. 


612  APPLIED    MECHANICS. 

4.  In  a  steam-engine  the  piston  at  the  beginning  of  its  stroke  is 
exposed  to  a  total  effective  steam  pressure  of  2,000  Ibs.,  but  the  inertia  of 
the  piston  is  such  that  the  thrust  of  the  piston-rod  is  only  1,600  Ibs. 
The  speed  of  the  engine  is  now  raised  until  it  becomes  half  as  great  again 
as  before,  while  the  steam-pressure  is  unchanged.     What  is  the  thrust  of 
the  piston-rod  ?  Ans.,  1,100  Ibs. 

5.  An  engine  is  making  150  revolutions  per  minute.     What  is  the 
acceleration  of  the  piston  at  the   commencement  of  each  stroke,  the 
connecting-rod  being  4  feet  long  and  the  crank  9  inches  P 

Ans.,  223  ;  153. 

6.  The   weight   of    the    reciprocating    parts   of    a    steam-engine    is 
equivalent  to  3  Ibs.  per  square,  inch  of  the  area  of  the  piston.     If  the 
length  of  crank  be  9  inches,  find  how  much  the  initial  effective  pressure 
is  reduced  by  the  inertia  of  the  reciprocating  parts  when  the  crank  makes 
70  revolutions  per  minute,  the  obliquity  of  the  connecting-rod  being 
neglected.  Ans.   3-8  Ibs.  per  square  inch. 


613 


CHAPTER    XXVIIL 

SPRINGS. 

511.  ANY  contrivance  which  can  store  energy   as  strain 
energy,  and  give  it  out  again  readily,  is  a  spring.     Hence,  any 
tie-bar  or  strut,  any  beam — in  fact,  any  object  whatsoever  is  a 
spring.     The  term  is,  however,  generally  applied  only  to  such 
objects  as  can  be  changed  in  shape  very  much  without  fracture. 

A  tie-rod  of  indiarubber  can  be  stretched  to  eight  times  its 
old  length,  again  and  again,  without  hurt ;  whereas  a  tie- rod 
of  the  best  steel  can  only  be  stretched  to  g^n  of  its  old 
length,  again  and  again,  without  hurt. 

Hence  the  indiarubber  tie-rod  or  a  strut  may  be  called 
a  spring,  just  like  the  spiral  spring  j  but  it  would  not  be  right 
to  speak  of  a  tie-rod  or  strut  of  steel  as  a  spring.  The  differ- 
ence is,  however,  only  one  of  degree,  and  indeed,  a  mine  cage 
suspended  by  a  steel  rope  half  a  mile  long  vibrates  up  and 
down  just  as  if  hung  from  any  ordinary  spring. 

512.  Springs  are  almost  always  used  as  reservoirs  of  energy 
— that  is,  as  hydraulic  accumulators  are  used,  or  fly-wheels  of 
steam-engines,  or  cisterns  of  water,  or  electric  accumulators. 

The  mainspring  of  a  clock  or  watch  takes  a  store  of  energy 
in  winding-up,  and  gives  it  out  gradually  for  about  twenty- 
four  hours  in  unwinding.  A  bow  gets  gradually  a  store  of 
energy,  which  it  gives  out  rapidly  when  the  arrow  is  set  free. 
A  buffer-stop  spring  stores  up  all  the  kinetic  energy  of  a  train, 
and  the  stiffer  it  is  the  more  quickly  will  it  store  the  energy, 
and  therefore  the  more  suddenly  will  the  train  be  brought  to 
rest.  Ordinary  buffer  springs  are  continually  storing  up  and 
giving  out  energy,  equalising  only  gradually  the  velocities  of 
tte  two  railway  carriages,  so  that  if  one  gets  a  sudden  change 
of  velocity,  the  other  shall  only  be  affected  gradually.  In  the 
same  way,  when  any  two  objects  have  a  springy  connection,  if 
one  of  the  objects  alters  its  velocity  suddenly,  the  other  alters 
its  velocity  in  consequence  only  gradually.  These  are  cases  in 
which  a  spring  is  used  to  prevent  shocks,  or,  as  we  may  put 
it,  a  spring  is  used  to  lengthen  the  time  of  a  blow,  and  there- 
fore to  diminish  the  average  force  of  a  blow. 

513.  Now  there  are  several  distinct  cases  here  to  consider. 


614  APPLIED    MECHANICS. 

I.  A  body  of  mass  B,  moving  with  velocity  v  ,  overtakes  a 
body  A,  moving  with  velocity  v  .     The  buffer  between  them  is 

strained    until   they   move  with  the  same   velocity.      If  the 
common  velocity  then  is  v, 

B(vb-v)=A  (v  -  va) 


The  energy  now  stored  up  in  the  spring  is  — 


In  case  the  bodies  B  and  A  are  themselves  elastic,  they  them- 
selves, to  a  greater  or  less  extent,  act  as  buffers,  and  less 
energy  is  shored  in  the  buffer  itself.  Also  as  changes  of  shape 
are  usually  accompanied  by  friction,  some  of  the  energy  is 
wasted.  If  no  energy  were  wasted,  the  two  objects  would 
keep  going  faster  and  slower  relatively  to  one  another,  as  the 
spring  was  compressed  and  extended,  and  to  some  extent  this 
does  go  on  after  the  blow  ;  but  these  vibrations  are  usually 
rapidly  stilled  as  the  energy  is  wasted  in  friction,  partly,  as 
we  have  already  said,  in  the  buffer  spring  itself,  and  partly  by 
friction  opposing  generally  the  motion  of  the  bodies. 

II.  If  one  of  the  bodies,  A,  is  fixed  to  the  earth,  then  the 
mass  of  A  may  be  regarded  as  infinite  ;  va  and  v  in  the  above 

calculation  become  0,.  but  the  same  reasoning  applies  as  before. 

III.  Two  carriages,  A  and  B,  are  at  rest  connected  by  a 
spring.     A  is  made  suddenly  to  move  through  a  distance  a. 
Now  if  the  spring  were  infinitely  stiff,  B  would  just  as  suddenly 
move  through  the  same  distance  a.     As  the  spring  is  less  and 
less   stiff,   B   moves  over   the  distance  a  with  less   and    less 
suddenness,  because  the  kinetic  energy  which  must  eventually 
be  given  to   B  is  suddenly  stored  in  the  spring,  but  is  only 
gradually  given  to  B  by  the  spring.     If  there  is  no  friction,  B 
will  be  left  vibrating.     If  there  is  no  friction  opposing  B'S 
motion,   but  there  is  friction  due  to  change  of  shape  of  the 
spring,  B  will  vibrate  and  gradually  come  to  rest. 

If  there  is  friction  opposing  B'S  motion,  but  there  is  no 
friction  in  the  spring,  the  store  of  energy  in  the  spring  is 
greater  than  before.  B  must,  of  course,  come  to  rest.  With 
friction  in  the  spring,  B  will  more  rapidly  come  to  rest. 


APPLIED    MECHANICS. 


615 


In  all  these  cases,  B  comes  to  rest  at  the  distance  a  from 
its  old  position. 

IV.  As  in  the  last  case — but  A  moves  suddenly  through  a 
distance  a,  and  after  a  short  time  T  is  moved  suddenly  back 
to  its  old  position. 

(1)  If  T  is  exceedingly  small,  B  does  not  move. 

(2)  If  T  is  great,  B  moves  as  described  in  III.,  and  repeats 
its  motion  in  the  reversed  direction. 

(3)  If  T  is  neither  too  great  nor  too  small,  B  has  a  motion 
intermediate  between  the  (1)  and  (*2)  described  motions. 

.  (4)  If  A'S  motion  is  quickly  vibrating  through  the  ampli- 
tude a,  B  gets  a  vibratory  motion  of  the  same  period;  but 
superimposed  on  this  is  the  natural  vibratory  motion  of  a 
period  which  depends  on  the  stiffness  of  the  spring  and  the 
mass  of  B;  the  natural  vibrations  will  die  away  if  there  is 
friction. 

514.  It  is  well  worth  while  for  a  student  to  illustrate  this  last 
case  by  means  of  a  model.  The  crank  Q,  Fig.  359,  is  turned 
round  regularly ;  if  this  is  done  h 

by  hand,  a  fly-wheel  ought  to  be 
used  to  give  steadiness  of  motion. 
By  means  of  a  connecting-rod,  p 
gets  an  up  and  down  motion 
which,  diminished  in  the  ratio 
OA/OP,  is  given  to  A ;  this  creates 
a  forced  vibration  in  B.  The 
natural  vibration  of  B,  when 
A  is  not  moving,  ought  previ-  Fig.  359. 

ously  to  have  been  studied — (1), 

when  there  is  very  little  friction,  and  B  goes  on  vibrating  with 
nearly  the  same  amplitude  for  a  long  time ;  (2)  when  there  is 
fluid  friction,  B  vibrating  in  a  vessel  of  water.  In  this  case 
there  is  really  a  change  of  inertia  difficult  to  calculate;  the 
water  being  set  in  motion.  The  amplitude  of  B'S  motion  gra- 
dually diminishes  because  of  friction.  When  A  suddenly  begins 
to  vibrate,  B  may  have  a  large  natural  vibration,  but  this 
gradually  gets  destroyed  by  friction,  just  as  if  there  were  no 
forced  vibration.  We  shall  now  speak  of  the  forced  vibration 
only,  and  assume  no  friction.  Let  A  vibrate  with  f  times  the 
frequency  of  the  natural  vibrations,  and  let  A'S  amplitude  be  1. 
Then  the  amplitude  of  B'S  motion  will  be  as  follows  :  when 
there  is  no  friction  B  moves  synchronously  with  A,  so  that  B 


616 


APPLIED    MECHANICS. 


is  at  the  top  when  A  is  at  the  top,  or  else 
(and  this  is  indicated  with  a  —  sign)  B  is  at 
the  bottom  when  A  is  at  the  top. 

To  prove  this,  let  w  be  the  weight  in 
Ibs.  of  B,  so  that  its  mass  is  w/^.  As 
before,  a  better  approximation  to  accuracy 
is  obtained  by  letting  w  include  one- 
third*  of  the  weight  of  the  spring.  Let 
the  spring  be  such  that  a  force  of  1  Ib. 

AV 

elongates  it  h  feet.  Then  -  x  the  accelera- 
tion is  the  force  in  the  spring;  and  if  x  is 
the  distance  in  feet  of  the  body  below  its 

position   of  equilibrium,  -  h  —  .  j^    is    the 

extra  elongation  of  the  spring  when  there  is 
this  acceleration.  But  if  B  is  a;  feet  below, 
and  if  A  is  y  feet  below  their  positions  of 
equilibrium,  then  x  -  y  is  this  extra  elonga- 
tion, and  hence 


or 


•-a 
4 


Letting  w2  =  -^,  we  see  that  n  divided  by 

2  IT  ^is  the  natural  frequency  of  vibration. 
As  in  Art.  19,  if  we  introduce  our  force  of 
friction  equal  to  b  .times  the  velocity,  using 
2  F  for  i^/w,  we  have 

By  putting  y  =  a  sin.  qt,  it  is  easy  to  find 
the  forced  vibration.  For  simplicity,  if  we 
let  F  =  o,  then 

Table  XII.  is  obtained  by  letting  q/n  be  called/. 

Note  in  the  model  and  from  the  tabled 
numbers  that  when  the  forced  frequency  is 
a  small  fraction  of  the  natural,  the  forced 
vibration  of  B  is  a  faithful  copy  of  the 
motion  of  the  point  of  support  A  ;  the  spring 

*  Let  the  student  prove  that  £  of  the  weight  of 
the  spring  is  to  be  t^ken  and  not  £  of  it. 


APPLIED    MECHANICS.  617 

and  B  move  like  a  rigid  body.  When  the  force'd  frequency 
is  increased,  the  motion  of  B  is  a  faithful  magnification  of 
A'S  motion.  As  the  forced  gets  nearly  equal  to  the  natural, 
the  motion  of  B  is  an  enormous  magnification  of  A'S  motion. 
There  is  always  some  friction,  and  hence  the  vibration  cannot 
become  infinite.  When  the  forced  frequency  is  greater  than 
the  natural,  B  is  always  a  half-period  behind  A,  being  at  the 
top  of  its  path  when  A  is  at  the  bottom.  When  the  forced  is 
many  times  the  natural,  the  motion  of  B  gets  to  be  very  small ; 
it  is  nearly  at  rest. 

515.  Men  who  design  Earthquake  recorders  try  to  find  a 
steady  point  which  does  not  move  when  everything  else  is 
moving.  For  up  and  down  motion,  observe  that  in  the  last 
case  just  mentioned  B  is  like  a  steady  point. 

When  the  forced  and  natural  frequencies  are  nearly  equal 
we  have  the  state  of  things  which  gives  rise  to  resonance  in 
acoustic  instruments ;  which  causes  us  to  fear  for  suspension 
bridges  or  rolling  ships. 

It  is  obvious,  then,  that  the  simple  statement  of  the 
problem,  "  How  do  springs  prevent  shocks  ? "  presents,  when 
we  consider  it  very  carefully,  many  quite  different  problems. 
It  is  worth  while  observing  carefully  the  up  and  down  motion 
of  the  body  B  of  a  waggon  on  the  street  when  the  wheels  A 
go  up  and  down  over  the  stones.  But  we  can  study  the  sub- 
ject better  perhaps  in  the  model.  We  see  that  although  a 
spring  connection  between  A  and  B  does  prevent  shocks,  the 
motion  of  B  may  be  dangerously  great.  Thus,  for  example, 
when  an  earthquake  occurs  it  does  not  always  do  to  merely 
have  an  elastic  connection  between  the  ground  and  the  house, 
as  the  earthquake  leaves  a  Japanese  house  vibrating  sometimes 
so  much  as  to  give  quite  a  sea-sick  feeling  to  the  inhabitants. 

We  see  that  in  all  cases,  unless  there  is  friction  opposing 
vibration  of  the  body  B — whether  this  friction  exists  in  the 
parts  of  the  spring  itself,  or  more  directly  opposes  the  motion 
of  B — there  will  be  vibrations,  sometimes  dangerously  large  in 
amplitude.  When  by  means  of  the  spring  connection  we  seek 
to  diminish  shocks,  friction  may  be  introduced  by  some  dash- 
pot  arrangement,  which  may  consist  of  a  porous  piston  moving 
in  a  cylinder  filled  with  water  or  oil ;  or  the  piston  may  be 
solid,  and  there  may  be  a  pipe  and  cock  connection  between 
the  opposite  sides  of  it.  Here  the  friction  is  mostly  fluid 
friction.  If  it  were  all  fluid  friction,  there  would  be  no 


618  APPLIED   MECHANICS. 

opposition  to  motion — there  would,  in  fact,  be  no  friction — if 
the  motion  were  slow  ;  the  friction  is  greater  and  greater  as  the 
motion  is  quicker  and  quicker.  The  friction  between  the  plates 
of  a  carriage  spring,  which  rub  together  every  time  the  spring 
is  changed  in  shape,  is  solid  friction,  which,  if  anything,  is 
probably  greater  for  slow  motions  than  for  quick  motions. 
In  all  cases  when  vibrations  are  to  be  stilled  it  is  better  that 
the  friction  should  be  of  the  nature  of  fluid  friction,  but  this 
is  not  always  convenient.  Its  effect  in  stilling  vibrations  may 
very  readily  be  studied  in  the  laboratory.  It  will  be  found 
that  by  adjusting  the  cock  of  this  dash-pot  we  can  vary  at 
will  the  rate  of  stilling  of  the  vibrations  of  even  very  large 
masses,  so  that  after  a  shock  through  the  spring  the  body  B 
may  vibrate  for  a  long  time  before  it  comes  to  rest,  or  it  may 
come  to  rest  after  one  slow  lurching  motion  only.  Now,  it  is 
necessary  to  understand  that  a  dash-pot  arrangement,  or  any 
other  arrangement  for  introducing  fluid  friction,  will  not 
affect  the  static  law  of  the  spring  in  any  way.  It  may  be 
introduced  on  spring  balances  which  are  required  to  measure 
forces  accurately,  for  example.  But  the  sort  of  friction  we 
find  in  carriage  springs  is  very  different.  Here  the  plates  of 
which  the  spring  is  made  rub  on  one  another,  and  there  is 
solid  friction,  and  such  a  spring  as  this  cannot  be  used  for 
spring  balances.  The  law  of  the  spring  is  altered ;  we  cannot 
depend  upon  the  spring  for  measuring  forces  if  there  is  any 
place  where  rubbing  of  solids  takes  place.  Such  springs  are 
never  used  for  purposes  of  measurement ;  they  are  only  used 
for  preventing  shocks. 

516.  Springs  are  of  many  different  forms ;  they  are  used  aa 
small  as  balance  springs  in  watches,  and  they  are  sometimes 
so  large  as  we  see  them  in  some  buffer  and  locomotive  springs. 
The  following  list  is  not  given  as  by  any  means  an  exhaustive 
one: — 

I.  Cylindric  spiral  springs,  subjected  to  axial  loading. — In 
these  it  will  be  found  that  the  wire,  whether  round  or  nearly 
square  in  section,  is  twisted  like  an  ordinary  revolving  shaft 
transmitting  power,  and  the  strain  is  one  of  torsion  in  the 
material. 

Examples. — Most  forms  of  spring  balance  and  dynamo- 
meters ;  the  springs  of  indicators ;  many  railway  carriage  and 
tram-car  springs ;  safety-valve  springs  for  marine  and  loco- 
motive boilers. 


APPLIED    MECHANICS.  619 

II.  Cylindric  spiral  springs,  subjected  to  a  torque  about 
the  axis  : — In  these  it  will  be  found  that  the  material  is  sub- 
jected everywhere  to  bending. 

Examples. — The  balance  springs  of  the  best  chronometers ; 
the  springs  used  as  elastic  joints  between  two  lengths  of 
small  shafting. 

III.  Flat  spiral  springs,  subjected  to  a  torque  about  the 
axis. — In  these  it  will  be  found  that  the  material  is  subjected 
to  bending. 

Examples. — The  main  and  balance  springs  of  watches  and 
many  clocks ;  the  springs  used  in  nearly  all  contrivances 
which  require  to  be  wound  up,  or  wherever  a  large  reservoir 
of  energy  is  required  in  a  small  space. 

IV.  Nearly  straight  strips  of  material  subjected  to  bend- 
ing.— These  are  used  in  a  great  variety  of  cases,  sometimes  in 
one  piece,  as  in  the  limb  of  a  tuning-fork. 

V.  More  or  less  flat,  or  corrugated,  circular,  or  of  other 
outline,  fixed  or  only  supported  at  three  or  more  points  at  the 


VI.  Indiarubber  springs,  usually  subjected  to  merely  ten- 
sion or  compression,   now  being  used  largely  for  tram-cars, 
In  all  the  above  cases  there  is  an  absence  of  solid  friction. 

VII.  The   Ayrton-Perry   spring,    used   in   indicating   the 
amount  of  a  force  by  a  large  relative  rotation  of  a  pointer. — 
It  will  be  found  that  the  material  in  these  springs  is  subjected 
to  a  combination  of  bending  and  twisting  strains.    The  Ayrton- 
Parry  twisted  strip,  in  which  a  very  small  elongation  is  accom- 
panied by  great  relative  rotation. 

VIII.  More  or  less  straight  strips  of  material  subjected  to 
bending  (like  IV.),  a  number  of   pieces   being  used  in  one 
spring,  these  pieces  rubbing  on  one  another.     When  bending 
occurs,  this  introduces  solid  friction. 

Examples. — Locomotive,  waggon,  and  carriage  springs, 
and  nearly  all  the  large  springs  used  for  minimising  shocks,  as 
buffer-stop  springs. 

IX.  Gases  in  closed  vessels,  the  volume  of  which  may  be 
altered,    as   in   the   air-chamber   of    some   force-pumps ;    the 
cylinder  and  piston  air-spring. 

It  is,  however,  obvious  that  the  functions  of  springs 
cannot  be  specified  in  a  few  words.  Springiness  comes  in 


620  APPLIED    MECHANICS. 

usefully  in  the  packing-rings  of  pistons,  and  in  packing 
generally,  to  produce  a  good  fit  between  pieces  which  rub 
on  one  another ;  in  spring  split  rings,  as  washers,  which 
prevent  a  nut  from  becoming  loose ;  when  a  springy  piece  of 
material  is  used  between  two  more  rigid  pieces  which  are 
bolted  together,  to  give  a  uniform  bearing  without  undue 
strains  in  the  nuts.  When,  in  fact,  we  discuss  the  elasticity 
of  material  generally,  we  see  that  everything  in  nature  is  a 
spring,  and  performs  most  of  its  functions  in  nature  by  means 
of  this  elastic  property. 

517.  The  main  uses  to  which  springs  are  put  are  these  :— 

1.  Lengthening  impacts,   so  as  to  diminish  the  forces  of 
blows,  and  therefore  absorbing,  and  in  these  cases  usually  also 
dissipating  energy. 

2.  Regulating  motion — that  is,  preventing  large  fluctua- 
tions in  speed  in  driven  pieces  of  machinery.     This  is  not  a 
common  use  of  springs,  because  of  the  waste  of  energy. 

3.  As  reservoirs  of  energy. 

4.  Regulating,  as  in  watches  and  clocks. 

5.  As  measurers  of  force. 

6.  As  measurers  of  distance. 

7.  As  measurers  of  angles. 

518.  The  Best  Materials  to  Use  in  Springs! — A  spring's 
usefulness  depends  primarily  on  its  being  a  reservoir  of  energy. 
In  the  first  two  cases  of  the  preceding  table  this  capacity  for 
storage  of  energy,  and  of  course  cheapness  and  ease  of  manu- 
facture, ought  to  settle  for  us  the  material  of  which  a  spring 
should  be  made.     In  the  other  cases  we  must  also  consider 
the  question  of  perfect  or  imperfect  elasticity  and  viscosity  of 
the  material.     First,  then,   as  to  the  energy  which  may  be 
stored. 

The  energy  which  must  be  given  to  distort  a  spring  before 
it  takes  a  permanent  set  is  called  its  resilience. 

.Now,  it  will  be  shown  in  Art.  535  that  in  all  springs  sub- 
jected to  bending,  as  springs  of  classes  2,  3,  4,  5,  7,  and  8,  the 
resilience  per  unit  volume  of  the  material  depends  upon  the 
resilience  per  cubic  inch  of  the  material  when  subjected  to 
compressive  or  tensile  force,  and  this  is  f 2  -j-  2  E  in  inch- 
pounds,  where  f  is  the  greatest  tensile  or  compressive  stress 
which  the  material  will  stand  without  taking  a  set,  and  E  is 


APPLIED    MECHANICS. 


621 


Young's  modulus  of  elasticity.      The   following   table  shows 
the  value  of  this  constant  for  various  materials  : — 

TABLE  XIII. — SPRING  MATERIALS  SUBJECTED  TO  BENDING. 


/ 

E 
In  Millions  of  Pounds 

/S-r  2B 

per  Square  Inch. 

Wrought  Iron 

24,000 

29 

10 

Mild  Steel     

35,000 

30 

20 

Mild  Steel,  Hardened 

70,500 

30 

83 

Cast  Steel,  Unhardened 

80,000 

30 

107 

Cast  Steel,  Hardened 

190,000 

36 

50  J 

Copper 

4,300 

15 

0-62 

Brass 

6,950 

9-2 

2-62 

Gun  Metal    ... 

6,200 

9-9 

2-00 

Phosphor  Bronze 

19,700 

14 

13-85 

Glass 

4,500 

8 

1-26 

The  numbers  in  Tables  XIII.  and  XIV.  are  subjected  to  very 
considerable  variations,  especially  in  the  cases  of  copper, 
brass,  gun  metal,  phosphor  bronze,  and  glass.  Indeed,  in  our 
opinion,  definite  statements  as  to  the  values  of  /2  -h  2  E,  or 
fi2  -r  2  N,  ought  not  to  be  made,  until  careful  experiments 
have  been  made  on  such  varieties  of  these  materials  as  are 
actually  used  in  spring-making,  and  this  has  not  yet,  we  be- 
lieve, been  done.  We  have  taken  the  values  of/,  E,  and  N  from 
Table  XXII.  In  the  best  spiral  steel  springs,  for  example, 
the  value  of  f±  (the  proof  shear  stress)  has  been  found  to  be 
rather  60,000  or  70,000  Ibs.  per  square  inch  than  the  145,000 
given  in  the  table. 

It  will  be  shown  in  Art.  535  that  in  all  springs  where  the 
material  is  subjected  to  twisting  merely,  as  springs  of  Class  1 
—  the  most  important  class,  probably,  for  the  use  of  mechanical 
engineers  and  instrument  makers  —  the  resilience  per  unit 
volume  of  the  material  depends  upon  the  value  of 


where  f^  is  the  greatest  shear  stress  which  the  material  will 
stand  without  taking  a  set,  and  N  is  the  modulus  of  rigidity  of 
the  material.  The  following  table  shows  the  value  of  this 
constant  for  various  materials  :  — 


622  APPLIED    MECHANICS. 

TABLE  XIV. — MATERIALS  FOR  CYLINDRIC  SPIRAL  SPRINGS. 


N 

/i 

In  Millions  of  Pounds 

/12-T-2N 

s 

per  Square  Inch. 

Wrought  Iron 

20,000 

10-5 

19 

Mild  Steel    ...         

26,500 

11 

32 

Mild  Steel,  Hardened 

53,000 

11 

128 

Cast  Steel,  Unhardened      .  . 

64,000 

11 

186 

Cast  Steel,  Hardened 

145,000 

13 

809 

Copper          ...         ...         .  . 

2,900 

5-6 

0-75 

Brass              ...         ...         .  . 

5,200 

3-4 

4-00 

Gun  Metal    

4,150 

3-7 

2-33 

Phosphor  Bronze     ... 

14,500 

5-25 

20 

The  numbers  given  in  Tables  XIII.  and  XIV.  are  supposed  to 
express,  then,  the  actual  relative  values  of  the  various 
materials  for  spring-making. 

It  will  be  observed  that  hardened  cast  steel  is  very  much 
better  than  any  other  material  for  spring-making  ;  hardening 
makes  it  five  times  more  valuable.  It  is  about  35  to  40 
times  more  valuable  than  phosphor  bronze ;  more  than  40 
times  more  valuable  than  wrought  iron  (which  is  not  so  good 
as  phosphor  bronze).  The  fact  that  phosphor  bronze  makes 
probably  the  best  non-magnetic  material  for  springs  has 
been  known  to  me  for  fifteen  years.  I  tested  this  result 
by  a  great  deal  of  experimenting  with  various  materials. 
But  this  is  not  the  only  virtue  of  phosphor  bronze.  In  quite 
a  remarkable  degree  it  is  free  from  many  of  the  vices  of  other 
metals — sub-permanent  set  after  small  loads,  effects  due  to 
fatigue,  etc.  It  is  worth  while  to  mention  that  by  the  nature 
of  the  process  of  manufacture,  the  material  may  have  initial 
strains  in  it ;  before  applying  it  in  an  instrument  it  ought  to 
receive  a  considerable  set  in  the  direction  in  which  it  will 
most  usually  be  strained.  Phosphor  bronze  springs  in  my 
electrical  instruments  receive  a  set  from  a  load  which  is  six 
times  as  great  as  the  greatest  load  ever  applied  to  the  spring 
when  it  is  in  use. 

The  numbers  in  the  tables  give  us  guidance,  but  we  must 
also  consider  special  conditions.  The  hardening  and  temper- 
ing of  steel  require  great  care ;  so  great  is  this  that  we  may 
almost  say  that  there  is  only  one  steel  spring  maker  in  the 


APPLIED    MECHANICS. 


623 


whole  of  England.  Now,  phosphor  bronze  and  brass  and 
copper  receive  their  greatest  hardness  by  drawing  through 
dies  or  rolling.  They  can,  in  fact,  be  hardened  very  uni 
formly  in  the  cold  state  quite  readily,  and  springs  of  them  are 
easy  to  make.  Again,  although  tempered  steel  has  usually  an 
oxide  of  iron  to  protect  it,  and  a  soft  iron  spring  of  any  kind 
can  also  be  given  such  an  oxide  by  BarfTs  process  as  a  cover- 
ing, yet,  on  the  whole,  steel  and  more  especially  iron  springs 
are  much  more  subject  to  rust  when  exposed  to  a  damp  atmo- 
sphere than  copper,  brass,  or  phosphor  bronze.  Again,  in 
certain  electrical  instruments  where  springs  are  used,  steel 
and  iron  must  not  be  used  because  of  their  magnetic  pro-r 
perties,  and  in  other  measuring  instruments  the  properties 
catalogued  in  the  tables  may  not  always  be  all-important. 

519.  The  following  property  is  not  of  importance  in  springs 
used  to  prevent  shocks,  as  in  buffer  and  carriage  springs  : — 

When  a  spring  is  loaded  with  even  a  small  load  it  may 
continue  to  lengthen  axially  slowly  if  the  load  is  kept  on ;  and 
afterwards,  when  the  load  is  taken  off,  it  may  not  immediately 
shorten  to  its  original  length,  but  needs  time.  This  is  usually 
called  sub-permanent  set,  and  is  greater  with  greater  loads. 
When  such  a  spring  is  unloaded  after  it  has  experienced  loads 
of  various  amounts  and  various  periods  of  rest,  it  will  not 
usually  go  back  to  its  old  length,  but  will  slowly  undergo 
slight  shortenings  and  lengthenings  of  various  amounts  de- 
pending on  its  previous  experiences.  I  sometimes  call  this 
property  the  "  creeping  "  property  of  the 
material.  Professor  Ayrfcon  and  I  have 
written  a  paper  concerning  this  property. 
A  material  possessing  much  of  it  is  quite 
unsuitable  for  the  springs  of  measuring 
instruments. 

520.  Spiral  Springs. — A  spiral  spring  is 
a  wire,  or  rod,  or  strip  of  any  constant  or 
varying  section  (we  shall  always  speak  of 
it  as  a  wire,  of  whatever  size  or  shape  its 
section  may  be),  coiled  so  that  the  centre 
line  of  the  wire  lies  everywhere  on  some 
surface  of  revolution.     In  most  cases  the 
wire  is  wound  on  a  cylindric  surface,  the 
winding  being  perfectly  regular — that  is,  the 
angle  made  by  the  centre  line  of  the  wire, 


Fig.  360. 


624  APPLIED    MECHANICS. 

with  a  plane  at  right  angles  to  the  axis  of  the  spiral,  is  constant. 
Many  cylindric  spiral  springs  in  use  have  wire  of  square  or  of 
elliptic  section.  In  another  form  which  is  used  as  a  buffer 
spring,  the  diameter  of  the  coils  varies  as  if  the  wire  had  been 

wound  on  the  half  of  a  very 
bulging  barrel-shaped  mandril ; 
and  we  see  that  under  pressure 
this  spring  can  get  very  short 
axially  without  the  coils  coining 
into  contact.  In  the  form  Fig. 
x  360,  the  mandril  was  of  a  coni- 
cal shape.  Lastly,  in  Fig.  361, 
we  have  the  flat  spiral  spring — 
what  the  conical  form  would  be- 
come if  it  were  squeezed  until  all 
its  coils  lay  one  plane. 

621.  As  an  example  of  the  bending  of  a  strip  of  material,  which 
might  have  been  considered  after  Art.  327,  let  us  take  the  case  of 
a  flat  spiral  spring,  such  as  the  main  or  balance  spring  in  watches. 
Let  N  P  M  (Fig.  361)  be  such  a  spring,  fastened  to  a  case  at  N,  and 
to  an  arbor  or  axle  at  M.  When  no  forces  are  acting  on  the  spring 
it  has  a  spiral  shape.  Suppose  that  in  this  case,  at  a  point  P,  the 
radiiis  of  curvature  is  r0,  and  that  when  the  spring  is  partly  wound 

up  there  is  at  P  a  radius  of  curvature  r,  then is  the  change 

of  curvature  at  P,  and  we  know  that  the  bending  moment  which 
produced  this  change  of  curvature  is  equal  to  E  i  ( V  where 

E  is  the  modulus  of  elasticity  of  the  material  and  i  is  the  moment 
of  inertia  of  the  cross-section.  (Thus,  taking  E  at  36,000,000,  if 
the  breadth  of  the  spring  is  0'2  inch  and  its  thickness  0'03,  then 
EI  is  16-2.)  Now  suppose  the  arbor  to  have  turned  through  the 
angle  x  o  Q  (which  we  shall  call  A)  from  the  unstrained  condition. 
What  are  the  forces  acting  on  P  M  and  the  arbor  ?  Whatever 
these  forces  may  be,  they  must  be  in  equilibrium.  If  these  forces 
were  changed,  there  would  be  an  alteration  in  the  shape ;  but  so 
long  as  these  forces  do  not  change,  the  shape  and  position  of  things 
do  not  alter.  This  is  why  we  can  apply  to  the  spring,  p  M,  and  the 
arbor  the  laws  of  forces  acting  on  rigid  bodies.  So  long  as  p  M  o 
does  not  alter  in  shape,  it  obeys  the  laws  of  rigid  bodies. 

522.  Now,  the  forces  acting  on  the  arbor  may  be  very  numerous 
— pressure  of  the  pivots,  pull  of  the  f  uzee  chain,  or  pressure  of  teeth 
of  wheels  ;  but  whatever  they  may  be,  we  know  that  they  can  be 
represented  by  one  force  acting  at  o,  the  centre,  together  with  a 
couple,  c.  If  the  spring  is  not  in  contact  with  the  top  or  bottom 
of  its  case,  and  if  the  coils  are  not  in  contact  with  one  another,  no 
other  forces  act  on  the  spring,  M  p,  except  at  p.  The  particles  of 


APPLIED    MECHANICS.  625 

steel  on  one  side  of  the  section  at  P  are  acting  on  the  particles  on 
the  other  side  ;  but  whatever  the  forces  at  each  of  the  particles 
may  be,  we  know  that  the  total  effect  at  P  is  the  same  as  that  of 
one  force  and  one  couple.  We  cannot  easily  say  what  the  force  is, 
but  if  r  is  the  radius  of  curvature  at  p,  and  if  r0  was  the  radius  of 
curvature  at  P  when  the  spring  was  unstrained,  then  the  couple  at 
p  is  what  we  have  already  called  the  bending  moment, 


12 


n        l\ 

\r  ~  fa)' 


Let  us  suppose,  for  simplicity,  that  the  spring  is  everywhere  of 
the  same  breadth  and  thickness,  and  let  us  use  the  letter  e  instead 

of  vy,  which  is  now,  of  course,  the  same  everywhere.  The  couple 
at  P  is  then  e  (  ---  Y  The  only  forces  acting  on  PMO  are: 

A  force  at  o,  of  amount  H,  in  the  direction  o  H,  say  ;  a  couple  at  o 
whose  moment  is  -  L  ;  a  force  at  P  ;  a  couple  at  p  whose  moment 
is  given  above  and  is  positive.  Now,  we  know  that  the  sum  of  the 
moments  of  all  the  forces  about  any  point  must  be  nothing.  Take 
all  the  moments  about  the  point  p.  The  force  at  p  has,  then,  no 
moment,  and  is  to  be  neglected,  and  we  have 

-    H    X    P  H  —  L   +   <?  (  ---  )  =  0. 

\f       r0/ 

In  fact,  e  (-  —  —  j  =  L  +  H  .  p  H. 

Let  P  Q  be  a  short  distance  measured  from  P  along  the  spring. 
Multiply  every  term  of  the  above  equation  by  P  a,  and  we  find 

(p  Q       P  oA 
-    I  =  L  .  P  Q   +   H.PH.PQ. 
r         r0  / 

Now,  when  p  Q  is  very  small  it  may  be  regarded  as  the  arc  of  a 
circle  whose  radius  is  r  ;  consequently  —  simply  means  the  angle 

between  the  radius,  or  normal  at  P,  and  the  normal  at  Q  ;  in  fact, 
it  means  the  small  angle  which  the  tangent  at  P  makes  with  the 

P  Q         P  Q, 

tangent  at  Q.     Thus  —  --   simply  means  the  change  which 

has  occurred  in  the  angle  between  the  direction  of  the  spring  at  p 
and  the  direction  at  Q.  If  now,  instead  of  considering  what  occurs 
at  the  point  p,  we  take  the  point  Q,  we  shall  get  just  a  similar 
equation  for  another  little  length  of  the  spring.  Suppose  we  do 
this  for  every  short  length  of  the  spring,  and  add  up  our  results  ; 

we  shall  find  that  the  sum  of  all  terms  such  as  —  —  —  means 

ro 

the  change  which  has  been  produced  in  the  angle,  between  the 
tangents  to  the  spring,  at  its  two  ends.  Thus,  suppose  the  arbor 
has  turned  through  the  angle  0,  and  suppose  that,  whether  or  not 
the  point  of  fastening  at  N  has  been  moved,  the  direction  of  the 


626 


APPLIED    MECHANICS. 

spring  at  N  has  on  the  whole  changed  through  an  angle  ft ;  then 
we  find  that  the  sum  of  all  the  above-mentioned  terms  amounts  to 
6  —  ft.  (d  may  be  called  the  amount  of  winding  up  of  the  spring  ; 
ft  may  be  called  the  amount  of  yielding  in  the  fastening  to  the 
case.)  Hence  the  sum  of  all  the  left-hand  sides  of  all  such  equa- 
tions as  the  above  is  e  (6  —.ft). 

Now  let  us  consider  the  right-hand  sides  of  the  equations. 
Evidently  the  sum  of  all  such  terms  as  L  x  P  Q  will  be  L  x  length 
of  spring  ;  say  L  I.  The  sum  of  all  such  terms  asn  x  PH  x  PQ 
is  (Art.  109)  equal  to  H  multiplied  by  the  length  of  the  spring 
multiplied  by  the  perpendicular  distance  of  the  centre  of  gravity 
of  the  spring  from  the  line  o  H.  This  is,  of  course,  the  length  of 
the  spring  multiplied  by  the  moment  of  the  force  H  about  the 
centre  of  gravity  of  the  spring.  Summing  up  our  results,  we  find 
that  if  the  force  on  the  arbor  through  the  pivots,  etc.,  has  a 
moment  about  the  centre  of  gravity  of  the  spring  of  the  amount  o, 
if  the  length  of  the  spring  is  I,  if  the  angle  turned  through  by  the 
arbor  from  the  unstrained  position  is  6,  and  if  ft  is  the  angular 
yielding  at  N,  and  L  is  the  couple  with  which  the  arbor  tends  to 

unwind  itself,  then  e  (0  —  0)  =  L  /  —  G  I,  or  L  =  -  (0  -  ft)  +  o. 

The  term  a  depends  on  the  position  of  the  centre  of  gravity  of  the 
spring. 

If  the  coils  are  numerous,  each  will  be  nearly  circular,  and  the 
centre  of  gravity  of  the  spring  will  nearly  be  at  o,  and  o  becomes 

insignificant ;  so  that  the  equation  becomes  L  =-  (0  —  ft).  If  the 
spring  is  so  rigidly  fastened  at  its  ends  that  there  is  no  change 
of  direction  relatively  to  the  barrel,  L  =r  -  0,  and  the  couple  exerted 

by  the  spring  in  trying  to  unwind  itself  is  simply  proportional  to 
the  amount  of  turning  of  the  arbor  or  the  amount  of  winding  up. 
If,  then,  the  centre  of  gravity  of  the  spring  always  remained  in 
the  centre  of  the  arbor,  and  if  the  spring  were  rigidly  fastened  at 
N  and  M,  we  should  have  the  couple  exerted  simply  proportional 
to  the  angle  of  winding;  and  this  is  the  condition  for  perfect 
isochronism  in  the  balance  spring  of  a  watch.  I  need  hardly  say 
that  this  condition  can  never  be  perfectly  satisfied.  If  we  use  a 
f uzee,  the  mainspring  may  be  fastened  as  we  please  ;  but  suppose 
we  want  the  couple  exerted  by  the  spring  to  be  nearly  constant  for 
various  amounts  of  winding  up,  it  is  evident  that  the  angle  ft 
ought  to  increase  as  fast  as  6 ;  that  is,  there  ought  to  be  a  very 
considerable  amount  of  yielding  in  the  fastening  of  the  spring  to 
its  case.  The  same  effect  will  be  produced  by  exerting  consider- 
able pressure  on  the  arbor  at  its  pivots,  or  in  some  way  causing 
the  arbor  and  its  case  to  be  not  quite  concentric  with  one  another. 
The  watchmaker's  usual  plan  to  get  moderately  good  iso- 
chronism is  to  make  one  of  the  above  errors  tend  to  correct 
another  ;  that  is,  by  allowing  a  greater  yielding  or  greater  stiffness 


APPLIED    MECHANICS, 


627 


of  the  outer  attachment  to  counteract  the  results  due  to  centre  of 
gravity  of  the  spring  not  remaining  exactly  in  the  axis  of  the 
balance. 

523.  Thus  we  see  that  by  applying  the  law  given  in  Art.  522 
to  a  flat  spiral  spring  fastened  to  a  case  at  its  outer  end,  x, 
and  to  an  arbor  or  axle  at  its  inner  end,  M,  we  find  that  if  the 
spring  is  riveted  firmly  both  at  N  and  M,  and  if  it  is  so  long  and 
its  coils  so  nearly  circular  that  its  centre  of  gravity  is  always 
nearly  in  the  centre  of  the  axle,  then,  when  partly  wound 
up,  the  spring  tends  to  unwind  itself  with  a  turning  moment 
which  is  proportional  to  the  amount  of  winding  up.  This  is 
the  case  in  the  balance  spring,  and  it  is  this  condition  that 
gives  to  the  balance  its  character  of  taking  almost  exactly  the 
same  time  to  make  a  small  swing  as  to  make  a  great  one. 
(See  Art.  455.)  When  the  end  N  is  not  riveted,  but  merely 
hinged  or  fastened  in  any  way  that  will  allow  it  to  turn  about 
N,  the  unwinding  tendency 
is  not  proportional  to  the 
amount  of  winding  up ;  it 
is  proportional  to  the  dif- 
ference between  the  angle 
of  winding  and  this  angu- 
lar yielding  at  N.  If  the 
strip  is  everywhere  of  the 
same  breadth  and  thick- 
ness, the  unwinding  tendency  is  proportional  to  the  moment  of 
inertia  of  its  own  section — that  is,  to  its  breadth  and  to  the 
cube  of  its  thickness  ;  it  is  also  proportional  to  the  modulus  of 
elasticity  of  the  material  used,  and  is  inversely  proportional  to 
the  total  length  of  the  strip.  Suppose  we  wind  a  cord  round 
the  barrel  or  case  containing  a  mainspring  of  a  watch  whose 
arbor  is  fixed  firmly,  and,  using  a  scale-pan  with  weights,  we 
find  the  turning  moment  of  the  spring  for  various  amounts  of 
winding  up.  If  we  plot  our  results  on  squared  paper,  we  shall 
find  that  the  points  lie  in  a  curve  like  A  o,  B  o,  c  o,  or  D  o  of 
Fig.  362,  whereas  for  a  balance  spring  we  should  get  nearly  a 
straight  line  through  o. 

In  Fig.   363  is  represented  an  instrument  which  I  have 
been  in  the  habit  of  using  in  my  laboratory,*  to  show  the  con- 
nection between  the  turning  moment  and  the  angular  wind- 
ing in  a  flat  spiral  spring.    Different  weights  used  at  the  end 
*  The  woodcutter  has  represented  too  large  a  weight  and  too  thin  a  spring. 


628 


APPLIED    MECHANICS. 


of  the  string  give  different  readings  of  the  pointer.  By  means 
of  such  an  apparatus  we  are  enabled  to  verify  the  laws  de- 
scribed above.  When  we  have  performed  one  set  of  experi- 
ments with  a  spring,  another  set  may  be  made  on  the  same 
spring  with  its  length  diminished  or  increased  by  means  of 
the  arrangement  for  clamping,  shown  in  Fig.  321.  In  this 
way  we  can  experiment  with  springs  of  different  breadths  and 
thicknesses,  as  well  as  of  different  materials. 

524.  The  flat  spiral  spring  just  considered  is  a  case  of  the  bending 
of  a  strip  of  steel  along  its  entire  length.  I  will  now  take  tip  a 
case  in  which  there  is  no  bending.  Fig.  364  shows  a  cylindric 
spiral  spring  whose  coils  are  very  flat.  Besides  its  own  weight,  it 


Pig.  363. 


is  acted  upon  by  two  equal  and  opposite  forces  in  the  direction  of 
its  axis,  the  supporting  force  at  N  and  a  weight  at  M.  Now  let  us 
consider  the  equilibrium  of  the  portion  of  the  spring  from  any 
point  P  to  M.  Suppose  the  wire  cut  at  P  by  a  plane  passing 
through  the  axis;  this  section  will  be  more  and  more  nearly  a 
cross-section  normal  to  the  axis  of  the  wire,  as  the  spirals  are 
more  and  more  nearly  horizontal.  Let  us  regard  it  as  a  normal 


APPLIED    MECHANICS. 


cross-section  of  the  wire.     Now,  whatever  may  be  the  stresses  at 

this  cross-section,  they  must  balance  all  the  other  forces  acting  on 

p  M — namely,  the  force  F  at  M,  which  is  axial,  and  the  weight  of 

p  M,  which  is  very  nearly  axial.     If  we  neglect  the  weight  of  p  M, 

we  have   only  to  balance  the  force  F  acting 

at  M.    To  do  so  we  evidently  need  a  shearing 

force,   F,   at    P,   distributed  over  the  section, 

and  a  twisting  torque  which  is  equal  to  F  .  p  H. 

It  is  easy  to  show  that  the  shear  is  of  much 

less  importance  than  the  torsion.     Indeed,  in 

many  ways  it  is  like  the  shearing  force  in 

beams  (see  Art.  369),  and  we  shall  neglect  it. 

Again,   since  P  H  is  the  same  for  every  part 

of  the  spring,  every  section  of  the  wire  is  acted 

on  by  the  same  twisting  couple,  just  as  the 

shaft  of  Fig.  191  or  the  wire  of  Fig.  186  and 

its  strength   is  calculated  in   the  same  \vay. 

Now,  what  is  the  amount  of  motion  at  M  in 

consequence  of  this  twist?     As  the  wire  is 

everywhere  twisted,  just  as  if  it  were  a  straight 

wire  fastened  at  one  end  whilst  at  the  other 

end  there  were  a  force,  F,  acting  at  the  end  of 

an  arm  whose  length  is  equal  to  P  H,  the  radius 

of  the  coils  of  the  spring,  the  amount  of  the 

motion  of  M  is  just  the  same  as  the  motion  of 

the  end  of  such  an  arm  attached  to  the  straight 

wire. 

525.  We  have,  then,  the  following  pretty 
illustration  (Fig.  365),  which  serves  to  keep 
the  rule  for  spiral  springs  in  our  memory. 
Let  two  pieces  of  the  same  wire  of  the  same 
length  be  taken ;  one  of  them  kept  straight, 
fixed  firmly  at  A,  and  fastened  at  B  to  the 
axis  of  a  pulley  which  can  move  in  roller 
bearings.  A  cord,  c,  fastened  to  the  rim  of 
this  pulley,  carries  the  upper  end  of  a  spiral 
spring,  D  E,  formed  of  the  other  piece  of  wire, 
the  diameter  of  its  coils  being  equal  to  the 
diameter  of  the  pulley.  Evidently,  if  a  weight, 
w,  is  placed  in  the  scale-pan,  a  point  E  gets 
just  double  the  motion  of  a  point  c,  for  E 
gets  c's  motion  as  well  as  the  lengthening  of 
the  spring.  The  scales  P  and  G  and  the  little 
pointers  are  for  the  purpose  of  making  exact  pi'  354 

measurements.     It  is  interesting  to  note  how 
accurately  the  law  is  fulfilled,  even  in  a  roughly-constructed 
piece  of  apparatus  such  as  anyone  may  easily  put  up  for  himself. 


630 


APPLIED    MECHANICS. 


Example. — A  spiral  spring  of  charcoal  iron  spring  wire,  0*1 
inch  diameter,  21*6  inches  long,  its  coils  having  a  radius  of 
1'3  inch,  is  extended  by  a  weight  of  10  Ibs.  Supposing  that 
a  piece  of  wire  of  the  same  material  1  inch  long  and  0'05  inch 


Pig.  &66. 


diameter,  gets  a  twist  of  24  •  2  degrees  with  a 
twisting  moment  of  2  inch-pounds,  what  is  the 
extension  of  the  spring?  We  see  that  if  the 
trial  wire  were  of  twice  the  diameter  the  twist 
would  be  24-2  -f-  16,  or  1'51  degrees,  and  with 
a  twisting  moment  of  13  inch-pounds,  which 
is  6 '5  times  as  great,  the  angle  would  be  9 '8  2 
degrees,  and  on  a  wire  21 -6  times  as  long  would 
be  212'  degrees,  or  3 '7  radians,  and  the  arc  of 
a  circle  whose  radius  is  1*3  inch,  subtending 
this  angle  is  3-7  x  1 '3  or  4 -8  inches,  the' 
answer. 

526.  In  designing  a  cylindric  spiral  spring  it  is 
very  important  to  know  the  greatest  elongation  it  will  bear  without 
taking  a  permanent  set.  If  the  material  has  internal  strains  given 
to  it  during  its  manufacture — and  this  it  is  very  difficult  to  prevent 
in  steel  springs,  unless  great  care  is  taken  in  tempering,  and  it  is 
almost  impossible  to  prevent  in  brass  springs,  because  the  elasticity 
added  in  manufacture  is  often  regarded  as  a  necessary  quality 
which  ought  not  to  be  destroyed  by  any  annealing  process — in  this 
case  the  reader  must  keep  in  mind  the  considerations  of  Art.  294a. 
Otherwise,  let  /  be  the  greatest  shearing  stress  per  square  inch 
which  the  material  can  resist  without  getting  a  permanent  set. 
Let  M  be  the  greatest  twisting  moment  which  a  round  wire  of 
diameter  d  can  bear  without  getting  a  permanent  set.  We  see 
from  Art.  296  that 

M  =  «P//16     .  .  .  (1). 

Now/ will  be  approximately  known  from  Table  XIV.,  or  M  may  be 
found  by  experiment  for  a  given  wire  by  any  person  who  wishes 
to  make  a  spring ;  and  whether  M  or  /  is  used  in  a  formula,  you 
now  know  how  to  calculate  one  when  given  the  other  and  the  size 
of  the  wire.  If,  then,  we  have  a  spring  made  of  wire  whosti 


APPLIED    MECHANICS.  631 

diameter  is  d,  and  if  the  radius  of  the  coils  as  measured  to  the 
centre  of  the  wire  from  the  axis  of  the  spring  is  r}  we  see  that 
when  w  is  the  greatest  weight  with  which  the  spring  may  be 
elongated  without  producing  a  permanent  set, 

W  =  -  =  veflf/16  r  .  .  .  .  (2), 

being  independent  of  the  length  of  wire  employed. 

From  Art.  295  we  see  that  if  N  is  the  modulus  of  rigidity  of  the 
material,  T'  the  greatest  angular  twist  in  radians  which  we  can 
give  to  a  wire  of  diameter  d  inches  and  length  1  inch,  and  m'  the 
twisting  moment  which  produces  this  twist,  then 


m'  being  what  we  have  previously  measured  or  calculated.  N  is 
approximately  known  for  a  material  from  Table  XIV.,  or  T'  may  be 
found  by  experiment  for  a  given  wire  ;  and  whether  T'  or  N  is 
used  in  a  formula,  you  now  know  how  to  calculate  one  when  given 
the  other. 

Putting  the  result  of  our  reasoning  in  Art.  524  into  an  algebraic 
form,  we  see  that  a  load,  w,  will  elongate  the  spring  by  the  amount 

' 


and  hence  the  greatest  elongation  which  can  be  given  to  the  spring 
without  its  getting  a  permanent  set  is 

32  Irm'        2lrf 

xf  =.  IT'I\  or  ---  -r,  or  —  f  ....  (5). 
IT  N  d*  '        N  d 

Combining  (2)  and  (5),  we  see  that  when  a  spring  is  stretched 
to  its  elastic  limit,  the  mechanical  energy  stored  up  in  it,  which  is 
called  it^"  resilience,"  being  half  the  product  of  w',  the  proof  load, 
into  the  proof  elongation,  is 


,  ..... 

527.  Many  interesting  methods  may  be  taken  to  express  in  words 
the  meanings  of  these  results.     Thus  the  second  expression  in  (6) 
shows  that  the  work  which  we  can  store  up  in  a  spiral  spring  is 
simply  proportional  to  the  weight  or  quantity  of  material  in  it.    It 
would  be  easy  to  show  that  we  can  store  more  energy  in  a  spring 
formed  of  wire  of  circular  section  than  in  one  of  equal  weight  of 
the  same  material  whose  wire  has  any  other  than  a  circular  section. 

528.  The  following  readings  of  our  formulae  may  prove  to  be 
useful  :  —  1st.  If  d,  the  diameter  of  the  wire,  and  r,  the  radius  of 
the  coils,  be  fixed,  the  elongation  produced  by  any  weight,  w,  will 
be  proportional  to  I,  the  length  coiled  up  to  form  the  spring.. 
2nd.  If  a  wire  of  a  certain  length  and  diameter  be  given  to  form 
a  spring,  the  elongation  produced  by  a  certain  weight,  w,  will  be 
proportional  to  the  square  of  the  diameter  which  we  may  adopt  for  the 
coil.     3rd.  If  the  diameter  of  the  wire  be  fixed,  and  the  axial  length 
of  the  spring,  when  closed,  so  that  the  coils  may  touch  one  another, 
or,  what  is  the  same,  the  number  of  coils  be  also  fixed,  I  must  be 
proportional  to  «,  and  therefore  the  elongation  due  to  a  weight,  w, 


632  APPLIED    MECHANICS. 

will  be  proportional  to  the  third  power  of  the  radius  which  we  may 
adopt  for  the  coil.  4th.  If  the  length  of  the  wire  and  the  radius 
of  the  coil  be  fixed,  the  elongation  due  to  a  weight,  w,  will  be 
inversely  proportional  to  the  fourth  power  of  the  diameter  of  the 
wire  which  we  may  adopt.  5th.  With  a  given  weight  of  metal 
and  a  given  radius  of  the  coil,  the  elongation  due  to  a  weight,  w, 
will  be  proportional  to  Z3,  or  inversely  to  d6,  since  /  must  be 

proportional  to  -^. 

We  see  that  the  ultimate  elongation  is  —  1st,  proportional  to 
the  length  of  the  wire,  if  the  diameter  of  the  wire  and  the  radius 
of  the  coil  be  fixed;  2nd,  proportional  to  the  radius  of  the  coil, 
if  the  length  and  the  diameter  of  the  wire  be  fixed  ;  3rd,  inversely 
proportional  to  the  diameter  of  the  wire,  if  the  length  of  the  wire 
and  the  radius  of  the  coil  be  fixed. 

It  will  be  found  that  a  weight  hung  at  M  (Fig.  364)  will  tend 
to  turn  as  the  spring  lengthens,  unless  the  coils  of  the  spring  are 
very  flat.  This  is  due  to  the  fact  that  the  cross  -sections  of  the  wire 
are  really  subjected  to  a  little  bending  as  well  as  torsion. 

529.  We  can  cause  the  strain  in  such  a  spring  to  consist  altogether 
of  bending,  if,  without  exerting  any  axial  force  such  as  I  have 
shown  in  Fig.  364,  we  exert  a  couple  about  the  axis  such  as  we 
exerted  on  the  wire  in  Fig.  186.  The  wire  in  Fig.  186  would  be 
twisted,  but  the  wire  in  Fig.  364  is  subjected  everywhere  to  bend- 
ing without  any  twisting,  or  with  only  a  very  little  twisting,  due 
to  the  fact  that  the  coils  are  not  perfectly  flat. 

If  a0  is  the  radius  of  the  coils  to  the  centre  line  of  the  wire 
when  unstrained,  and  the  length  of  the  coiled  wire  is  I,  then  the 
number  of  coils  multiplied  by  the  circumference  of  each  is  the  total 
length,  so  that  the  number  of  coils  is  /  -j-  2  va0.  If  now  the 
moment  of  inertia  of  the  cross-section  of  the  wire  about  the  axis 
through  its  centre  about  which  it  bends  is  i,  and  if  M  is  the  moment 
which  acts  at  the  unfixed  end  of  the  spring  to  twist  it,  then  the 
new  radius,  a,  of  every  coil  is  obtained  from  o  ur  knowledge  of  the 
fact  given  in  Art.  325. 

M  =  E  i  X  change  of  curvature, 

*=!-!..  ..(1), 
BI       a       a0 

E  being  the  modulus  of  elasticity  of  the  material.  I  is  Id3  -5-12 
for  a  wire  of  rectangular  section,  d  being  the  dimension  in  inches 
of  the  section,  measured  radially  out  from  the  axis  of  such  a  spring  ; 
i  is  vd4  -i-  64  for  a  wire  of  circular  section  of  diameter  d. 

Now  I  -T-  2  ira  would  be  the  number  of  windings  ;  so  that,  if 
n  is  the  new  number  and  n0  the  old  number  of  windings,  we  have 


But  one  additional  winding  means  2  IT  radians  ;  so  that  if  0  is  the 
amount  of  winding  up  corresponding  to  2  v  (n  —  «0),  we  have 


APPLIED    MECHANICS.  633 

We  see  that  it  does  not  depend  upon  the  radius  of  the  coils,  and  is, 
therefore,  the  same  formula  as  given  in  Art.  522  for  a  flat  spiral 
spring  whose  radius  varied  continually.  E  i  is  called  the  flexural 
rigidity  of  the  wire,  being  E  ^3/12  for  a  strip  of  thickness  / ;  being 
B  ird*/64:  for  a  circular  section  of  diameter  d ;  being  E  s*/l2  for  a 
square  of  side  *.  The  strength  is  as  that  of  a  beam  subjected  to 
the  bending  moment  M. 

530.  From  these  considerations  it  is  evident  that  a  spiral 
spring  like  Fig.  364,  when  it  lengthens  under  the  action  of  a 
weight,  has  all  its  wire  subjected  to  torsion.      The  spring 
itself   is  extended,   but   the  wire   of   the  spring   is   twisted. 
Again,  if  we  subject  the  spring  to  torsion  as  a  whole,  the 
strain   really    going   on    in   the   wire   is    a    bending   strain. 
Usually,  a  spiral  spring,  as  its  coils  are  not  perfectly  flat,  has 
its  wire  subjected  to  torsion  principally,  and  a  little  bending 
as  well,  when  the  spring  is  extended ;    and  when  the  spring  is 
twisted  as  a  whole  its  wire  is  mainly  subjected  to  bending,  but 
there  is  also  a  little  twist  in  it.      The  extension  of  a  spiral 
spring  is  proportional  to  the  pulling  force,  and  also  to  the 
length  of  the  wire  and  to  the  square  of  the  diameter  of  the 
coils ;    it  is  inversely  proportional  to  the  fourth  power  of  the 
diameter  of  the  wire  if  the  wire  is  round.     The  twist  given  to 
a  spiral  spring  as  a  whole  is  proportional  to  the  moment  of  the 
twisting  forces — it  does  not  depend  on  the  size  of  the  coils ; 
it  is  proportional  to  the  length  of  wire,  and  inversely  pro- 
portional to  the  fourth  power  of  the  diameter  of  the  wire  if 
the  wire  is  round. 

531.  We  have  taken  up  at  length  the  two  cases  of  spiral 
springs  in  which  the  angle  of  spiral  is  0.     It  will  be  found 
that  when  the  angle  is  not  0,  the  stiffness  of  the  cylindric 
spiral  spring  follows  much  the  same  law  as  for  springs  of 
small  angle,  but  it  is  necessary  to  take  up  the  general  case. 

632.  The  theory  of  the  cylindric  spiral  spring  is  enough  to 
study,  because  each  small  portion  of  any  spiral  spring  may  be  re- 
garded as  part  of  a  cylindric  spiral  spring.  A  rough  model  will 
help  a  student  to  understand  the  work  better. 

We  shall  imagine  the  upper  end  of  a  vertical  cylindrio  spiral 
spring  to  be  held  fixed,  and  that  at  the  other  end,  by  means  of  a 
rigid  arm  coming  in  from  the  wire  to  the  axis  of  the  spring,  we  are 
able  to  apply  an  axial  force  F,  by  means  of  a  weight,  tending  to 
elongate  the  spring,  and  also  a  couple,  L,  about  the  axis  of  the 
spiral  tending  to  increase  the  number  of  coils ;  the  directions  of 
elongation  axially  and  of  greater  winding  up  are  our  positive 
directions  of  motion.  Let  the  axial  elongation  produced  be  called 
<r,  and  the  angular  rotation  produced"  be  called  <f>.  We  shall 


634  APPLIED    MECHANICS. 

neglect  the  weight  of  the  spring  itself  as  it  will  be  quite  easy 
afterwards  to  correct  for  this. 

Let  r  be  the  radius  of  the  coils—  that  is,  the  distance  of  the 
centre  of  the  wire  everywhere  from  the  axis.  Let  the  whole  length 
of  wire  be  I,  the  angle  of  the  spiral  a.  Let  B  be  the  flexural 
rigidity  of  the  wire  in  the  osculating  plane  of  the  spiral.  B  =  E  i 
where  E  is  Young's  modulus  of  elasticity,  and  i  is  the  moment  of 
inertia  of  the  section  of  the  wire  about  the  line  through  its  centre 
of  gravity  which  touches  the  cylindric  surface,  and  is  at  right 
angles  to  osculating  plane.  The  bending  moment,  divided  by 
the  flexural  rigidity,  gives  the  change  of  curvature  produced 
by  bending.  Let  A  be  the  torsional  rigidity  of  the  wire. 
The  twisting  moment  applied  to  a  wire  divided  by  A  gives  the 
angle  of  twist  produced  per  unit  length  of  wire.  In  Table  XV.  a 
number  of  values  of  A  and  B  are  given  for  sections  of  wire  which 
are  in  common  use  in  springs.  A  is  the  torsional  rigidity  of  the 
wire,  being  the  twisting  moment  required  to  produce  unit  angle  of 
twist  per  unit  length,  u  is  the  flexural  rigidity  of  the  wire  in  tha 
osculating  plane  of  the  spiral,  being  the  bending  moment  required 
to  produce  unit  change  of  curvature  in  that  plane.  The  line  P  Q 
represents  the  axis  of  the  spiral  relatively  to  the  wire. 

N  is  the  modulus  of  rigidity,  and  E  the  Young's  modulus  for 
the  material.  In  the  last  two  cases  t  is  supposed  to  be  small  in 
comparison  with  b.  Notice  that  A  in  the  elliptic  sections  becomes 
NirDd3/16  when  d  is  small  in  comparison  with  D,  and  it  becomes 
N£tf3/3  in  the  rectangular  sections  when  t  is  small.  Coulomb's 
wrong  assumption  was  that  A  =  N  i  when  i  is  the  moment  of 
inertia  of  a  section  about  its  centre.  Now,  for  the  elliptic  section 

I  =  57,  (ifld  +  D^3),  and  hence 


Coulomb's  * 


*i  +  D»  =     /        i 

D3  d6  *\         x 


correct  A 

if  a?  is  i>/d.  Coulomb's  value  is  correct  when  the  section  is  circular, 
and  we  see  that  it  is  more  and  more  wrong  as  the  section  gets 
flatter  and  natter.  The  true  value  of  A  is  always  easy  to  calculate 


APPLIED    MECHANICS. 
TABLE    XV. 


635 


I 
j 


A. 

Torsional 
rigidity. 


NTT^4 

~32~ 


NIT 

16"  D2 


NTT       D8^3 
"16"  D2   + 


0-14058  N*4 


3  V^TP 


I2  +  ?:' 


B. 

Flexural 
rigidity. 


W. 

Axial  load  when  /is 
greatest  shear  stress. 


64 


12 


12 


12 


r  =  radius  of  coils) 


636  APPLIED   MECHANICS. 

in  the  case  of  an  elliptic  section.  Its  value  for  a  rectangular 
section  is  calculated  from  an  infinite  series,  and  it  is  therefore 
very  important  to  know  that  Cauchy  has  proved  that  the  torsional 
rigidity  of  a  rectangle  bears  (approximately)  to  the  torsional 
rigidity  of  an  inscribed  ellipse  the  proportion  of  their  moments  of 
inertia  in  the  case  when  D  is  several  times  d  (see  Art.  313). 

Consider  the  portion  of  spring  below  any  cross-section  p.  This 
portion  is  in  equilibrium.  Hence  the  molecular  forces  exerted  on 
it  at  the  section  p  must  balance  F  and  L,  and  these  are  the  only 
conditions  which  we  find  it  necessary  to  consider.  Let  Fig.  366 
represent  the  elevation  of  a  portion  of  the  wire  below  the  section  p. 
Let  T  T  be  the  elevation  of  the  axis  of  the  spring.  Then  about  the 
axis  PM  we  know  that  the  moment  due  to  the  force  P  must  be 
balanced  by  a  torque  of  molecular  forces  whose  amount  is  F  r ;  and 
about  the  axis  PT  there  must  be  a  torque  of  molecular  force  of 
amount  L.  Resolving  these  in  the  directions  p  s  and  p  u  in  the 
plane  of  the  paper,  which  is  tangential  to  the  cylindric  surface  at  p, 
we  have  about  p  s  the  torque 

F  r  sin.  a  -  L  cos.  a, 
and  about  p  u  we  have 

»  r  cos.  a  +  L  sin.  a. 

Now  the  moment  about  PS  means  a  bending  moment  which 
produces  an  angular  change  per  unit  length  whose  amount  is 

rr  sin.  a  —  L  cos.  a  ... 

~~T~ 

and  the  moment  about  p  u  means  a  twisting  moment  which  pro- 
duces an  angle  of  twist  per  unit  length  of  the  amount 

F  r  cos.  a  +  L  sin.  a  /ox 

— —        -  ....  (2). 

I^et  us  now  consider  how  these  two  angular  changes  in  unit 
length  of  the  wire  cause  elongation  and  rotation  at  the  bottom  end 
of  the  spring.  If  the  spring  is  sufficiently  long,  it  is  obvious  that 
an  axial  elongation  x  and  a  rotation  <J>  will  occur  at  the  free  end  of 
the  spring,  for  we  can  then  imagine  that  the  lateral  motions  due  to 
all  the  elements  of  the  wire  exactly  counteract  one  another.  It  is 
therefore  only  necessary  for  us  to  obtain  from  the  above  expressions 
those  elements  which  produce  x  and  <J>.  Now  any  rotation  of  the 
body  below  p  about  any  axis  can  be  resolved  into  equivalent 
rotations  about  other  axes,  according  to  the  laws  for  the  resolution 
of  vectors  generally. 

The  rotation  (1)  about  the  axis  PS  is  equivalent  to  (1)  multiplied 
by  cos.  o  about  p  T,  and  to  (1)  multiplied  by  sin.  a  about  p  M.  The 
rotation  (2)  about  the  axis  p  u  is  equivalent  to  (2)  multiplied  by 
sin.  o  about  the  axis  P  T,  and  to  (2)  multiplied  by  cos.  a  about  p  M. 
Adding,  we  get,  on  the  one  hand,  the  rotation  about  p  T,  produced 
by  the  flexural  and  torsional  strains  in  unit  length  of  the  wire ; 
and  this  multiplied  by  /,  the  total  length  of  the  wire,  gives  <J>. 
Adding,  we  get,  on  the  other  hand,  the  rotation  about  p  M,  produced 
by  these  same  flexural  and  torsional  strains ;  and  it  is  obvious  that 
this  rotation  causes  a  point  on  the  axis  of  any  portion  of  the  spring 


APPLIED    MECHANICS.  637 

below  p  to  be  lowered  by  a  distance  which  is  obtained  by  multiply- 
ing the  rotation  by  r,  the  radius  of  the  spiral.  Multiplying  by  /, 
we  obtain  the  whole  axial  lengthening  of  the  spring  —  that  is, 
putting  our  answer  in  its  simplest  shape, 

d>  /I       1\  /sin.2a       cos.2a\ 

*  =  Pr  sm.  a  cos.  a  (;  -  -)  +  L  (—  +  _)  .  .  .  .  (3). 

x  /cos.2a       sin.2o 

_  =  rr^__r  +  _ 

<f>  and  x  being  the  rotation  and  elongation  produced  in  a  spiral  spring 
by  an  axial  force  F  and  a  couple  L  acting  together.  In  the  theory 
<f>  and  x  are  assumed  to  be  very  small.  If  we  use  8  </>  and  8  x  for 
them,  using  <f>  for  2  v  n  where  n  is  the  number  of  turns,  and  x  for 
the  axial  length  of  the  spring,  so  that  xjn  is  what  we  sometimes 

call  the  pitch  of  the  spiral  ;    then  -  -f-  2  ir  r  =  tan.  <f>  ....  (5) 

and  sin.  <£>  =  x\l  .  .  .  .  (6),  and  by  means  of  integration  we  can  find 
accurately  the  effect  of  alteration  in  r  and  a  as  the  spring  changes 
in  shape.  Taking  (3)  and  (4)  as  they  stand,  however,  I  find  that 
young  students  can  have  no  better  easy  mathematical  exercises 
than  are  to  be  obtained  by  working  problems  on  them. 

For  example,  let  I,  r,  o,  A  and  B  for  a  spiral  spring  be  given,  we 
can  calculate  x  and  <f>  when  given  F  and  L,  or  if  given  x  and  <p  we 
can  calculate  P  and  L. 

The  very  common  case  of  a  small,  say  o  =  0,  leads  to 
<t>  =  L//B  .  .  .  .  (7) 

*  =  *&*/  A....  (8). 

Thus  if  the  spirals  are  very  flat,  whatever  the  nature  of  the  section 
of  the  wire  may  be,  the  spring  when  subjected  to  axial  force  has  no 
tendency  to  rotate  at  its  end,  and  the  spring  when  subjected  to  a 
couple  merely  has  no  tendency  to  alter  in  axial  length. 

Again,  in  any  given  spring,  suppose  we  are  informed  that  the 
rotation  of  the  end  is  prevented,  and  that  a  force  F  acts,  putting 
0  =  0  we  find  L  in  terms  of  F  from  (3),  and  using  this  in  (4)  we 
have  x  in  terms  of  F.  Again,  as  in  chronometer  springs,  if  the 
axial  elongation  is  prevented  and  only  a  torque  L  acts,  put  x  =  0 
in  (4)  and  find  F  in  terms  of  L  ;  use  this  in  (3)  to  find  <f>  in  terms  of  L. 


Let  a  =  ^  or  45°.     (3)  and  (4)  become 


;-;    ....(10). 

If  we  apply  to  (9)  and  (10)  the  first  condition  given  above,  that 
is,  </>  =  0,  and  an  axial  force  F  acting  (9)  gives  L  =  -  F  r  B  ~  A 
so  that  (10)  gives 


638 


APPLIED    MECHANICS. 


Again,  if  we  apply  to    (9)  and  (10)  the  second  condition  given 
above,  that  is,  x  —  0  and  a  torque  L  acting,  (10)  gives 

B  -  A 

fr  =  -  L 


so  that  (9)  gives 


B  +  A 


B  +  A* 

.  .  .  (12). 


In  estimating  for  general  purposes  the  effect  of  altering  the 
angle  o  ;  or  the  effect  of  constraint,  such  as  preventing  rotation  of 
the  end  of  a  spring  when  an  axial  force  is  applied ;  or  the  effect  of 
change  of  shape  of  section,  etc.,  it  is  useful  to  remember  that  for 
most  substances  we  may  take  it  that  approximately  E  =  2-5  N.  It 
is  very  interesting  to  study  the  formulae  (8),  (10),  (12),  using  the 
Table  XV.  for  the  values  of  A  and  B  for  various  forms  of  section. 
We  shall  leave  this  study  to  each  student  for  himself — working  out 
only  the  following  interesting  examples. 

Students  will  please  work  carefully  the  following  fifteen 
exercises  :—Take  N  =  10-2  x  106,  E  =  25-5  x  106.  Apply  an 
axial  load  of  1  Ib.  and  find  the  elongation  in  each  case.  There  are 
10  springs  of  wire,  100  inches  long ;  diameter  of  coils  2  inches.  The 
first  has  round  wire  of  0-1  inch  diameter.  All  the  others  are  of 
such  sizes  of  section  that  there  is  the  same  total  volume  of  material. 
Half  the  spirals  have  an  angle  of  45°.  "When  the  exercises  are 
finished,  divide  all  the  elongations  by  that  of  the  first  spring,  and 
enter  the  results  on  such  a  table  as  the  following : — 

TABLE  XVI. — RELATIVE  ELONGATIONS  FOR  SAME  AXIAL  LOADS. 


a  =  o 

o=  45° 

Whether  the  end 
is  allowed  freedom 
to  rotate  or  not. 

End  free  to 
rotate. 

End  restrained 
from  rotating. 

Solid  round  wire. 

1 

0-9 

0-889 

Hollow  circular  section, 
thickness  of  tube  ^th 
of  outside  diameter. 

0-196 

0-197 

0-195 

Square  wire. 

1-132 

0-948 

0-912 

Rectangular  wire,  as  in 
Fig.  6,  Table  XV. 
Breadth  4  times  thick- 

2-029 

1-108 

0-349 

Rectangular  wire,  as  in 
Fig.  7.  Breadth  4  times 
thickness. 

2-029 

2-544 

2-440 

APPLIED    MECHANICS.  639 

In  the  theory  of  the  spiral  spring  I  considered  the  hend  and 
twist  given  to  each  small  portion  of  a  spring,  and  assumed  that  the 
resultant  action  at  the  bottom  of  the  spring  was  a  rotation  and  an 
axial  elongation — that  is,  that  the  lateral  motions  produced  on  the 
bottom  by  all  angular  motions  of  portions  of  the  spring  exactly 
balanced  one  another.  As  a  matter  of  fact,  however,  it  is  only  when 
the  spring  is  very  long  that  these  lateral  actions  balance.  To  take 
the  lateral  actions  into  account  is  of  no  practical  importance  in  such 
springs  as  we  have  been  considering  (cylindric  spiral  springs) ;  but 
when  the  spires  of  a  spring  rapidly  change  their  character,  particu- 
larly in  the  case  of  flat  spiral  springs,  the  lateral  actions  are  of 
importance  (see  Art.  522). 

533.  If  any  important  object  were  to  be  served,  I  would  calculate 
the  relative  strengths  of  all  the  springs  under  the  various  condi- 
tions here  mentioned.  It  is  only  necessary  to  know  the  load  P, 
which,  when  producing  a  bending  moment  F  r  sin.o  and  a  twisting 
moment  prcos.a  in  a  wire  of  the  particular  sec-tion  at  the  same 
time,  shall  be  capable  of  just  producing  permanent  set.  For  a 
circular  section  this  is  well  known — that  is,  it  is  well  known  that 
for  a  circular  section  the  above  two  moments,  acting  together,  are 
equivalent  to  a  twisting  moment 

rr  sin.a  +  x/F2r2  sin.2a  +  r2^2  cos.2a. 
or 

F  r  (1  +  sin.  a) 

if  the  material  has  equal  strength  to  resist  shearing  and  tensile 
stresses.  This  is  merely  (3)  of  Art.  298.  We  see,  then,  that 
for  a  round  wire  the  breaking-load  for  &  spring  of  angle  a 
is  less  than  for  one  of  the  same  size  of  wire  and  length  of  wire 
and  diameter  of  coils,  but  quite  flat  in  its  spirals,  in  the  ratio 
1  :  (1  +  sin.o.)  Thus  a  spring  with  a  =  45°  has  a  strength  only  a 
little  over  one-half  of  a  spring  with  flat  coils.  This  is  the  reason 
why,  when  a  weight  has  been  hung  from  a  spring  which  produces 
permanent  set,  the  spring  so  very  rapidly  gets  completely  spoilt, 
oven  although  there  is  a  counteracting  influence  due  to  the  coils 
becoming  smaller  in  diameter. 

It  is  sufficient  for  most  practical  purposes,  then,  to  say  that  if 
the  load  to  produce  permanent  set  in  a  round  wire  spring  is  1  when 
the  coils  are  flat,  the  load  which  will  produce  permanent  set  when 
the  coils  are  not  flat  is 

1  -T-  (1  +  sin.  a). 

Again,  in  springs  with  flat  coils,  all  of  radius  1",  the  loads  to 
produce  permanent  set  are  simply  equal  to  the  twisting  moments 
which,  when  applied  to  the  wire,  produce  permanent  set;  and 
hence  the  following  table  is  very  useful. 

w,  in  the  last  column  of  Table  XV.,  gives  the  load  which  will 
produce  the  maximum  shear  stress /in  each  of  the  sections  of  wire 
in  the  case  of  cylindric  spiral  springs,  the  angle  of  the  spiral  a 
being  0.  Hence,  if  /  is  the  proof  shear  stress  of  the  material,  w 
is  the  proof  load.  The  student  will  calculate  the  numbers  of 
Table  XV.  as  an  exercise 


640 


APPLIED    MECHANICS. 


In  the  following  table  the  load  to  produce  permanent  set  and 
the  resilience  are  given  for  cylindric  spiral  springs,  so  that 
students  who  are  in  the  habit  of  only  calculating  the  strength  and 
stiffness  of  springs  with  round  wire  may  be  able  to  arrive  easily  at 
the  strength  and  stiffness  of  springs  made  with  other  kinds  of  wire. 

Comparison  of  the  loads  which  will  produce  permanent  set  in 
springs  of  the  same  diameter  of  coils  and  same  area  of  cross-section 
of  wire :  in  all  of  them  a  =  0 — that  is,  the  coils  are  supposed  to 
be  as  flat  as  possible.  This  is  really  a  comparison  of  the  torsional 
strength  moduli  of  sections  of  the  same  area.  Let  students 
calculate  the  various  numbers  here  given. 

TABLE    XVII. 


Load  to  Produce 
Permanent  Set. 

Resilience  per  Cubic  Inch. 

1 

1 

Hollow    circular,    thickness  of  } 
tube  JQ  of  outside  diameter     / 

Square 

2-733 
0'740 

SMore  and  more  nearly 
2,  as  tube  is  thinner 
and  thinner. 

0-fiSft 

Rectangular  wire,  as  in  Fig.  6.  ) 
Breadth,  4  times  thickne*  ...  J 

0-556 

0-628 

Rectangular,     as     in    Fig.     7.  ) 
Breadth,  4  times  thickness  ...  j 

0-556 

0-628 

641 
CHAPTER    XXIX. 

RESILIENCE     OP     SPRINGS. 

034  An  examination  of  our  formulae  will  show  that  for  nearly 
round  wire,  even  when  o  is  considerable,  the  rotation  <£  depends  almost 
altogether  on  the  couple  L,  and  the  elongation  x  depends  almost 
altogether  on  the  axial  force  p.  But  when  the  section  of  the  wire 
is  very  different  from  circular,  the  angle  a  enters  very  materially 
into  the  calculation.  Many  examples  of  great  interest  may  be 
taken  up.  Nearly  flat  coils  are  presumed. 

1.  If  r  alters  in  a  spring,  and  if  we  want  the  resilience  of  the 
material  to  be  the  same  everywhere  ;  that  is,  the  wire  being  round, 
if  we  want  the  material  to  be  equally  ready  to  break  everywhere, 

we  must  make  —  constant.    That  is,  if  the  coils  get  twice  as  large, 

the  diameter  of  the  wire  becomes  ^/  2,  or  1'26  times  as  large. 

2.  Round  wire  all  of  same  diameter  d,  but  the  coils  capable  of 
lying  just  within  one  another  if  the  spring  is  compressed  by  axial 
force. 

Let  n  be  the  number  of  coils  to  any  point  starting  from  the  end 
of  the  wire  at  the  small  coil  side,  where  the  radius  is  r0.  We  may 
take  r  =  r0  +  ndy  I  =  r02im  +  4  iPrfid  ;  and  adding  together  the 
elongations  produced  on  each  elementary  length  into  which  we  may 
imagine  the  wire  divided,  we  find 


The  strength  of  this  spring  is  to  be  calculated  as  if  all  the  coils 
were  equal  in  size  to  the  largest. 

Some  students  may  perhaps  be  interested  in  working  the 
problem  :  —  Find  the  cylindric  spiral  spring  which,  made  of  the 
same  wire  with  the  same  number  of  turns,  will  receive  the  same 
axial  elongation  or  compression  from  the  same  axial  force. 

The  answer  is,  if  n  is  the  total  number  of  turns,  R  is  the  radius 
of  coils  in  the  new  spring, 


Thus,  as  a  numerical  example,  if  the  first  and  last  coils  were  4  and 
9  inches  in  radius,  d  =  1  inch,  n  =  5  turns,  R  is  6'8  inches. 

535.  Resilience.  —  The  average  resilience  per  cubic  inch  of 
a  spring  is  the  whole  resilience  divided  by  the  volume.  It  is 
only  in  the  case  of  uniform  stress  and  strain  in  the  material 
everywhere  that  we  have  maximum  resilience  for  the  whole 
spring.  In  actual  cases  this  only  occurs  in  tie-rods  and  struts, 
and  in  spiral  springs  made  of  a  wire  which  is  a  thin  tube. 


642  APPLIED    MECHANICS. 

|  M  B  is  the  resilience  per  unit  length  of  a  wire,  if  M  is  the 
proof  twisting  moment  and  6  is  the  proof  angle  of  twist,  or  if 
M  is  the  proof  bending  moment  and  6  is  the  proof  change  of 
curvature  produced.  It  is  easy,  therefore^  to  work  out  the 
following  table  of  values,  and  every  student  ought  to  do  this 
as  an  exercise  : — 

TABLE  XVIII.— RESILIENCE  PER  UNIT  VOLUME  IN  INCH-POUNDS. 

Simple  compression  or  extension.     Only  convenient  in  1    ,/2 ,QQ 

springs  made  of  indiarubber    ...         ...         ...         ...  J  ~E  ~ 

Bending  if  the  maximum  stress   is  reached  in   every  ^ 

section,  and  section  is  rectangular.     As  in  beams  of  |  ^2 

uniform  strength,  in  well-made  carriage  springs,  and  }>  £  —  =     133 
in  spiral  springs  subjected  to  a  torque  only ;  also  [ 

C-springs.     Art.  370     j 

Bending.     Uniform  strip  fastened  at  one  end  and  loaded  )  ^2 

at  the  other,  or  supported  at  the  ends  and  loaded  in  >  ^  —  =      44 

middle;     ...        ...     '  ...        )  E 

Simple  shearing.     Possible  in  springs   of  indiarubber ;  )  ^ 

also  possible  in  spiral  springs  made  of  thin  tubes  >  \  —  =1,000 

circular  in  section          )  N 

Torsion.    As  in  spiral  springs  of  round  wire  subjected  to  \  i  /^  __    500 

axial  loads          }  *  ^ 

Torsion.     As  in  spiral  springs  of  square  wire  subjected  \  .ic^/^ 469 

to  axial  forces.  }          N  ~~~ 

Torsion.  As  in  spiral  springs  of  oval  or  rectangular  wire  subjected 
to  axial  forces,  anything  less  than  for  circular  depending  on  ratio  of 
diameters  or  of  breadth  to  thickness. 

536.  The  student  may  not  yet  have  been  sufficiently  im- 
pressed with  the  importance  of  knowing  the  resilience  per  unit 
volume  of  the  material  in  all  the  springs. 

The  resilience  per  unit  volume  tells  us  the  value  of  a 
particular  shape  of  spring. 

Suppose  that  a  spiral  spring  of  any  given  material  is  to  be 
used  for  any  purpose. 

The  greatest  load  is  stated,  and  the  elongation  or  com- 
pression due  to  that  load.  Then  whatever  other  conditions 
may  be  given  as  to  the  coils  being  of  the  different  sizes, 
etc.,  we  know  that  the  spring  may  be  made  of  any  shaped 
section  of  wire,  but  that  the  thin  tubular  circular  section 
is  the  very  best  and  the  solid  circular  section  is  half  as 
good,  and  any  other  sections  are  not  half  as  good,  and  that  the 
spiral  spring  form  is  better  than  any  other. 

In  any  spring,  if  a  force  F  produces  a  known  motion  x  in 


APPLIED    MECHANICS.  643 

its  own  direction,  then  the  resilience  per  unit  volume,  multi- 
plied by  the  volume  of  material,  gives  the  proof  load  F'  multi- 
plied by  half  the  motion  x'  produced  by  it.  This  principle 
enables  easy  calculations  to  be  made. 

Again,  if  in  any  spring  a  torque  L  produces  a  known 
angular  motion  0  in  its  own  direction,  then  the  resilience  per 
unit  volume,  multiplied  by  the  volume  of  material,  gives  the 
proof  load  I/  multiplied  by  half  the  angular  motion  <f>'  produced 
by  it. 

In  nearly  any  case  we  consider,  it  will  be  found  that  the 
important  thing  looked  for  when  we  choose  a  particular  shape 
of  spring  is  total  resilience — total  energy  stored  up. 

Thus  for  example :  We  want  a  spring  to  exert  a  force  p' 
and  only  to  alter,  say,  w  lb.,  for  a  change  of  shape  indicated  by 
a  motion  in  the  direction  of  P'  by,  say,  b  inches.  Evidently  here 
the  law  of  stiffness  of  the  spring  is  b  oc  w,  say  b  =  k  w,  where 
k  is  a  known  number. 

And,  therefore,  if  x'  is  the  greatest  motion,  x'  =  -  p' ;   and 

the  total  resilience  is  -=-  p'  x'  or  «  —  p'2. 
2  2  w 

That  is,  we  are  given  p',  w,  and  b,  and  therefore  the  total 
resilience  of  the  spring,  and  if  we  know  the  type  of  the  spring 
we  can  find  the  volume  of  the  substance  required  and  therefore 
its  weight.  What  people  generally  mean  by  the  "springi- 
ness "  of  a  spring  means  that  it  shall  not  change  much  in  the 
force  with  which  it  acts,  for  a  considerable  amount  of  alteration 

in  shape.     Now  b  is  the  change  of  shape  and  —7  is  the  fractional 

p 

change  of  shape,  so  that  —  p'  represents  the  springiness  of  a 
spring.  But  the  value  of  a  spring  also  depends  upon  the 
greatest  force  it  can  exert.  That  is,  its  value  depends  on  -  p'&; 

that  is,  on  its  resilience. 

Looked  at  from  almost  any  point  of  view,  we  see  that  the 
value  of  a  particular  form  of  spring  is  represented  mainly  by 
its  resilience.  Of  course,  its  shape  and  cost  of  manufacture  are 
also  of  importance.  We  may  know  that  a  tubular  spiral  spring 
may  be  the  best  for  weight,  and  yet  a  C-spring  shape  may  be 
best  suited  to  the  use  to  which  it  is  to  be  put,  and  we  put  up 


644  APPLIED    MECHANICS. 

with  the  disadvantage  of  using  from  twenty-two  to  twenty-eight 
times  the  weight  of  metal  because  of  some  other  convenience. 

537.  If  the  angle  of  the  spiral  is  a,  the  length  of  each  turn,  instead 
of  being  2?rr,  as  we  take  it  in  approximate  calculations,  is  really 
2  TT  rf  sin.  o.  If  the  axial  length  of  a  spring  is  x,  the  angle  a  is 
such  that  cos.  a  =  x  \l.  If  the  axial  length  changes  to  x  +  x,  the 
angle  changes  to  a1,  such  that  cos.  o1  =  (x  +  z)/l.  We  have  not 
taken  these  matters  into  account,  assuming  that  our  elongation? 
were  small  or  that  our  calculations  were  to  be  only  approximate. 
The  student  who  knows  a  little  calculus  may  use  ?  x  and  5  <£  instead 
of  x  ana  </>  in  Art.  532,  and  by  integration  obtain  general  and 
accurate  expressions. 

A  conical  spiral  spring  of  round  wire  of  diameter  d,  whose 
spires  vary  gradually  from  greatest  radius  r1  to  smallest  r0 ;  the 
proof  load  is  evidently  to  be  calculated  from  r^  or  w1  =  ird3f/l6ri. 
The  elongation  for  a  given  load  w  ought  to  be  calculated  for  short 
lengths  and  added  up.  Mathematically  it  is  evident  that 


0 

Taking  it  that  if  s  is  distance  from  the  end  of  the  wire  where  r  =  r0, 
r  =*  r0  +  s  r*  ~  r°,  and  it  is  easy  to  show  that 


In  fact,  we  see  that  in  a  conical  spring,  instead  of  r2  for  a  cylindric 
spring  we  take 


EXERCISES. 

1.  If  a  conical  spring  is  formed  of  round  wire  0*2  inch  diameter,  40 
inches  long,  the  coils  varying  from  rl  =  2  inches  to  r0  =  1  inch,  what  is 
the  proof  load,  and  the  axial  shortening  with  this  load  ? 

Ans.,  47  Ibs.  if/=  60,000;  0-4  inch. 

2.  An  Ayrton-Perry  spring  is  made  of  strip  section,  as  in  Table  XV., 
which  nearly  covers  a  cylindric  surface,  and  a  =  45°,  so  that  Ib  =  2  *r  x  if 
x  is  the  axial  length,  and  x  =  I  sin.  a,  or  If  V%.     Hence  Ib  =  2-jrrl/  Vz. 
b  =  it  </2r.     If,  then,  r  =  O'l  inch,  b  =  -44  inch.     If  t  =  -001  inch,  and 
we  have  been  able  to  obtain  very  broad  strips  of  steel  of  this  thinness, 
A  =  -0019  B  =  -00132  according  to  Table  XV.     Hence  a  load  F  will  pro- 
duce an  elongation  x  =  12-8  Fl  and  a  rotation  0  =  —  11-5  F/.    It  is  to  be 
noticed  that  the  strip  section  of  wire  in  a  spiral  spring  (as,  for  example, 
in  what  are  sometimes  called  volute  springs  for  buffers)  is  very  wasteful 
of  metal  for  ordinary  purposes. 

3.  Spring  of  strip  4  inches  parallel  to  the  axis  0-25  inch  thick.     If  a 
is  taken  to  be  0,  and  the  coils  lie  inside  one  another,  touching ;  if  there 
are  n  turns,  the  smallest  of  radius  r0  and  the  largest  of  radius  rj,  the 


APPLIED    MECHANICS.  645 

thickness  of  the  strip  "being  t,  then  the  average  radius  is  ^  (TI  -f-  r0),  the 
length  is  I  =  IT  (rj  +  »*0)»  and  w£  =  rl  -  r0,  so  that   I  =  —  (r^  -  r02) 

Hence,  as  when  i  is  small  compared  with  b,  w  of  column  3,  Table  XV., 
becomes  bfif/3r,  we  have  in  this  case 

wi  =  ^2//3r!  ....  (1). 
(8)  of  Art.  532  gives  for  the  shortening  under  a  load  F,  x  =  F  lt*/-z  bt3, 

O 

and  as  in  the  case  of  our  conical  spring  we  must  take  instead  of  the 
constant  r2  the  value  £  (r^  +  »'ir0  +  r02),  we  have 

»  =  F|  (n«  -  r.")  i  ft"  +  rxr0 


EXERCISES  ON   SPRINGS. 

Take  proof  /=  140,000  Ibs.  per  square  inch,  N  =  13  x  106  Ibs.  per 
square  inch  for  the  best  spring  steel. 

1.  A  spring  of  coils  4  inches  in  radius  is  of  round  steel,  1  inch  in 
diameter  and  12  feet  long.      What  are  its  proof  axial  load,  its  proof 
shortening  or  lengthening,  and  its  resilience  ? 

2.  A  spring  of  round  steel  wire  the  radius  of  whose  coils  is  2  inches 
is  to  be  20  inches  long  when  its  coils  lie  close  together  ;  it  is  to  elongate 
2  inches  for  a  load  of  400  Ibs.,  and  this  is  to  be  its  proof  load.     Give  its 
dimensions. 

Here,  if  n  is  the  number  of  coils,  n  2  trr  =  I  nearly,  nd  =  20,  so  that 

1/2  irr  =  20/d  ____  '(I), 

400  =  Trrf3  x  14  x  104/16r  .  .  .  .  (2), 

2  =  32  x  400JrYirNd4  ----  (3). 


Wo  have  only  to  find  J,  r,  d  from  equations  (1),  (2),  and  (3). 

3.  A  safety-valve  spring  of  square  steel  wire  is  to  shorten  0-4  inch 
when  the  pressure  of  steam  is  150  Ibs.  per  square  inch,  the  effective  area 
of  the  valve  then  being  12  square  inches.  The  radius  of  the  coils  is 
2  inches.  A  load  of  400  Ibs.  per  square  inch  on  the  valve  produces  the 
proof  load  on  the  spring.  Find  I  and  d. 


646 


CHAPTER   XXX. 

CARRIAGE    SPRINGS. 

538.  Carriage  springs  usually  consist  of  strips  of  steel  in 
contact,  as  shown  in  Fig.  367.     They  are  fastened  at  the  ends 
and  middle  to  the  objects  through  which  the  loads  are  applied, 
in  various  ways,  which  must  be  examined  by  the  student  in 
actual  examples.     The  ends  of  the  strips  are  usually  shaped, 
as   in    Fig.    368    or   Fig.    369.      These   plates,    before   being 
tempered,  are  curved  very  accurately  to  a  template;    whilst 
dark  red  at  the  end  of  this  shaping  they  are  dipped  into  cold 
water,  the  two  ends  entering  the  water  first  and  then  the  whole 
plate  being  lowered  beneath  the  surface,  so  that  if  there  is  a 
variation  in  the  hardness  at   different   places   it  shall  be  a 
symmetrical  variation.     Each  plate  has  a  little  pin  or  feather 
which  enters  a  slot  in  its  neighbour,  so  that  the  plates  shall  not 
be  laterally  displaced  relatively  to  one  another.     Sometimes 
this  is  effected  by  making  a  short  wrinkle  or  corrugation  length- 
wise at  the  end  of  each  plate.     A  bolt  passes  through  holes  in 
the  middles  of  all  the  plates,  and  when  its  nut  is  tightened  up 
the  plates  are  made  to  lie  closely  against  one  another.     When 
springs  of  the  general  shape  of  Fig.  367,  or  of  half  of  it,  are 
made  of  one  solid  piece  instead  of  a  number  of  plates,  the 
calculation  of  strength  and  stiffness  is  made  by  the  method  of 
Art.  339. 

539.  Carriage  springs  are  usually  tested  by  means  of  a 
hydraulic  press  or  a  steam  press  which  forces  them  to  become 
quite  straight  three  or  more  times,  and  a  spring  is  thought  to  be 
satisfactorily  made  if  after  this  it  is  found  not  to  have  taken 
any  permanent  set.     When  the  student  works  the  following 
exercises,  he  will  see  that  if  the  greatest  load  is  that  which  pro- 
duces straightness  in  all  the  strips,  their  initial  curvatures 
ought  to    be    the    same.      This  will   be   found  also  to  give 
sufficient  tightness  when  the  plates  are  bolted  up. 

I  think  that  carriage  spring  makers  and  buyers  are  too 
particular  in  their  wishing  to  have  all  the  plates  lying  very 
tightly  together  when  bolted  up.  There  is  too  much  of  such 
tightness.  Usually,  too,  in  actual  use  a  carriage  spring  never 
becomes  unloaded,  and  its  plates  are  therefore  always  very 


APPLIED    MECHANICS.  647 

much  more  tightly  pressing  together  than  when  the  spring  is 
examined  unloaded.  If,  however,  such  tightness  is  really 
thought  to  be  necessary,  and  if  to  obtain  it  we  must  have  such 
great  differences  in  curvature  as  I  have  found  in  springs  by 
the  best  makers,  it  would  be  advisable  to  make  the  shorter 
plates  thinner  than  the  rest. 

But  if  all  the  plates  are  of  the  same  thickness,  they  ought 
certainly  to  have  the  same  curvature  before  bolting  up.  If  the 
student  draws  a  set  of  such  plates,  he  will  see  that  when  they 
are  bolted,  up  they  will  be  sufficiently  tight,  and  if  such  a 
spring  is  straightened  the  stresses  in  all  the  plates  will  be 
exactly  the  same. 

540.  The  following  exercises  on  bending  are  to  be  worked 
by  the  student  to  lead  him  to  the  theory  of  these  springs. 
These  exercises  deal  almost  altogether  with  the  best  construc- 
tion of  carriage  springs. 

EXERCISES. 

1.  A  beam  of  constant  strength  ;  what  is  its  curvature  everywhere  ? 
Ans.,  the  strength,  modulus  z  is  i  -^  \d  if  d  is  the  depth  at  any  place 

and  i  is  the  moment  of  inertia  of  the  cross-section  there.  Hence  —  d  =/, 
a  constant  for  all  sections,  being  the  greatest  stress  in  all,  or  —  =*  —  • 

El        tid 

and  this  is  the  curvature. 

2.  If  a  beam  has  had  the  same  change  of  curvature  everywhere,  show 
that  if  it  is  of  constant  strength  it  must  be  of  constant  depth. 

3.  If  a  beam  of  rectangular  section  of  constant  breadth  and  variable 
depth  d  is  loaded  at  one  end  and  fixed  at  the  other,  what  must  its  depth 
everywhere  be  for  the  curvature  to  be  constant  ? 

Ans.,  if  a;  is  distance  from  the  load  w,  —  -        must  be  constant,  and 


therefore  d  oc  z^/lc. 

4.  In  the  last  example,  what  is  the  greatest  stress  in  each  section  ? 

-—j-p  =/,  so  that  /«  jp  or  /«  —  ,   or  a  x>3-.     In  fact,  /  is  simply 

proportional  to  the  depth,  if  the  depth  is  proportional  to  the  cube  root 
of  x.  If  we  compare  the  answers  to  exercises  (2)  and  (4),  we  see  that  the 
overlapping  parts  in  carriage  springs  are  most  economically  made  of 
constant  depth  but  varying  breadth. 

5.  n  strips,  each   of   thickness  £,  make  up   a  carriage  spring  if   the 
length  of  the  top  strip  (or  the  length  of  the  curve  A  c  B)  is  2  I  inches  ;  the 
overlap  being  A,  I  =  n\.     If  o  c  is  small  compared  with  ri,  the  radius  of 
curvature  of  tbe  top  strip,  00  =  ^/2  r\  nearly.     Hence  change  in  o  c  = 
£  I2  x  change  of  curvature  of  top  strip. 

V* 


648  APPLIED    MECHANICS. 

6.  If  when  loaded  the  radius  of  the  top  strip  is  R,,  instead  of  Iteing 
infinite,  what  is  the  radius  of  every  strip  when  the  spring  is  taken  apart 

if  the  resilience  is  to  be 
uniform?  Also  what 
is  rlt  the  radius  of  the 
top  strip  in  the  un- 
loaded spring  ?  This 

pig.  867.  is      a      more     tedious 

exercise,    but    is    easy 
enough, 
Ant.  First,  if  the  radius  of  the  top  strip  when  separate  is  pi,  pi  is 

known,  because  ---  =  2//B  t,  and  /is  the  maximum  stress. 

pi         HI 

J._I,_i__i  m 

RI      pi      RI  +  st      Ps°       '  v  ' 

where  ps  is  the  radius  of  the  sth  strip  when  separate.  We  now 
have  the  interesting  problem  :  When  a  number  of  strips  whose 
radii  when  free  are  known,  are  fastened  together  to  form  a  spring, 
what  will  n  now  be  ?  The  radius  of  the  sth  strip  was  Ps  when 
free,  and  is  now  r\  +  st,  and  so  its  change  of  curvature  is 

—  -.      Hence  the  force  vrt   at  its    end,  producing  the 

bending  moment  Ws  A,  must  be  such  that  —  —  produces  this  change 
of  curvature.  Hence 


We  must,  therefore,  write  out  equation  (1)  for  every  strip  to  find 
the  values  of  pa  for  each,  in  terms  of  pi,  which  is  known,  and  also 
equation  (2),  thus  calculating  every  value  of  ws,  the  unknown  TI 
being  in  every  term.  Now  we  make  the  statement  that  the  sum  of 
all  the  values  of  ws  is  0,  and  this  enables  r\  to  be  found.  Thus  (2) 
and  (1)  give 

EI/      1  1        1  1 


and  we  must  find  i\  by  inserting  1,  2,  3,  etc.,  for  *  in  the  equation 


If  st  is  small  compared  with  r\  and  RI,  we  find  that,  if  there  are  n 
strips  altogether, 


a  quadratic  to  find  r\. 

It  is  not  necessary  to  pursue  the  subject  farther.     Students 
have  the  means  of  working  all  sorts  of  practical  problems, 


APPLIED  MECHANICS.  649 

541.  Tempering  Carriage  Springs. — In  the  last  part  of  tho 
work  on  the  strips,  a  plate  being  red  hot,  by  an  interesting 
manipulation  between  several  pairs  of  smiths'  tongs  held  by  two 
workmen  it  is  fitted  to  lie  everywhere  close  to  a  curved  piece  of 
metal,  so  that  it  receives  a  definite  shape.     It  is  now  a  dark 
red,  and  the  workman  dips  the  plate  into  water,  letting  the 
two  ends  go  in  simultaneously  and  the  middle  of  the  plate  last. 
The  plate  is  now  placed  in  an  air  furnace,  where  it  gradually 
heats ;  it  is  taken  out  once  or  twice  and  rubbed  with  a  piece  of 
partly  charred  wood.     The  experienced  workman  can  tell  by 
the  nature  of  the  smoke  coming  from  the  wood  whether  the 
temperature  of  the  plate  is  high  enough ;  when  it  has  been 
heated  to  a  sufficiently  high  temperature  it  is  withdrawn  from 
the  air  furnace  and  allowed  to  cool,  lying  on  a  metal  table  with 
others  which  have  preceded  it. 

542.  Tempering  Spiral  Springs. — The  processes  employed  by 
Messrs.  Salter  are,  unfortunately,  unknown  to  me.    In  all  cases, 
however,  they  probably  consist  in  winding  wire  or  rod  in  the 
red  hot  state  on  a  smooth  iron  mandrel,  shaping  the  end  parts 
of  the  spring,  if  the  wire  is  small,  by  pliers ;  if  large,  by  special 
tools  which  can  readily  be  designed  for  special  cases.     The 
formed  spring  is  now  heated  to  a  dull  red  heat  in  an  air 
furnace  and  plunged  into  hot  oil,  just  of  such  a  temperature  as 
will  produce  the  blue  colour  peculiar  to  spring  steel.     It  is  in 
this  dipping  of  the  hot  steel  into  the  oil  bath  that  any  trade 
secret  can  exist.     If  all  the  steel  could  from  the  same  instant 
and  at  the  same  rate  lose  its  heat,  so  that  the  hardened  steel 
would  be  uniformly  hard  everywhere,  the  process  would  be 
perfect.     Large  spiral  springs  for  safety-valves  are  dipped  so 
that  their  axes  remain  vertical.     To  lay  them  sideways  into 
the  bath  might  cause  the  lower  half  of  each  coil  to  cool  more 
quickly  than  the  upper  half.     It  is  quite  possible  that  good 
spring  makers  may  not  only  dip  their  springs  axially  but  also 
give  to  them  a  rotatory  motion  round  their  axes  as  they  enter 
the  bath. 

543.  Very  small  spiral  springs  for  the  balances  of  chrono- 
meters and  watches  require  special  care,  whether  they  are  cylin- 
dric  or  flat  spiral  springs.     This  is  on  account  of  the  thinness 
of  the  material  and  the  rapidity  with  which  it  may  cool.     They 
are  usually  enclosed  in  a  box — very  small  springs  are  enclosed 
in  a  platinum  box — either  on  their  mandrels  if  cylindric,  or 
coiled  up  together  with  others  if  flat,  surrounded  by  powdered 


650  APPLIED    MECHANICS. 


charcoal.  The  box  is  heated  to  redness  for  a  sufficient  time  to 
ensure  the  springs  inside  being  all  red  hot,  and  the  box  is  then 
immersed  in  water.  The  little  box  is  manipulated  by  means  of 
a  long  handle.  It  is  now  placed  upon  an  iron  pan  placed  over 
a  flame,  and  lying  beside  the  box  on  the  pan  is  a  piece  of 
brightened  steel.  As  the  pan  gets  heated  the  brightened 
piece  of  steel  takes  different  colours,  and  it  is  presumed  that 
the  steel  springs  have  about  the  same  temperature.  When  the 
dark  blue  colour  is  reached,  which  is  so  characteristic  of  spring 
steel,  the  box  is  removed  from  the  pan  and  allowed  to  cool. 

544.  Tempering  Flat  Spiral  Springs. — In  many  cases  these 
springs  are  not  shaped  before  tempering.      Great  lengths  of 
steel   strip   are    passed    through    an    air    furnace   from    one 
set  of  rollers  to  another,  at  a  rate  which  depends  upon  the 
particular  purpose  for  which  the  strip  is  designed,  upon  its 
breadth  and  thickness,  upon  whether  it  is  Bessemer  or  crucible 
steel.     Just  when  in  the  red  hot  state  and  leaving  the  air 
furnace,  the  strip  is  passed  through  a  vessel  through  which 
cold  water  is  kept  circulating,  and  is  consequently  made  very 
hard.     It  then  passes  between  pieces  of  cotton  waste  kept  well 
soaked  in  oil,  and  then  over  a  flame  which  keeps  the  oil  ignited ; 
and  on  leaving  this  region  the  strip  is  gradually  allowed  to 
cool  before  being  wound  upon  a  roller.     If  allowed  to  move  too 
slowly,  the  strip  and  the  oil  covering  its  surface  are  for  too 
long  a  time  subjected   to  the  heating  effects  of  the  second 
furnace,  and  the  steel  becomes  softer.     I  have  seen  strips  which 
were  said  to  be  continuous  and  a  mile  long  which  had  been 
tempered  in  this  way. 

The  very  fine  strip  steel  which  is  used  in  the  Waterbury 
Watch  is  tempered  in  this  way.  Great  quantities  of  cheap 
Bessemer  steel  strip  for  use  in  ladies'  corsets  and  numerous 
other  purposes  are  also  tempered  in  this  way. 

545.  I  have  never  seen  this  process  carried  out  by  electrical 
heating  as  a  substitute  for  the  air  furnace,  but  it  is  obviously 
quite  easy  to  make  the  substitution.     The  strip  or  wire  to  bo 
tempered  receives  an  electrical  current  from  a  certain  roller  A, 
over  which  it  moves,  and  another  roller  B  at  a  short  distance 
from  the  first ;  and  at  B,  or  even  before  it  reaches  B,  the  steel 
passes  into  an  oil  bath.     In  this  case  we  have  no  sudden  great 
hardening  and  then  a  softening  process ;  the  tempering  is  all 
done  in  one  operation,  a  cooling  from  red  heat  to  the  tempera- 
ture of  hot  oil.     As  the  wire  or  strip  is  still  kept  heated  when 


APPLIED    MECHANICS.  651 

in  the  oil,  the  cooling  is  gradual,  and  obviously  the  softness  of 
the  Steel  will  depend  on  how  much  of  it  is  kept  heated  in  the 
oil  and  on  the  average  temperature  of  the  oil. 

546.  In  well-made  carriage  springs  the  resilience  is 

;r  —  per  cubic  inch. 
6    E  r 

and  in  designing  a  spring  a  knowledge  of  this  fact  enables  us 
somewhat  to  shorten  the  work. 

Thus,  suppose  a  spring  is  wanted  to  take  a  proof  load  of 
3  tons,  with  a  deflection  of  3  inches  just  making  it  straight. 
The  total  resilience  is 

£  x  3  x  2,240  x  3,  or  10,080  inch  pounds. 

If/'  =  30,000  and  E  =  30,000,000,  the  resilience  per  cubic 
inch  may  be  5  inch  pounds,  so  that  the  volume  of  steel  required 
for  the  spring  is  2,016  cubic  inches.  Now  if  there  are  n  plates, 
this  volume  may  be  taken  as 

n  b  tl  (I  being  half  the  length  of  the  longest  strip), 

mnbtl=  2,016  ....  (1), 
which  is  one  equation  towards  the  solution  of  some  problem. 

EXERCISES. 

1.  What  ought  the  initial  curvature  of  a  |-inch  plate  to  be  if  the 
proof  stress  of  the  material  is  30,000  Ibs.  per  square  inch  (exists  when  the 
plate  is  straight)  and  E  is  30,000,000  P 

Here,  by  formula  of  Art.  540, 

30,000=15,000,000  x  £«>, 
w  =  '004. 

The  radius  of  curvature  is  -^^  =  250  inches. 

2.  If  the  above  plate  is  36  inches  long,  what  is  its  initial  dip,  oct  in 
Fig.  367  ?     Using  the  approximate  formula  oc  =  I2  •*•  2  r,  we  have 


0-64  inch. 

3.  If  the  spring  is  formed  of  six  plates,  the  longest  being  36  inches, 
the  overlap  on  one  side  of  plate  on  its  neighbour  is 

36  -j-  12,  or  3  inches. 

4.  If  the  breadth  of  each  plate  is  3  inches, 

D  =  -000259  w. 

5.  The  load  w'  which  will  produce  the  deflection  0*64  inch — that  is, 
which  will  straighten  the  spring — is 


652  APPLIED    MECHANICS. 

6.  How  many  plates  of  f  inch  thick  and  3  inches  hroad,  the  longest 
being  30  inches  in  length,  of  the  ahove-mentioned  steel  will  he  required 
for  a  spring  whose  proof  load  is  to  he  1  ton  ? 

2,240  =  >>x>,x*x 

o    X    oU 

or    n  =  8. 
Hence  the  spring  ought  to  he  made  of  eight  plates. 

547.  The  answers  to  the  above  exercises  guide  us  as  to  the 
best  way  of  constructing  carriage  springs  so  that  the  strips' 
throughout  shall  have  the  strength  of  the  overlapping  parts. 
The  following  rules  are  based  upon  the  dimensions  of  these 
overlapping  parts,  which  are  merely  little  cantilevers. 

If  \  is  the  overlap,  Fig.  367,  and  I  is  the  half-length  of  the 
longest  strip,  so  that  n  being  the  number  of  strips  n\  =  I , 
then  t  being  thickness  of  each  strip,  b  its  breadth,  d  the  deflec- 
tion of  the  spring,  f  the  maximum  stress,  E  Young's  modulus, 
w  the  load  at  A  and  at  B,  there  being  a  load  2  w  at  the  middle, 
it  is  evident  that 

/=  6  w  I  +  n  b  P,  D  =  6  w  P/n  E  b  ts. 

If  y  is  the  proof  stress,  the  resilience  in  a  well-made  spring 
is/"2  -7-  6  E  inch  pounds.  Carriage  springs  are  usually  made  of 
Bessemer  steel,  and  we  may  take  f  =  30,000  Ibs.  per  square 
inch  and  E  =  30,000,000. 

Examples. — 1.  If  the  overlap  X  is  2  inches,  and  there  are 
10  strips  each  f  inch  thick,  so  that  the  half-length  of  the  whole 
spring  is  I  =  20  inches,  t  =  f ,  and  if  b  =  2  J  inches  ;   find  the 
load  which  will  deflect  the  spring  2  inches.     Here  D  =  2 
2  =  6  w  20*  -T-  10  x  3  x  107  x  2J  x  (1)3,  and  hence 
w  =  5-926  x  105  Ibs. 

If  the  plates  were  not  of  the  same  initial  curvature — that 
is,  if  the  spring  was,  in  this  respect  only,  badly  made,  then  the 
law  for  deflection  is  still  true  for  the  badly  made  spring  ;  but 
the  rules  for  strength  are  untrue.  We  only  calculate  the 
increase  in  f  due  to  a  load  w. 

548.  In  the  above  theory  I  have  assumed  no  friction  between 
the  plates.    This  friction  makes  the  bending  to  be  less  for  a  given 
load  if  the  load  has  been  increasing,  but  if  the  load  has  been 
diminishing  the  bending  will  be  greater  than  the  formula  (1) 
gives,  on  account  of  friction.  When  we  test  a  carriage  spring, 
the  effect  of  friction  in  causing  the  deflections  to  be  greater  with 
diminishing  than  with  increasing  loads  is  very  noticeable.     I 
have  already  pointed  out  the  unsuitableness  (on  this  account) 


APPLIED    MEC&AtflCS.  653 


of  these  springs  for  all  measuring  purposes,  and  how  suitable 
they  are  for  carriages  and  under  other  conditions  where  vibra- 
tions must  be  rapidly  stilled  by  frictional  forces. 

549.   In  a  spring  such  as  Fig.  367,  if  2  w  is  the  upward  load 


Fig.  368.  Fig_  m 

applied- at  the  middle,  the  downward  supporting  forces  at  A  and 
B  are  each  w,  and  the  end  of  each  plate  may  be  regarded  as 
applying  to  the  plate  above  it  this  force  w.  Thus  in  Fig.  368, 
if  A  B  is  the  overlap  of  one  plate  on  another,  every  part  of  the 
plate  except  the  overlap  part  A  B  at  each  end  is  subjected  to  a 
bending  moment  w  .  A  B  everywhere,  so  that  the  change  of 
curvature  everywhere  is  the  same. 

Also  the  part  A  B  ought  to  be  fashioned  like  a  beam  of 
uniform  strength  and  curvature  fixed  at  B  and  loaded  at  A. 
We  find  from  Exercise  2,  Art.  540,  that  we  can  make  it  of 


f 


Fig.  370.  Pig.  371. 

uniform  strength  and  uniform  curvature  at  the  same  time  by 
varying  the  breadth  of  the  plate  but  not  its  thickness,  the 
proper  shape  being  shown  in  Fig.  370.  This  is  roughly 
approximated  to  in  many  springs  where  the  ends  are  shaped 
as  in  Fig.  368. 

If  we  keep  the  breadth  constant,  as  in  Fig.  369,  but  alter 
the  thickness  so  as  to  have  constant  curvature  everywhere 
(so  that  the  end  part  of  one  plate  may  fit  properly  against 
another  plate),  we  see  from  Exercises  3  and  4  that  the  thick- 
ness ought  to  vary  as  the  cube  root  of  the  distance  from  A  ; 
but  it  is  not  possible  to  have  also  uniform  strength  in  this 
part  of  the  plate. 


APPENDIX  I. 

TABLE   XIX.— USEFUL  CONSTANTS. 

Time.  One  sidereal  day  =  86,164  seconds. 
Mean  solar  day  =  86,400  seconds. 
One  year  =  365 '24224  mean  solar  days. 

Length.     British  standard,  the  yard  =  3  feet  =  36  inches. 
1  chain  =  66  feet  ==  100  links  =  4  perches. 
1  mile  =1,760  yards  =  5,280  feet  =  80  chains  =  8  furlongs. 
1  nautical  mile  =  6,080  feet  (average). 

=  10  cables  =  1,000  fathoms  (nautical). 
Telegraph  poles  220  feet  apart. 
1  fathom  =  6  feet. 

French  standard,  the  metre  =  39*37  inches  =  3-2809  feet. 

1  kilometre    =  '6214  miles. 
1  inch  =  -0254  metre  =  2-54  centimetres. 
1  foot  =  30-48  centimetres. 
1  yard  =  '9144  metres. 
1  mile  =  1,609  metres  =  1-609  kilometres. 

Surface.     1  square  inch  =  6 '451  square  centimetres. 
1  acre  =4,840  square  yards. 

1  square  mile  :=  640  acres  =  2-59  square  kilometres. 

Volume.     1  cubic  inch  =  16-387  cubic  centimetres. 

1  cubic  foot  =  -02832  cubic  metre  =  28-31  litres. 
1  litre  =  -22  gaUon. 

For  many  practical  calculations  it  is  sufficiently  correct  to  take  :— 
Length.     1  inch  =  25  millimetres,  but  25-4  is  more  correct. 

1  metre          =  3  feet  3  inches  and  |ths  of  an  inch. 
10  metres       =11  yards. 
20  metres       =  1  chain. 
8  kilometres  =  5  miles. 

Surface.     1  square  inch    =  6£  square  centimetres. 
6  square  yards  =  5  square  metres. 
1  acre  =  4,000  square  metres. 

Volume      4  cubic  yards  =  3  cubic  metres. 
1  gallon  =  4J  litres. 

4  litres  =  7  pints. 

100  litres         =  22  gallons. 
1  cubic  foot     =  28-3  litres. 

Weight.     1  gramme  =  15  £  grains. 

10  kilogrammes        =  22  pounds. 
50  kilogrammes        =  1  hundredweight. 
1,000  kilogrammes  =  1  English  ton. 

Speed.     1  knot=:l  nautical  mile  per  hour  =  1-15  miles  per  hour  =1-7 
ft.  per  sec.  =  101  ft.  per  min.  =51 -5  centimetres  per  sec. 


APPLIED    MECHANICS.  655 

60  miles  per  hour  =  88  feet  per  second. 
Weight  and  Force.     British  engineer's  unit  of  force  =  the  weight  of  the 

standard  pound  in  London. 

Weight  of  lib.  =  16  ozs.  =7,000   grains  =  453-6    grammes  = 
445,000  dynes  =  -4536  kilogrammes.     1  oz.  =  28-35  grammes. 
Weight  of  1  grain  =  63-57  dynes. 
1  kilogramme  =  2-2046  Ibs. 
1  ton  =  2,240  Ibs.  =  1,016  kilogrammes. 

1,000  kilogrammes  =  1  ton  (metric)  =  2,205  Ibs.  =  '9842  ton  (British), 

g  =  981  centimetres  per  second  per  second. 
=  32-2  feet  „  „ 

Value  of  g  at    London  =  32-182  feet  per  second  per  second. 

„      Equator  =  32-088     „ 

„         „   the  Poles  =  32-253     „  „  „ 

Inertia  or  mass  of  a  body  =  weight  in  Ibs.  at  London  -j-  32-2. 

1  gallon  of  water  at  62°  F.  weighs  10  Ibs.  by  English  law. 
1  cubic  foot  of  water  weighs  62-3  Ibs. 

1  cubic  foot  of  air  at  0°  0.  and  1  atmosphere  weighs  -0807  Ib. 
1  cubic  foot  of  hydrogen  „  „  „  „     '00557  Ib. 

For  academic  calculations  we  usually  take  the  weights  in  pounds  of 
1  cubic  foot  of  each  of  the  following  to  be  : — 
Brickwork,  112;  concrete,  150;  grindstone  and  Portland  stone, 
131  ;  granite  and  marble,  164  ;  dry  oak,  58 ;  dry  fir,  47 ;  cork, 
15;  coal,  80;  clay,  120. 

Weights  per  cubic  inch  of  each  of  the  following  materials,  in 
pounds — cast  iron,  '26  ;  wrought  iron,  -28  ;  steel,  '28  ;  brass,  '3  ; 
copper,  -32;  bronze,  '3;  lead,  -4;  tin,  -27;  zinc,  -26. 

Work,  Energy.     1  erg  =  1  dyne  x  1  centimetre. 
1  gramme- centimetre  =  981  ergs. 
1  foot  pound  =  1  -356  x  107  ergs. 

1  kilogramme  metre     =  7  "2  3  3  foot  pounds. 
1  horse-power  =33, 000  ft.  Ibs.  per  min.  =  550  ft.  Ibs.  per 

sec.  =  7  '46  x  109  ergs  per  sec.  =  746  watts  (watts  = 

volts  x  amperes). 

1  kilo- watt  =  1,000  watts  =  1-34  horse  power. 
1  force -de-cheval  =  -9863  horse-power. 

Energy  obtainable  from  1  Ib.  of  coal  =  8,500  Centigrade  heat 
units  =  15,000  Fahrenheit  heat  units  =  12  x  106  foot  pounds. 
Joule's  equivalent — 

1  pound  Fahrenheit  unit  of  heat  =  780  foot  pounds. 
1  pound  Centigrade  unit  of  heat  =r  1,400  foot  pounds. 
See  note,  p.  42. 

1  horse-power  hour  =  1*98  x  106  =  2  x  106  ft. -Ibs.  nearly. 

1  Board  of  Trade  electric  unit  =  1,000  watts  for  1  hour  = 

1000 

-=^j  horse-power  hour  =  2-654  x  10*  ft.-lbs. 

The  base  of  the  Napierian  logarithms  is  E  =  2 •  7 1 8.  To  convert  com- 
mon logarithms  into  Napierian  logarithms,  multiply  by  2*303. 


656 


APPLIED    MECHAttlCfe. 


TABLE  XX. — MODULI  OF  RIGIDITY. 


Substance. 


N  =  Modulus  of  Rigidity  in  Million? 
of  Pounds  per  Square  Inch. 


Steel  Plates,  ^  per  cent.  Carbon 
»         »       2         »  » 

»>         »        *         »  »> 

Steel  Boiler  Plates        

Cast  Steel  (tempered)   ... 
„       „      (untempered) 

Soft     ,,      (unhardened) 
„       „      (hardened)   ... 

Cast  Iron  ...         ...         ,.\ 

Iron  Boiler  Plates 

Wrought  Iron  Bars 

„  „     Plates 

Soft  Iron  

Brass       , 

Copper 

Lead  ' 

Zinc 

Tin          

Gold        

Silver       :   ... 

Platinum  

Aluminium 

Delta  Metal  (rolled)      

Gun        „  

Phosphor  Bronze 

Glase       

Wood      

Granite 

Marble     ...         

Slate 


13 

13-5 

14-0 

12-0 

11-0 

11-0 
5-0  to  7'6 

14-0 

10-5 

9-5 

10-8  to  11 

5  to  5-5 

5-6  to  67 

•27 
5-1  to  5-4 

2-2 
4-0  to  5-6 

3-8 

8-9  to  9-4 

3-4  to  4-8 

5-26 

3-7 

5-2 

3-3.  to  3-9 
•1  to  '17 

1-8 

1-7 

3-2 


APPLIED    MECHANICS. 


657 


TABLE  XXI. — MODULI  OP  COMPRESSIBILITY. 


Substance. 

Modulus  of 
Compressibility 
in  Pounds  per 
Square  Inch. 

Temperature. 

Authority. 

Steel         

21-6  x   106 

Amagat. 

Steel        

267  x  106 

Thomson's  "Elasticity." 

Iron 

21-1  x  106 

tt                  » 

Brass        

15-3  x   10« 

»                   » 

Copper     

17-1  x  106 

Buchanan. 

Copper     ...         ^.V, 

24-4  x  106 

Thomson's  "  Elasticity." 

Delta  Metal        «W 

14-4  x  106 

Amagat. 

Lead        

5-3  x  106 

M 

Glass        ...       'i; 

5-8  x  106 

» 

Distilled  Water... 

3-2  x  105 

15°  C. 

Paglianni. 

Alcohol    ...        A, 

1-76  x  105 

0°C. 

Amaury  and  Deschamps 

Alcohol    ...        ^ 

1-62  x  105 

15°  C. 

»                                n 

Ether       ...         «. 

1-35  x  105 

0°  C. 

>»                        »» 

Ether       ...        •>£ 

1-15  x  105 

15°  C. 

»»                        » 

Carbon  Bisulphide. 

2-32  x  105 

14°  C. 

» 

Glycerine.,.         ... 

5-85  x  105 

20-5°  C. 

Quincke. 

Patroleum 

2-11  x  105 

16-5°  C. 

Martini. 

Mercury  ...         ... 

7-85  x  106 

15°  C. 

Amaury  and  Deschamps. 

Mercury  ...         ... 

4-35  x  106 

0°C. 

Colladon  and  Sturm. 

Mercury  ... 

3-74  x  106 

0°C. 

Amagat. 

658 


APPLIED    MECHANICS. 


TABLE 


Breaking  Stress, 

Weight 

in  tt>s.  per 

MATERIAL. 

Melting 
Point 

Specific 
Grsvitv 

of 
One  Cubic 
Foot 

square  inch. 

(Fahr.). 

in 

Pounds. 

Tensile. 

Soft  Steel  (unhardened)        } 
Soft  Steel  (hardened)             ( 

2,400° 
to 

7-85 

490 

60,  000  to  100,  000 
120,000 

Oast  Steel  (untempered)         f 
Cast  Steel  (tempered)            ) 

3,000° 

90,  000  to  150,000 

Steel  Plates      

60,000  to  80,000 

Steel  Bars         

100,  000  to  130,  000 

Cast  Steel  (drawn)      

77 

480 

120,000 

Steel  Wire  (English),  drawn... 
Steel  Wire  (Common),  tern-  ) 
pered  blue            ...         { 

7'4 

460 

120,  000  to  140,000 
330,000 

Steel  Pianoforte  "Wire,  English 

7-73 

480 

340,000 

Manganese  Steel,  Cast 

85,000 

Nickel  Steel,  unhardened 

75,000 

Nickel  Steel,  hardened 

190,000 

Wrought  Iron  Bars  and  Bolts  ^ 

55,000  to  70,000 

Wrought  Iron  Plates,  with 

3,000° 

fibre  
Wrought  Iron  Plates  (across  j 
fibre)           

'to 
3,300° 

77 

480 

51,000 
46,000 

Wrought  Iron  Plates  (mean)./ 

48,500 

Cast  Iron  -1 

2,000° 
to  2,500° 

7-2 

450 

14,000  to  30,000 

Aluminium        ;  ... 

1,300° 

2-6 

162 

33,000 

Aluminium  (annealed) 

13,500 

Brass,  YeUow   \ 

1,700° 

7'8 

490 

17,500 

Brass,  Sheet     J 

to  1,850° 

to  8-4 

to  525 

30,000 

Brass,  Tube      

80,000  to  100,000 

Brass,  Cast       

8-0 

500 

20,000 

Brass,  Wire      

50,000 

Copper,  Wrought        

2,000° 

8-8 

550 

33,000 

Copper,  Cast     

20,000 

Copper  Wire,  hard-drawn     ... 

8-9 

555 

58,000 

Copper  Wire,  annealed 

47,000 

Zinc,  Cast         

750° 

7-0 

436 

7,500 

Zinc,  Sheet       

7'2 

450 

30,000 

Lead        

615° 

11-4 

712 

1,900 

Lead,  Cast        

11-2 

700 

3,000 

Tin                    

450° 

7  -4 

462 

4,600 

Platinum  Wire  

3,300° 

21-5 

1,340 

50,000 

Gold  (drawn)    

2,200° 

19-2 

1,200 

38,000  to  41,000 

Silver  (drawn)  ... 

1,850° 

10-4 

650 

42,000 

Phosphor  Bronze  (cast  ) 
Phosphor  Bronze  Wire  (hard)... 
Phosphor  Bronze  Wire  (an-  ) 
nealed)        j 

55,000 
100,000  to  150,  000 

50,000  to  60,000 

APPLIED    MECHANICS. 


659 


XXII. 


Breaking  Stress, 
in  Ibs.  per 
square  inch. 

Stress  which  produces 
Permanent  Set, 
in  Ibs.  per  square  inch. 

Safe  Limit  of 
Stress, 
in  Ibs.  per  square  inch. 

Young's 
Modulus 
of  Bias- 
ticity,  in 
millions 
of  Ibs.  per 
square 
inch. 

Com- 
pressive. 

Shear. 

Tensile. 

Com- 
>ressive. 

Shear. 

Tensile. 

Com- 
>ressive. 

Shear. 

35,000 

26,500 

17,700 

17,700 

13,000 

30 
30 
30 
36 

29  to  42 
27 
28 
26 
29 

50,000 

50,000 

45,000 
35,000 
56,000 
24,000 

24,000 

20,000 

10,000 

10,000 

7,800 

29 

OK 

»  50,000  to 
(   120,000 

28,500 

20,000 
10,500 

20,000 
21,000 

15,000 
8,000 

10,000 
3,500 

10,000 
10,500 

7,800 
2,700 

27 

26 

14  to  23 

10,500 

7,000 

5,000 

3,600 

2,700 

9 

58,000 

4,300 

4,000 

3,000 

3,600 

3,200 

2,300 

9'2 
14-2 
16 

3,200 

7,300 

1,500 

72 

20,000 

14,500 

10,000 

7,400 

6 
24 
12 
11 
14 

i 

660 


APPLIED    MECHANICS. 


TABLE 


Weight 

Breaking  Stress, 
in  Ibs.  per 

MATERIAL. 

Melting 
Point 

Specific 
Gravity. 

of 
One  Cubic 
Foot 

square  inch. 

(Fahr.). 

in 

Pounds. 

Tensile. 

Aluminium  Bronze,                \ 
Cu95A15J 

8-25 

515 

60,000 

Aluminium  Bronze,                 \ 
Cu90A110J 

77 

480 

100,000 

Manganese  Bronze      

65,000  to  85,000 

Delta  Metal  (cast)       

47,000 

Delta  Metal  (rolled)   

74,000 

Muntz  Metal     

45,000 

Sterro  Metal     

60,000 

Gun  Metal        

1,900° 

8'6 

536 

25,000  to  50,000 

Ebony     

17 

73 

Oak,  European  

•93 

58 

14,500 

Mahogany,  Spanish     
Ash 

•85 
•8 

53 
50 

15,000 
17,500 

Pitch  Pine         

•7 

44 

12,000 

Red  Pine           

10,500 

Birch         .         ..          

'54 

34 

14,500 

Beech                           

•7 

44 

11,500 

Fir,  Larch         

•53 

33 

11,000 

Fir,  Spruce 

•54 

34 

12,500 

Hornbeam         

•76 

47 

15,000 

Teak,  Indian    

•78 

49 

15,000 

Lancewood 

•95 

59 

20,000 

Elm,  British     

•55 

34 

14,000 

Lignum  Vitae    

•65 
to  1-33 

41  to  83 

16,000 

Sycamore          

13,000 

Cedar  of  Lebanon        

•59 

37 

11,400 

Granite             

2-7 

170 

Marble 

2-8 

175 

700 

Limestone         

2-8 

175 

Sandstone 

2-3 

144 

800 

Slate                   

2-8 

175 

11,000 

Brick,  Red        1 
Brick,  Fire        j 

2  to  2-2 

125  to  137 

300 

Brickwork         

1-8 

112 

Concrete  | 

1-9 
to  2-2 

119  to  137 

Leather  

4,000 

Hemp  Rope  (in  ordinary  state) 
Glass,  Plate 

1-3 
27 

170 

10,000 
2,700 

Ice 

•917 

57 

Quartz  Fibre  (Professor  Boys') 

140,000 

APPLIED    MECHANICS. 


661 


XXII.  (continued). 


Breaking  Stress, 
in  Ibs.  per 
square  inch. 

Stress  which  produces 
Permanent  Set, 
in  Ibs.  per  square  inch. 

Safe  Limit  of 
Stress, 
in  Ibs.  per  square  inch. 

Young's 
Modulus 
of  Elas- 
ticity, in 

Com- 
pressive. 

Shear. 

Tensile. 

Com- 
pressive. 

Shear. 

Tensile. 

Com- 
pressive. 

Shear. 

millions 
of  Ibs.  pei 
square 
inch. 

17,000 

12 

50,000 

13 

10 

18,000 

10,000 

2,300 

1-5 

1-4 

1-6 

6,000 

650 

1-4 

1-65 

1-35 

11 

1-6 

12,000 

2'4 

7,000 

10,000 

1D4 

•48 

17,000 

7 

6,000 

6 

5,000 

5,000 

2 

15,000 

13 

800 

2,000 

500 

2,000 

•025 

26,000 

8 

8-5 

662 


APPLIED    MECHANICS. 


LOGARITHMS. 


TABLE 


0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

123 

456 

789 

10 
IT 
12" 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0331 

0374 

4  9  13 
4  8  12 

7  21  26 
6  20  24 
5  19  23 
5  19  22 

30  34  38 
28  32  37 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

4  8  12 
4  7  11 

27  31  35 
26  30  33 

0792 

0828 

0864 

0899 

0934 

0969 

1303 
1614 

1004 
1335 

1038 
1367 

1072 

1106 

3  7  11 
3  7  10 
3  7  10 

3  7  10 

4  18  21 
4  17  20 

25  28  32 
24  27  31 

13 

1139 

1173 

1206 

1239 

1271 

1399 
1703 

1430 

3  16  20 
2  16  19 
2l5l8 
2  15  17 

23  26  30 
22  25  29 

14 

1461 

1492 

1523 

1553 

1584 

1644 

1673 

1732 

369 
369 

21  24  28 
20  23  26 

15 

TT 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

369 
358 

1  14  17 
1  14  16 

20  23  26 
19  22  25 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

358 
358 
35  8 

2  5  7 

1  14  16 
0  13  15 

19  22  24 
18  21  23 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

0  13  15 
0  12  15 

18  20  23 
17  19  22 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

2  5  7 
257 

9  12  14 
9  11  14 

18  19  21 
16  18  21 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

2  4  7 
246 

9  11  13 
8  11  13 

16  18  20 
15  17  19 

20 

~2T 
22 
23 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

246 

8  11  13 

WWT2 
8  10  12 
7  9  11 

15  17  19 

3222 
3424 
3617 
3802 
3979 
4150 
4314 
4472 
4624 
477T 

3243 
3444 
3636 

3263 
3464 
3655 

3284 
3483 
3674 

3304 
3502 
3692 
3874 
4048 
4216 

3324 
3522 
3711 

3892 
4065 
4232 

3345 
3541 
3729 

3365 
3560 
3747 

3385 
3579 
3766 

3404 
3598 
3784 
39(52 
4133 
4298 

246 
246 
246 

14  16  18 
14  15  17 
13  15  17 

24 
25 
26 

3820 
3997 
4166 
T330 
4487 
4639 
4786 

3838 
4014 
4183 

3856 
4031 
4200 

3909 
4082 
4249 

3927 
4099 
4265 

3945 
4116 

4281 

245 
235 
235 
235 
235 
1  3  4 

7  9  11 
7  9  10 
7  8  10 

12  14  16 
12  14  15 
11  13  15 

27 
28 
29 

4346 
45C2 
4654 

4362 
4518 
4669 
4814 

4378 
4533 
4683 

4393 
4548 
4698 

4409 
4564 
4713 

4425 
4579 

4728 

4440 
4594 
4742 

4456 
4609 
4757 

689 
689 
679 

11  13  14 
11  12  14 
10  12  13 

30 

4800 

4829 

4843 

4857 

4871 

4886 

4900 

1  3  4 

679 

0  11  13 

31 
32 
33 
~34~ 
35 
36 

4914 
5051 
5185 

4928 
5065 
5198 
5328 
5453 
5575 

4942 
5079 
5211 

4955 
5092 
5224 

4969 
5105 
5237 

4983 
5119 
5250 
5378 
5502 
5623 

1997 
5132 
5263 

5011 
5145 

5276 

5024 
5159 
5289 

5038 
5172 
5302 

134 
134 
1  3  4 
13  4 
124 
1  2  4 
1  2  3 
1  2  3 
123 

678 
578 
568 

id  11  12 

9  11  12 
9  10  12 
9  10  11 
9  10  11 
8  10  11 

5315 
5441 

5563 

5340 
5465 

5587 

5353 
5478 
5599 
5717 
5832 
5944 

5366 
5490 
5611 

5391 
5514 
5635 

5103 
5527 
5647 

5416 
5539 

5658 

5428 
5551 
5670 
5786 
5899 
6010 

568 
567 

5  6  7 

37 
38 
39 

5682 
5798 
5911 
6021 
t)128 
6232 
6335 

5694 
5809 
5922 

5705 
5821 
5933 

5729 
5843 
5955 

5740 
5855 
5966 

5752 
5866 
5977 

5763 

5877 
5988 

5775 

5888 
5999 

5  6  7 
567 
457 
456 

8  9  10 
8  9  10 
8  9  10 

40 

0031 

(5042 

6053 

6064 

6075 

6085 

0096 

6107 

6117 

123 

8  9  10 

41 

42 
43 

6138 
6243 
6345 

6149 
6253 
6355 

6160 
6263 
6365 

6170 
6274 
6375 

6180 
6284 
6385 

6191 
6294 
6395 

6201 
6304 
6405 

6212 
6314 
6415 

6222 
6325 
6425 
6522 
6618 
6712 
6803 
6893 
6981 

123 
123 

1  2  3 

456 
456 
456 

789 
789 
789 

44 
45 
46 
47~ 
48 
49 
W 

6435 
6532 
6628 
6721 
6812 
6902 

6444 
6542 
6637 

6454 
6551 
6646 

6464 
6561 
6656 

6474 
6571 
6665 

6484 
6580 
6675 

6493 
6590 
6684 

6503 
6599 
6693 
6785 
6875 
6964 

6513 
6609 
6702 

123 
123 
123 

4  5  C 
456 
456 

789 

789 
778 

6730 
6821 
6911 

6739 

6830 
6920 

6749 
6839 
6928 

6758 
6848 
6937 

6767 
6857 
6946 

6776 
6866 
6955 

6794 

6884 
6972 

123 
123 
1  2  3 

4  5 
4  4 
4  4 
3  4~ 

678 
678 
678 

69SO 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

123 

678 
678 
6  7  7 
667 

51 
52 
53 
~5T 

7076 
7160 
7213 

7084 
7168 
7251 

7093 

7177 
7259 

7101 
7185 
7267 

7110 
7193 
7275 

7118 
7202 

7284 

7126 
7210 
7292 

7135 

7218 
7300 

7143 

7226 
7308 

7152 
7235 
7316 

123 
1  2  2 

1  2  2 

3  4 
3  4 
3  4 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

1  2  2 

3  4 

667 

XXI1L 


APPLIED    MECHANICS. 
LOGARITHMS. 


663 


0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

123 

456 

789 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

122 

345 

567 

56 
57 
58 

7482 
7559 
7634 

7490 
7566 
7642 

7497 
7574 
7649 

7505 
7582 
7657 

7513 

75S9 
7604 

7520 
7597 

7672 

7528 
7604 
7679 

7536 
7612 
7686 

7543 
7619 
7694 

7551 
7027 
7701 

122 
122 
112 

345 
345 
344 

567 
567 
567 

59 
60 
61 

7709 

7782 
7853 

7716 

7789 
7SCO 

7723 

7796 
7S68 

7731 
7S03 
7875 

7738 
7810 

78S2 

7745 
7818 
7889 

7752 
7825 
7896 

7760 
7S32 
7903 

7767 
7839 
7910 

7774 
7846 
7917 

112 
112 
112 

3  4 
3  4 
3  4 

567 
566 
566 

62 
63 
64 

7924 
7993 
8062 

7931 
8000 
8069 

7938 
8007 
8075 

7945 
8014 
80S2 

7952 
8021 
8089 

7959 
8028 
8096 

7966 
8035 
8102 

7973 
8041 
8109 

7980 
8048 
8116 

7987 
8055 
8122 

1  1  2 
112 
112 

3  3 
3  3 
3  3 

566 
556 
556 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

1  1  2 

3  3 

556 

66 
67 
68 

8195 
8261 
8325 

8202 
8267 
8331 

8209 
8274 
8338 

8215 
8280 
8344 

8222 
S287 
8351 

8228 
8293 
8357 

S'235 
8299 
8363 

8241 
8306 
8370 

8248 
8312 
8376 

S254 
8319 
8382 

112 

1  1  2 
112 

3  3 
3  3 
3  3 

556 
556 
456 

69 
70 
71 

8388 
8451 
8513 

8395 
8457 
8519 

8401 
8463 
8525 

8407 
8470 
8531 

8414 
8476 
8537 

8420 

8482 
8543 

8426 
8488 
8549 

8432 
8494 
8555 

8439 
8500 
8561 

8445 
8506 
8567 

112 
112 
112 

2  3 
2  3 
2  3 

456 
456 
455 

72 
73 
74 

8573 
8633 
8692 

8579 
8639 
8698 

8585 
8645 
8704 

8591 
8651 
8710 

8597 
8657 
8716 

8603 
S663 
8722 

8609 
8069 
8727 

8615 
8675 
8733 

8621 
8681 
8739 

8627 
868.6 
8745 

112 
1  1  2 
112 

234 

234 
234 

455 
455 
455 

75 

8751 

8756 

8762 

8768 

8774 

8779 

87S5 

8791 

8797 

8802 

112 

233 

455 

76 
77 
78 

8808 
8865 
8921 

8814 
8S71 
8927 

8820 
8876 
8932 

8825 
8882 
8938 

8831 

88S7 
8943 

8837 
8893 
8949 

8842 
8899 
8954 

8848 
8904 
8960 

8854 
8910 
8965 

8859 
8915 
8971 

112 
112 

112 

233 
233 
233 

455 
4    5 
4    5 

79 
80 
81 

8976 
9031 
9085 

8982 
9036 
9090 

8987 
9042 
9096 

8993 
9047 
9101 

S998 
9053 
9106 

9004 
9058 
9112 

9009 
9063 
9117 

9015 
9069 
9122 

9020 
9074 
9128 

9025 
9079 
9133 

112 
112 
1  1  2 

233 
233 
233 

4    5 

4    5 
4    5 

82 
83 
84 

9138 
9191 
9243 

9143 
9196 
9243 

9149 
9201 
9253 

9154 
9206 
9258 

9159 
9212 
9263 

9165 
9217 
9269 

9170 
9222 
9274 

9175 
9227 
9279 

9180 
9232 
9284 

9186 
9238 
9289 

112 

112 
112 

233 
233 
233 

445 
445 

445 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

112 

233 

445 

86 
87 
88 

9345 
P395 
9445 

9350 
9400 
9450 

9355 
9405 
9455 

9360 
9410 
9460 

9365 
9415 
9465 

9370 
9420 
9469 

9375 
9425 
9474 

9380 
9430 
9479 

9385 
9435 
9484 

9390 
9440 
94  89 

112 
Oil 
0  1  1 

233 
223 
223 

445 
3  4 
3  4 

89 
90 
91 

9194 

9542 
9590 

9499 
9547 
9595 

9504 
9552 
9600 

9509 
9557 
9605 

9513 
9562 
9609 

9518 
9566 
9614 

9523 
9571 
9619 

9528 
9576 
9624 

9533 
9581 
9628 

9538 
9586 
9633 

Oil 
Oil 
Oil 

223 
223 
223 

3  4 
3  4 
3  4 

92 
93 
94 

9638 
»6S5 
9731 

9643 
96S9 
9736 

9647 
9694 
9741 

9052 
9699 
9745 

9657 
9703 
9750 

9661 
970S 
9754 

9666 
9713 
9759 

9671 
9717 
9763 

9675 
9722 
9768 

9680 

9727 
9773 

Oil 
Oil 
0  1  1 

223 
223 
223 

3  4 
3  4 
344 

95 

9777 

9782 

9786 

9791 

9795 

9SOO 

9805 

9S09 

9814 

9818 

0  1  1 

223 

344 

96 
97 
98 

99 

9823 

9S6S 
9912 

9956 

9827 
9872 
9917 

9961 

9832 
9877 
9921 

9965 

9836 
9881 
9926 

9969 

9S41 
9SS6 
9930 

9974 

9S45 
9S90 
9934 

9978 

9850 
9894 
9939 

9983 

9854 
9S99 
9943 

9987 

9859 
9903 
9948 

9991 

9863 
9908 
9952 

9996 

0  1  1 
0  1  1 
0  1  1 

0  1  1 

223 
223 
223 

223 

344 
344 
344 

334 

664 


APPLIED    MECHANICS. 
ANTILOGARITHMS. 


TABLE 


0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

1  2  3 

456 

789 

•oo 

1000 

1002 

1005 

1007 

1009 

1012 

1014 

1016 

1019 

1021 

001 

111 

222 

•01 

•02 
•03 

1023 
1047 
1072 

1026 
1050 
1074 

1028 
1052 
1076 

1030 
1054 
1079 

1033 
1057 
1081 

1035 
1059 
1084 

1038 
1062 
1086 

1010 
1064 
1089 

1042 
1067 
1091 

1045 
1069 
1094 

001 
001 
001 

111 
1  1  1 
111 

222 
222 
222 

•04 
•05 
•06 

1096 
1122 
1148 

1099 
1125 
1151 

1102 
1127 
1153 

1104 
1130 
1156 

1107 
1132 
1159 

1109 
1135 
1161 

1112 
1138 
1164 

1114 
1140 
1167 

1117 
1143 
1169 

1119 
1146 
1172 

Oil 
Oil 
Oil 

112 
1  1  2 
112 

222 
222 
222 

•07 
•08 
•09 

1175 
1202 
1230 

1178 
1205 
1233 

1180 
1208 
1236 

1183 
1211 
1239 

1186 
1213 
1242 

1189 
1216 
1245 

1191 
1219 
1247 

1194 
1222 
1250 

1197 
1225 
1253 

1199 
1227 
1256 

Oil 
Oil 
Oil 

112 
112 
1  1  2 

222 
223 
223 

10 

1259 

1262 

1265 

1268 

1271 

1274 

1276 

1279 

1282 

1285 

Oil 

112 

223 

11 
12 
13 

1288 
1318 
1349 

1291 
1321 
1352 

1294 
1324 
1355 

1297 
1327 
1358 

1300 
1330 
1361 

1303 
1334 
1365 

1306 
1337 
1368 

1309 
1340 
1371 

1312 
1343 
1374 

1315 
1346 
1377 

0  1  1 
Oil 
0  1  1 

122 
122 
122 

223 
223 
233 

14 
•15 
•16 

1380 
1413 
1445 

1384 
1416 
1449 

1387 
1419 
1452 

1390 
1422 
1455 

1393 
1426 
1459 

1396 
1429 
1462 

1400 
1432 
1466 

1403 
1435 
1469 

1406 
1439 
1472 

1409 
1442 
1476 

Oil 
0  1  1 
Oil 

122 
1  2  2 
122 

233 
233 
233 

17 
18 
19 

1479 
1514 
1549 

1483 
1517 
1552 

1486 
1521 
1556 

1489 
1524 
1560 

1493 
1528 
1563 

1496 
1531 
1567 

1500 
1535 
1570 

1503 
1538 
1574 

1507 
1542 
1578 

1510 
1545 
1581 

Oil 
0  1  1 
Oil 

122 
122 
122 

233 
283 
333 

•20 

1585 

1589 

1592 

1596 

1600 

1603 

1607 

1611 

1614 

1618 

0  1  1 

122 

383 

•21 
•22 
'23 

1622 
1660 
1698 

1626 
1663 
1702 

1629 
1667 
1706 

1633 
1671 
1710 

1637 
1675 
1714 

1641 
1679 
1718 

1644 
1683 
1722 

1648 
1687 
1726 

1652 
1690 
1730 

1656 
1694 
1734 

Oil 
Oil 
Oil 

222 
222 
222 

333 
383 
334 

•24 
•25 
•26 

1738 
1778 
1820 

1742 
1782 
1824 

1746 
1786 
1828 

1750 
1791 
1832 

1754 
1795 
1837 

1758 
1799 
1841 

1762 
1803 
1845 

1766 
1807 
1849 

1770 
1811 
1854 

1774 
1816 

1858 

Oil 
Oil 
Oil 

222 
222 
223 

334 
3  3 
3  3 

•27 
•28 
•29 

1862 
1905 
1950 

1866 
1910 
1954 

1871 
1914 
1959 

1875 
1919 
1963 

1879 
1923 
1968 

1884 
1928 
1972 

1888 
1932 
1977 

1892 
1936 
1982 

1897 
1941 
1986 

1901 
1945 
1991 

Oil 
Oil 
Oil 

223 
223 
223 

334 
3  4 
344 

•30 

1995 

2000 

2004 

2009 

2014 

2018 

2023 

2028 

2032 

2037 

Oil 

223 

344 

•31 
•32 
•33 

2042 
2089 
2138 

2046 
2094 
2143 

2051 
2099 
2148 

2056 
2104 
2153 

2061 
2109 
2158 

2065 
2113 
2163 

2070 
2118 
2168 

2075 
2123 
2173 

2080 
2128 
2178 

2084 
2133 
2183 

Oil 
Oil 
Oil 

223 
223 
223 

344 
344 
344 

•34 
•35 
•36 

2188 
2239 
2291 

2193 
2244 
2296 

2198 
2249 
2301 

2203 
2254 
2307 

2208 
2259 
2312 

2213 
2265 
2317 

2218 
2270 
2323 

2223 
2275 
2328 

2228 
2280 
2333 

2234 
2286 
2339 

112 
1  1  2 
112 

233 
233 
233 

445 
445 
445 

•37 
•38 
•39 

2344 
2399 
2455 

2350 
2404 
2460 

2355 
2410 
2466 

2360 
2415 
2472 

2366 
2421 
2477 

2371 
2427 
2483 

2377 
2432 
2489 

2382 
2438 
2495 

2388 
2443 
2500 

2393 
2449 
2506 

112 
112 
1  1  2 

233 
233 
233 

445 
445 
455 

•40 

2512 

2518 

2523 

2529 

2535 

2541 

2547 

2553 

2559 

2564 

112 

234 

455 

•41 
•42 
•43 

2570 
2630 
2692 

2576 
2636 
2698 

2582 
2642 
2704 

2588 
2649 
2710 

2594 
2655 
2716 

2600 
2661 
2723 

2606 

2667 
2729 

2612 
2673 
2735 

2618 
2679 
2742 

2624 
2685 
2748 

112 
4  1  2 

1  1  2 

234 
234 
334 

455 
456 
456 

•44 
•45 
•46 

2754 
2818 
2884 

2761 
2825 
2891 

2767 
2831 

2897 

2773 
2838 
2904 

2780 
2844 
2911 

2786 
2851 
2917 

2793 
2858 
2924 

2799 
2864 
2931 

2805 
2871 
2938 

2812 
2877 
2944 

112 
112 
112 

334 
334 
334 

466 
556 
556 

•47 
•48 
•49 

2951 
3020 
3090 

2958 
3027 
3097 

2965 
3034 
3105 

2972 
3041 
3112 

2979 
3048 
3119 

2985 
3055 
3126 

2992 
3062 
3133 

2999 
3069 
3141 

3006 
3076 
3148 

3013 
3083 
3155 

112 
112 
112 

334 
344 
344 

556 
566 
566 

XXIV. 


APPLIED    MECHANICS. 
ANTILOGARITHMS. 


665 


0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

1  2  3 

456 

789 

•50 

3162 

3170 

3177 

3184 

3192 

3199 

3206 

3214 

3221 

3228 

1  1  2 

344 

567 

•51 
•52 
•53 

3236 
3311 
3388 

3243 
3319 
3396 

3251 
3327 
3404 

3258 
8334 
3412 

3266 
3342 
3420 

£2/3 
3350 
3428 

3281 
3357 
3436 

3289 
3365 
3443 

3296 
3373 
3451 

3304 
3381 
3459 

122 
122 
1  2  2 

3  4 
3  4 
3  4 

5  fi  7 

567 
667 

•54 

•55 
•56 

3467 
3548 
3631 

3475 
3556 
3639 

3483 
3565 
3648 

3491 
8573 

3656 

3499 
3581 
3664 

3508 
3589 
3673 

3516 
3597 
3681 

3524 
3606 
3690 

3532 
3614 
3698 

3540 
3622 
3707 

3793 
3882 
3972 

122 
122 
123 

3  4 
3  4 
3  4 

667 
677 
678 

•57 
•58 
•59 

3715 
3802 
3890 

3724 
3811 
3899 

3733 
3819 
3908 

3741 
3828 
3917 

3750 
3837 
3926 

3758 
3846 
3936 

3767 
3855 
3945 

3776 
3864 
3954 

3784 
3873 
3963 

123 
1  2  3 
123 

3  4 
4  4 
4  5 

678 
678 
678 

•60 

3981 

4074 
4169 

4260 

3990 

4083 
4178 

4276 

3999 

4009 

4018 

4027 

4121 
4217 
4315 

4036 

"4130~ 
4227 
4325 

4046 

4055 

4064 

4159 
4256 
4355 

123 

456 

6'  7  8 

•61 
•62 
•63 

4093 
4188 
4285 

4102 
4198 
4295 

4111 

4207 
4305 

4140 
4236 
4335 

4150 
4246 
4345 

1  23 
123 
1  2 

456 
456 
456 

789 
789 
789 

•64 
•65 
•66 

4365 
4467 
4571 

4375 
4477 
4581 

4385 
4487 
4592 

4395 
4498 
4603 

4406 
4508 
4613 

4416 
4519 
4624 

4426 
4529 
4634 

4430 
4539 
4645 

4446 
4550 
4656 

4457 
4560 
4667 

123 
123 
123 

456 
456 
456 

789 
789 
7  9  10 

•67 
•68 
•69 

4677 
4786 
4898 

4688 
4797 
4909 

4699 
4808 
4920 

4710 
4819 
4932 

4721 
4831 
4943 

4732 
4842 
4955 

4742 
4853 
4966 

5082 

4753 

4864 
4977 

4764 
4875 
4989 

4775 
4887 
5000 

123 
123 
123 

457 
467 
567 

8  9  10 
8  9  10 
8  9  10 

•70 

5012 

5129 
5248 
5370 

5023 

5035 

5047 

5058 

5070 

5093 

5105 

5117 

124 

567 

8  9  11 

•71 

•72 
•73 

5140 
5260 
5383 

5152 
5272 
6395 

5164 
5284 
5408 

5176 
5297 
5420 

5188 
5309 
5433 

5200 
5321 
5445 

5212 
5333 
5458 

5224 
5346 
5470 

5236 
5358 
5483 

124 
1  2  4 
134 

567 
567 
568 

8  10  11 
9  10  11 
9  10  11 

•74 
•75 
•76 

5495 
5623 
5754 

5888 
6026 
6166 

5508 
5636 
5768 

5521 
5649 
5781 

5916 
6053 
6194 

5534 
5662 
5794 

5546 
5675 
5808 

5943 
6081 
6223 

5559 
5689 
5821 

5572 
5702 
5834 

5970 
6109 

6252 

5585 
5715 
5848 

5598 
5728 
5861 

5610 
5741 
5875 

134 
134 
134 

568 
578 
578 

9  10  12 
9  10  12 
9  11  12 

•77 
•78 
•79 

5902 
6039 
6180 

5929 
6067 
6209 

5957 
6095 
6237 

5984 
6124 
6266 

5998 
6138 
6281 

6012 
6152 
6295 

134 
134 
134 

578 
678 
679 

10  11  12 
10  11  13 
10  11  13 

•80 

6310 

6324 

6339 

6353 

6368 

6383 

6397 

6412 

6427 

6442 

134 

679 

10  12  13 

•81 
•82 
•83 

•84 
•85 
•83 

6457 
6607 
6761 

6471 
6622 
6776 

6486 
6637 
6792 

6501 
6653 
6808 

6516 
6668 
6823 

6531 
6683 
6839 

6546 
6699 
6855 

6561 
6714 
6871 

6577 
6730 
6887 

6592 
6745 
6902 

235 
235 
235 

689 
689 
689 

11  12  14 
11  12  14 
11  13  14 

6918 
7079 
7244 

6934 
7096 
7261 

7430 
7603 
7780 

6950 
7112 

7278 

7447 
7621 
7798 

6966 
7129 
7295 

6982 
7145 
7311 

6998 
7161 
7328 

7499 
7674 
7852 

7015 
717S 
7345 

7031 
7194 
7362 

7047 
7211 
7379 

7063 
7228 
7396 

235 
235 
235 

6  8  10 
7  8  10 
7  8  10 

11  13  15 
12  13  15 
12  13  15 

•87 
•88 
•89 

7413 

7586 
7762 

7464 
7638 
7816 

7482 
7656 
7834 

7516 
7691 
7870 

7534 
7709 
7889 

7551 
7727 
7907 

7568 
7745 
7925 

235 
245 
245 

7  9  10 
7  9  11 
7  9  11 

12  14  16 
12  14  16 
13  14  16 

•90 

7943 

7962 

7980 

7998 

8017 

8035 

8054 

8072 

8260 
8453 
8650 

8091 

8110 

246 

7  9  11 

13  15  17 

•91 
•92 
•93 

8128 
8318 
8511 

8147 
8337 
8531 

8166 
8356 
8551 

8185 
8375 
8570 

8770 
8974 
9183 

8204 
8395 
8590 

8222 
8414 
8610 

8241 
8433 
8630 

8279 
8472 
8670 

8872 
9078 
9290 

8299 
8492 
8690 

246 
246 
246 

8  9  11 
8  10  12 
8  10  12 

13  15  17 
14  15  17 
14  16  18 

•94 
•95 
•96 

8710 
8913 
9120 

8730 
8933 
9141 

9354 
9572 
9795 

8750 
8954 
9162 

8790 
8995 
9204 

8810 
9016 
9226 

8831 
9036 
9247 

8851 
9057 
9268 

8892 
9099 
9311 

246 
246 
246 

8  10  12 
8  10  12 
8  11  13 

14  16  18 
15  17  19 
15  17  19 

'97 
•98 
•99 

9333 
9550 
9772 

9376 
9594 
9817 

9397 
9616 
9840 

9419 
9638 
9863 

9441 
9661 
9886 

9462 
9683 
9908 

9484 
9705 
9931 

9506 
9727 
9954 

9528 
9750 
9977 

247 
247 
257 

9  11  13 
9  11  13 
9  11  14 

15  17  20 
16  18  20 
16  18  20 

666 


APPLIED    MECHANICS. 
TABLE  XXV. 


Angle. 

Radians. 

Sine. 

Tangent, 

Co-tangent. 

Cosine. 

0° 

0 

0 

0 

cc 

1-0000 

1-5708 

90° 

1 

•0175 

•0175 

•0175 

57-2900 

•9998 

1-5533 

89 

2 

•0349 

•0349 

•0349 

28-6363 

•9994 

1-5359 

88 

3 

•0524 

•0523 

•0524 

19-0811 

•9986 

1-5184 

87 

4 

•0698 

•0698 

•0699 

14-3006 

•9976 

1-5010 

86 

5 

•0873 

•0872 

•0875 

11-4301 

•9962 

1-4835 

85 

6 

•1047 

•1045 

•1051 

9-5144 

•9945 

1-4661 

84 

7 

•1222 

•1219 

•1228 

8-1443 

•9925 

1-4486 

83 

8 

•1396 

•1392 

•1405 

7-1154 

•9903 

1-4312 

82 

9 

•1571 

•1564 

•1584 

6-3138 

•9877 

1-4137 

81 

10 

•1745 

•1736 

•1763 

5-6713 

•9848 

1-3963 

80 

11 

•1920 

•1908 

•1944 

5-1446 

•9816 

1-3788 

79 

12 

•2094 

•2079 

•2126 

4-7046 

•9781 

1-3614 

78 

13 

•2269 

•2250 

•2309 

4-3315 

•9744 

1-3439 

77 

14 

•2443 

•2419 

•2493 

4-0108 

•9703 

1-3265 

76 

15 

•2618 

•2588 

•2679 

3-7321 

•9659 

1-3090 

75 

16 

•2793 

•2756 

•2867 

3-4874 

•9613 

1-2915 

74 

17 

•2967 

•2924 

•3057 

3-2709 

•9563 

1-2741 

73 

18 

•3142 

•3090 

•3249 

3-0777 

•9511 

1-2566 

72 

19 

•3316 

•3256 

•3443 

2-9042 

•9455 

1-2392 

71 

20 

•3491 

•3420 

•3640 

2-7475 

•9397 

1-2217 

70 

21 

•3665 

•3584 

•3839 

2-6051 

•9336 

1-2043 

69 

22 

•3840 

•3746 

•4040 

2-4751 

•9272 

1-1868 

68 

23 

•4014 

•3907 

•4245 

2-3559 

•9205 

1-1694 

67 

24 

•4189 

•4067 

•4452 

2-2460 

•9135 

1-1519 

66 

25 

•4363 

•4226 

•4663 

2-1445 

•9063 

1-1345 

65 

26 

•4538 

•4384 

•4877 

2-0503 

•8988 

1-1170 

64 

27 

•4712 

•4540 

•5095 

•9626 

•8910 

1-0996 

63 

28 

•4887 

•4695 

•5317 

•8807 

•8830 

1-0821 

62 

29 

•5061 

•4848 

•5543 

•8040 

•8746 

1-0647 

61 

30 

•5236 

•5000 

•5774 

•7321 

•8660 

1-0472 

60 

31 

•5411 

•5150 

•6009 

•6643 

•8572 

1-0297 

59 

32 

•5585 

•5299 

•6249 

•6003 

•8480 

1-0123 

58 

33 

•5760 

•5446 

•6494 

•5399 

•8387 

•9948 

57 

34 

•5934 

•5592 

•6745 

•4826 

•8290 

•9774 

56 

35 

•6109 

•5736 

•7002 

•4281 

•8192 

•9599 

55 

36 

•6283 

•5878 

•7265 

•3764 

•8090 

•9425 

54 

37 

•6458 

•6018 

•7536 

•3270 

•7986 

•9250 

53 

38 

•6632 

•6157 

•7813 

•2799 

•7880 

•9076 

52 

39 

•6807 

•6293 

•8098 

•2349 

•7771 

•8901 

51 

40 

•6981 

•6428 

•8391 

•1918 

•7660 

•8727 

50 

41 

•7156 

•6561 

•8693 

•1504 

•7547 

•8552 

49 

42 

•7330 

•6691 

•9004 

•1106 

•7431 

•8378 

48 

43 

•7505 

•6820 

•9325 

•0724 

•7314 

•8203 

47 

44 

•7679 

•6947 

•9657 

1-035 

•7193 

•8029 

46 

45 

•7854 

•7071 

1-0000 

1-0000 

•7071 

•7854 

45 

Cosine 

Co-tangent 

Tangent 

Sine 

Radians 

Anglb 

667 


APPENDIX    II. 


THE  NOTES  ARE  TO  BE  BEAD  IN  CONNECTION  WITH  THE  TEXT 
ON  THE  PAGE  REFERRED  TO. 

Page  3.— Sometimes  I  start  my  students  on  an  interesting  competi- 
tion as  to  their  powers  of  "  hefting  "  or  judging  of  weights  of  objects 
held  in  the  hand ;  their  power  to  estimate  distances  from  £  inch  to 
10  inches ;  their  power  to  estimate  distances  from  30  to  100  ft.  ; 
measurement  of  rooms  by  stepping.  Power  to  judge  of  a  fractional 
part  of  a  small  distance  is  best  acquired  by  the  use  of  scales  in 
drawing. 

Page  4. — Work  as  many  of  these  exercises  as  you  find  easy  with  a 
slide  rule,  and  note  the  accuracy  with  which  you  are  able  to  use  a  slide 
rule. 

Page  8.— The  area  of  the  curved  surface  of  a  spherical  segment  is 
equal  to  the  curved  surface  of  a  cylinder  of  the  same  height  as  the 
segment,  its  base  being  a  great  circle  of  the  sphere. 

Page  12.  Ex.  27. — If  the  length  of  his  pace  is  32-73  inches  show 
that  a  man's  speed  in  miles  per  hour  is  equal  to  the  number  of  his 
paces  in  4  seconds. 

Ex.  28. — If  each  length  of  railway  rail  is  22  feet,  show  that  the 
number  of  bumps  in  15  seconds  is  the  speed  in  miles  per  hour. 

Page  81. — In  fact,  viscosity  gives  stability.  The  flow  is  stable  in  a 
stream  whose  solid  boundaries  converge  and  tends  to  be  unstable  if  the 
boundaries  diverge.  A  curved  stream  is  stable  if  the  velocity  is 
greater  with  greater  radius,  and  tends  to  be  unstable  in  the  reverse 
case.  A  stream  flowing  through  still  water  tends  to  be  unstable. 

Page  84. — The  critical  velocity  in  all  similar  cases  of  fluid  motion 
seems  to  be  proportional  to  n/dp,  where  /i  is  viscosity,  p  is  density,  d  is 
dimension  ;  for  example,  diameter  of  pipe  in  the  present  case. 

Page  84. — In  using  the  mnemonic  of  the  last  eight  lines  there  is 
something  which  puzzles  a  student.  The  total  force  of  fluid  friction 
in  a  pipe,  or  past  an  immersed  object,  is  roughly  proportional  to  Aow2, 
where  A  is  wetted  area,  u  is  density  of  fluid,  and  u  is  relative  speed.  It 
is  important  to  know  about  w  in  the  case  of  gases.  Keeping  «  in  the 
argument  it  will  be  found  that  the  loss  of  energy  per  pound  of  fluid  in 
passing  along  a  pipe  or  round  a  bend  is  not  dependent  upon  «.  The 
loss  of  pressure  due  to  friction  is  proportional  to  «,  but  this  is  not  loss 
of  energy. 

Page  87.— Everybody  has  noticed  the  ease  with  which  a  belt  may 
be  slipped  from  a  revolving  pulley  as  compared  with  a  pulley  at  rest. 
There  are  cases  in  which  friction  or  resistance  to  sliding  at  right  angles 
to  other  enforced  sliding  seems  to  be  quite  destroyed. 


668  APPLIED    MECHANICS. 

Page  90. — In  the  above  crane  it  is  easy  to  show  that  the  efficiency 
e  is  1  I  (  R  -f  1-716  J,  so  that  however  great  R  may  be  the  efficiency 

cannot  exceed  1/1-716. 

Page  125. — Exercises  in  finding  the  resultants,  &c.,  of  forces  are 
difficult  to  set  unless  some  such  convention  as  the  following  is 
adopted  :— O  X  is  supposed  to  be  a  line  in  the  plane  in  which  all  the 
forces  are  supposed  to  act.  Let  the  force  A  P  be  F  Ib.  Let  it  cut  O  X 
in  the  point  A.  Let  O  A  =r  o.  Let  the  angle  X  A  P,  measured  anti-clock- 
wise, be  0.  Let  the  arrow  head  point  from  A  to  P.  Such  a  force  would  be 

F 
specified  as  a     0.    Thus  518330  is  a  force  equal  and  opposite  to  5182]3°. 

Page  141. — The  last  paragraph  of  p.  409  ought  to  be  inserted 
here. 

Page  279.— Quite  green  or  artificially  wetted  timber  has  only  about 
80  per  cent,  of  the  strength  of  well-seasoned  timber.  It  may  be  taken 
as  roughly  correct  that  the  strengths  of  various  timbers  in  the  same 
state  of  dryness  are  in  proportion  to  their  densities.  Strength  seems 
to  be  a  linear  function  of  the  amount  of  moisture  present. 

Page  302. — When  a  specimen  has  taken  a  set  through  overloading, 
Hooke's  law  is  not  quite  true  for  it,  even  for  small  loads,  and  there  is 
much  creeping  with  time.  Also  there  is  much  hysteresis.  Its  elasticity 
is  partially  restored  by  resting ;  it  is  quickly  restored  by  immersion  in 
boiling  water. 

Page  310. — Chains  of  cranes  need  to  be  annealed  after  several 
years'  use. 

Page  324. — The  assumption  that  p  +  %  is  constant  is  correct  for 
any  long  cylinder,  whether  or  not  there  is  internal  fluid  pressure, 
because  the  stress  on  sections  parallel  to  the  axis  will  be  constant. 
qa-pfi  is  the  fractional  diminution  of  the  radius,  only  if  there  is  no 
stress  parallel  to  the  axis,  and  we  only  need  to  use  this  in  the  manu- 
facture of  a  gun,  when  there  is  no  stress  parallel  to  the  axis. 

Page  325. — Perhaps  the  most  useful  approximation  is 

t=p  D/O+2/), 
where  D  is  the  outside  diameter. 

Page  339.— The  friction  between  the  plates  caused  by  the  contrac- 
tion of  the  rivets  in  cooling  gives  additional  strength,  which  is  usually 
neglected,  because  it  is  of  unknown  amount. 

Page  339. — The  old  careless  boiler-shop  methods  led  to  non-agree- 
ment of  holes  ;  modern  methods  are  scientific ;  we  now  use  drilling 
machines,  hydraulic  riveters,  edge-planing  machines,  &c.,  and  all  work 
is  done  to  templates. 

Page  370«. — Friction  of  wheels  against  an  atmosphere.  Such 
experiments  as  have  been  made  show  that  the  loss  of  power  by  friction 
when  similar  wheels  rotate  is  proportional  to  the  density  of  the  atmos- 
phere, the  cube  of  the  angular  velocity,  and  the  5th  power  of  the 
diameter.  Openings  as  between  vanes  of  a  turbine  wheel  increase  the 
frictional  loss.  Enclosing  the  wheel  generally  reduces  the  friction. 
For  discs  in  air  or  superheated  steam  we  may  take  the  wasted  horse- 
power P  to  be 

P  =f  w  D6  w3, 


APPENDIX.  669 

where  w  is  the  weight  of  air  or  steam  in  pounds  per  cubic  foot,  D  is 
diameter  of  the  disc  in  feet,  n  being  revolutions  per  minute, /being 

about  10"  .  In  wet  or  nearly  dry  steam /may  be  taken  as  T3  x  10"  • 
or  30  per  cent,  greater.  A  roughly  correct  theory  assumes  the  disc  to 
act  as  a  fan  receiving  air  at  its  middle,  and  delivering  it  to  the  atmo- 
sphere at  its  circumference.  The  lengths  of  path  of  similarly  placed 
particles  are  oc  D,  and  the  hydraulic  mean  depths  of  similar  channels 
are  also  oc  D,  so  that  the  loss  of  energy  by  friction  per  pound  of  air  is 
independent  of  D,  but  proportional  to  v2  if  v  is  the  speed  of  the  fluid. 
(See  Art.  69.)  Now,  differences  of  pressure  produced  by  fans  are  pro- 
portional to  w  n2  D2  (see  Ex.  4,  page  219),  and  if  V  is  volume  of  air  per 
second  the  work  done  per  second  is  proportional  to  V  w  nz  D3. 
Differences  of  pressure  divided  by  w  represent  loss  of  energy  per 
pound,  so  that  v2  oc  »2  D2.  Now  V  oc  D2f,  or  V  oc  n  D3,  so  that  power 
lost  oc  wo5  n3. 

In  the  Laval  turbine  CD  is  exceedingly  small,  and  there  is  not  much 
loss  of  power  by  friction  in  spite  of  the  high  speed.  In  some  parts  of 
some  other  steam  turbines  a>  is  perhaps  100  times  as  great  as  in  the 
Laval.  Notice  that  if  the  circumferential  velocity  is  fixed  the  frictional 
loss  is  proportional  only  to  the  square  of  the  diameter. 

Page  379.— My  students  have  occasionally  found  E,  Young's 
modulus,  for  annealed  iron  and  steel,  by  bending,  and  also  by  mere 
elongation  of  the  self-same  bar.  In  no  case  did  they  find  differences 
greater  than  what  might  be  due  to  errors  of  experiment.  Indirectly 
this  is  a  proof  of  the  correctness  of  the  engineers'  theory  of  bending. 

Page  404. — Instead  of  the  last  three  lines  we  might  say,  more 

r e 

logically  : — Letting^?  be  represented  by  ^  (0  -  <£),  r2    /  ^  (<f>)  sin  <ft  d<f>  = 

J  o 
Yr.  (1  -  cos  0) (3). 

Differentiating  with  regard  to  6  we  have 

r2  «//  (0)  sin  e  =  ~Fr  sin  0, 
or  ty  (0)  =  F/r  a  constant,  so  that^?  =  F/r  a  constant. 

Page  409. — The  last  paragraph  belongs  to  page  141.  Substitute 
here  the  following : — In  the  table,  page  398.  the  strength  modulus  of 
each  section  is  given.  Z  is  I  divided  by  the  greatest  y  of  the  section. 
It  is  evident  that  if /is  the  greatest  stress  in  the  material  the  bending 
moment  is  Z/. 

Page  442. — And  also  with  a  load  of  50  Ib.  at  the  end. 

Page  461. — I  give  the  preceding  investigation  with  some  misgiving  ; 
it  cannot  be  approximately  correct  except  in  long  beams,  because  the 
engineers'  theory  of  bending  is  only  correct  for  long  beams,  and  the 
result  is  quite  unimportant  except  in  short  beams. 

Page  471.— In  the  usual  unscientific  treatment  of  this  subject  the 
deflection  y  is  assumed  not  to  depend  upon  p,  or  to  be  what  (3)  gives  if 
P  is  o.  That  is,  the  usually  assumed  deflection  must  be  divided  by 
1  -  F/u  to  get  the  true  deflection.  Students  may  find  the  error 
negligible  in  short  struts,  but  they  must  be  on  their  guard  against  it 
in  long  struts.  Exercise — Prove  that  the  ordinary  treatment  is 
sufficiently  correct  if  E  I  ir2  >  44  P  Z2. 


670  APPLIED    MECHANICS. 

Page  477. — For  information  as  to  the  critical  speeds  of  shafts  with 
one  or  many  wheels  on  various  kinds,  and  numbers  of  supports, 
students  are  referred  to  a  paper  by  Prof.  Dunkerley  in  the  Phil.  Trans. 
for  1884,  page  281,  Vol.  185,  amplified  by  a  paper  in  the  Proc.  Physical 
Society  of  London,  by  Dr.  Chree,  Vol.  19,  July,  1904,  page  114. 

In  Art.  379  I  consider  a  rotating  shaft  under  applied  forces  due  to 
its  own  weight  and  an  endlong  thrust.  At  the  end  I  take  rotation 
alone,  and  arrive  at  what  is  given  below  for  that  case. 

Neglecting  the  mass  of  a  vertical  shaft  of  length  Z,  on  which  at  the 
middle  there  is  a  wheel  of  weight  w,  let  the  centre  of  gravity  of  w  be 
h  feet  from  the  centre  of  the  shaft  (7t  being  small).  When  rotating  let 
the  centre  of  gravity  be  y  +  h  from  the  axis,  y  being  the  deflection  of 
the  shaft  considered  as  a  beam,  the  moment  of  inertia  of  whose 
cross-section  is  I,  loaded  at  the  middle  with  the  centrifugal  force 

F  =  —  (y  +  li)  o2.       A  load  v  produces    the   deflection  FZ3/48  EI    or 

FZ3/192  EI  depending  on  whether  the  ends  of  the  shaft  are  free  to  change 
from  the  vertical  direction  or  not.     Keeping  to  free  ends 

73         w 

J— (y  +  K)  d? 


48  El     g 
Finding  y  from  this,  as  the  greatest  bending  moment  is  \  FZ,  this  is 

M  _  Za2W  /  h \ 

4<7      \  1  -  wZ3aa/48^  EI  / 

Note  that  however  small  h  may  be,  if  the  denominator  is  0  (this 
gives  critical  a)  we  have  fracture  of  the  shaft.  Taking  any  value  of  A, 
and  any  particular  case,  it  is  useful  to  see  how  M  increases  rapidly,  as 
the  critical  value  of  o,  say  a1,  is  being  approached.  Now,  imagine  such 
a  shaft  to  be  increased  in  speed  so  rapidly  from  0  to  values  greater 
than  a1  that  it  has  no  time  to  get  broken  when  passing  through  the 
critical  speed.  It  will  be  found  that  for  much  greater  values  of  o  than 
a1  we  have  small  bending  moment,  and  a  tendency  for  the  centre  of 
gravity  of  the  wheel  to  approach  the  centre  of  rotation. 

Taking  I  as  the  length  of  the  shaft,  and  not  21  as  in  Art.  379, 

letting  w  mean  (-—)*  where  w  is  weight  of  shaft  in  pounds  per 

\ff  EI  / 

foot  of  length,  I  the  moment  of  inertia  of  the  cross-section  about  a 
diameter,  there  are  the  following  results  : — 

1.  Shaft  alone.     Fixed  as  to  direction  at  one  end, 

the    other    end    free   and  unsupported,  the 

critical  speed  is  given  by          ...         ...         ...     ml  =  I '87 

2.  Shaft  alone.     Supported  at  two  free  ends       ...     ml  =  TT 

3.  Shaft  alone.    Fixed  as  to  direction  at  one  and 

free  at  the  other  end      ml  =  3'927 

4.  Shaft  alone.     Fixed  as   to  direction  at  both 

ends          ...         ...         ...          ..         ...         ...     ml  =  4*745 

5.  Shaft  of  no  mass,  ends  supported,  but  free.        

Wheel  w  at  middle  as  above,  critical  a  =  A/  — £^;3 


APPENDIX.  671 

6.  Shaft  of  no  mass,  ends  fixed  as  to  direction          

/\\Y2g  e  I 
Wheel  w  at  middle  as  above,  critical  a  =  A/ 75 — 

We  assume  that  a  wheel  is  attached  really  at  the  centre.  A  long 
well-fitting-  boss  makes  the  shaft  stiffer  and  the  critical  speed  greater. 

For  other  cases  students  must  be  referred  to  the  papers  already 
mentioned.  Dunkerley  established  by  his  experiments  an  empirical  rule 
which  may  be  taken  to  be  fairly  true  for  all  cases  likely  to  come  before 
the  engineer.  If  the  critical  speed  be  a,  then  27r/a  might  be  called  the 
critical  time  of  one  revolution,  T.  If  a  shaft  be  supported  at  one  or 
many  places,  and  loaded  with  one  or  many  wheels,  spaced  anyhow ; 
if  T?  be  the  critical  time  of  the  shaft,  assuming  no  wheels ;  if  Tx  be  the 
critical  time,  assuming  the  shaft  to  be  massless,  and  only  one  wheel, 
which  I  call  the  first,  is  on ;  if  T2  be  the  critical  time,  assuming  the 
shaft  to  be  massless,  and  only  one  wheel,  which  I  call  the  second,  is 
on,&c.  ;  then  T,  the  critical  time"  of  the  real  shaft  with  its  wheels,  is 


T  =  i      T02  +  T,2  +  T22  +  &c. 

There  is  another  important  general  rule,  true  for  shafts  alone,  but 
only  true  when  there  are  wheels  if  we  might  neglect  the  moment  of 
inertia  of  a  wheel  about  a  diameter  through  its  centre  of  gravity, 
sufficiently  true  for  many  practical  cases.  If,  when  a  shaft  with  wheels 
upon  it,  revolving  at  n  turns  per  second,  has  also  elastic  lateral  oscilla- 
tions^? per  second,  and  if  p0  is  the  value  of  p  when  n  =  o,  then 

&=p*  -  »» (1). 

The  critical  speed  is  when  n  =P0  or  when  the  time  of  a  revolution  is 
equal  to  the  periodic  time  of  lateral  vibration  of  the  shaft  when  not 
revolving.  It  is  important  to  put  the  result  as  Dr.  Chree  has  done  in 
the  shape  (1),  because  there  are  cases  of  rotating  shafts  being  subjected 
to  forced  lateral  vibrations,  and  it  is  then  p  and  not  pQ  which  is  of 
importance. 

I  quote  from  Dr.  Chree  :  "  Ordinarily,  when  a  shaft  held  at  one  or 
both  ends  is  acted  on  by  forces  tending  to  bend  it,  on  the  removal  of 
these  forces  it  tends  to  return  to  its  original  straight  position  ;  in  doing 
so  it  overshoots  the  mark  and  vibrates  to  and  fro  laterally.  The 
velocity  of  its  approach  to  the  equilibrium  position,  and  the 
frequency  of  the  vibrations  subsequently  executed,  are  greater  the 
larger  the  elastic  stresses  produced  in  the  bar  by  a  given  lateral 
displacement.  When  the  bar  is  rotating  round"  its  longitudinal  axis, 
and  is  displaced  laterally,  the  elastic  stresses  tend,  as  before,  to  bring 
it  back  to  the  undisturbed  position  ;  but  the  'centrifugal  forces'  have 
exactly  the  opposite  tendency.  They  thus  reduce  the  righting  forces, 
and  so  diminish  the  frequency  of  vibration."  The  student  ought  to 
take  two  cases  and  show  that  (1)  is  correct.  First  for  a  shaft  alone. 
Second  for  the  above  wheel  on  a  massless  shaft 

EXERCISES. 

1.  A  shaft  3  inches  in  diameter,  6  feet  long,  is  supported  in  each  of 
the  ways  (1),  (2),  (3),  (4).  Find  the  critical  speeds  in  revolutions  rer 
second.  Antwirs,  16-15,  45»6,  71  27,  101-1. 

W 


672 


APPLIED    MECHANICS. 


2.  A  wheel  of  10  cwt.  at  the  middle  of  the  above  shaft,  as  in  (5) 
and  (6).    Find  the  critical  speeds  in  revolutions  per  second,  neglecting 
the  mass  of  the  shaft.  Answers,  23-16,  11  '58. 

3.  Taking  the  mass  of  the  shaft  into  account,  find  the  answer  to 
question  2.  Answers,  22-6,  11-22. 

4.  What  is  the   frequency  for  lateral  vibrations  for  the  shaft  of 
question  3,  fixed  at  the  ends,  when  it  rotates  10  times  per  second  ? 

Ansuer,  20'27. 

5.  Consider  the  wheel  of  question  2,  the  shaft  being  fixed  at  the 
ends  to  be  O'Ol  foot  out  of  balance  (or  h  =  O'Ol).     Find  the  greatest 
bending  moment  at  the  following  speeds  o  : — 

6.  Kepeat  (5)  when  h  =  0*001. 
Answers  to  questions  (5)  and  (6)  : — 


M  =  Bending  mt.  in  Ib.  inches. 

a 

When  h  =  -01 

When  h  =  -001 

80 

2393 

239-3 

100 

4940 

4940 

120 

11740 

1174-0 

130 

21830 

2183  0 

140 

68950 

6895-0 

145-5 

sc 

oc 

160 

-  32120 

-  3212 

180 

-  15980 

-  1598 

200 

-  11730 

-  1173 

220 

-  9820 

-  982 

CRITICAL  SPEEDS,  CRANK  SHAFTS. 

I  shall  here  consider  a  different  kind  of  critical  speed.  A  crank 
shaft  is  subjected  to  variable  turning  moments,  and  when  it  is  the 
shaft  of  an  electrical  alternator  it  is  subjected  to  variable  resisting 
moments.  Should  the  periods  of  these  happen  to  approach  the  natural 
period  of  vibration  of  the  shaft  there  is  danger  of  fracture.  This 
article  may  be  regarded  as  a  continuation  of  Art.  514. 

Let  there  be  masses  m0,  mv  m2,  ms,  &c.,  connected  by  springs  whose 
stiffnesses  are  a^  cz,  cs,  &c.  Let  "there  be  forces  FO,  Flt  &c.,  acting  on 
the  masses  all  in  the  same  straight  line,  and  let  #0,  xv  #2,  &c.,  be  the 
displacements  all  in  the  same  direction  of  the  masses  from  their  posi- 
tions of  equilibrium.  Or  let  there  be  rotating  masses  whose  moments 
of  inertia  are  w0,  mit  &c.,  connected  by  shafts  all  in  one  line  whose 
stiffnesses  are  cr  c2,  &c.,  and  let  clockwise  couples  F0,  F1?  &c.,  act  on  the 
rotating  masses ;  let  #0,  arlf  &c.,  be  the  angular  displacements  clockwise 


APPENDIX.  673 

of  the  rotating  masses  in  advance  of  their  mean  positions.     Neglecting 
the  masses  of  springs  and  shafts,  it  is  evident  that  if  9  stands  for  djdt 
m    0 


m2  0%  +  cs  (a?2  -  *3)  -  c  2  (^  -  a?2)  =  P2 
ws  0%a  +  6'4  (aJ3-a?4)-03  («2-^)  =  FS.  &c. 
If  there  are  only  four  masses  oss-x4  =.  o. 

If  we  know  FO,  Fa,  &c.,  we  can  calculate  the  motions  of  all  the 
masses  ;  or  if  we  are  given  any  one  of  the  motions  and  all  the  F'S  but 
one.  In  fact,  we  have  the  means  of  working  many  problems. 

Thus,  for  example  :  let  there  be  a  fly-wheel  of  moment  of  inertia 
»?1  =  i  equidistant  between  two  wheels,  each  of  moment,  of  inertia 
m0  =  <ai.2  =  s  I,  and  let  FO  and  F2  be  equal,  but  in  quadrature,  as  if  there 
were  two  cranks  at  right  angles.  As  we  need  not  study  the  average 
angular  velocity,  let  F0  =  F  =  a  sin.  qt,  F2  =  a  cos.  qt,  or  F2  =  6  F/q. 
For  symmetry  Itt  us  take  xQ-xl  =  a,  %%  -  x-^  =  ft,  tl  en  we  get  from  our 
equations 

a  (c  +  sc  +  s  I  02)  +  Be  ft  =  F 

a  So  +  j8  (c  -f  s<?  +  S  I  02)  =  F2 
/  80    \ 

V  4-  stf-s  i  q2  -  —  e)p 

The  solution  is  a  =3  -  i  -  • 
c*  +  2S6-3  +  S2  12  q*  -  2  S  Iq*  (c  +  SC) 

We  see  that  if  F  is  a  sine  function,  or  sum  of  sine  functions  of  the 
time  we  can  find  a.  Taking  it  as  above,  if  o0  is  the  amplitude  of  a  and 
F0  of  F,  if  we  write  a0c  =  F0l  and  q-  s  i/o  =  z. 


jV         v/  (1  +  s  -  zy2  +  sa 

F0  —  1  -f-  2  s  +  z'2  -  2  z  (1  -1-  s) 

If,  then,  we  calculate  the  strength  of  the  shaft  in  the  ordinary 
statical  manner  from  F0  we  see  that  the  maximum  stress  is  greater  than 
we  assumed  in  this  ratio.  It  is  obvious  that  the  ratio  gets  critically 
great  as  we  approach  the  condition  that  the  denominator  is  o,  that  is 
when  z  is  1  or  1  +  2  s.  Thus — 

#i2  —  —    or     •—  (  —  +  2)  are  the  two  critical  conditions. 

IS  I     \ S  ' 

This  critical  q1  is  2  TT/ where/ is  the  natural  frequency  of  vibration 
of  the  arrangement,  and  we  see  that  it  has  two  values.  Now,  the 
moment  F  is  a  sin.  qt  where  q  is  twice  the  angular  velocity  of  the 
crank,  but  F  contains  also  lerms  in  2qt,  3qt,  &c.,  and  even  if  q  itself 
does  not  approach  the  critical  value  q1  its  multiples  may.  There  is, 
there'ore,  no  certain  safety  for  a  crank  shaft  unless  q  exceeds  q1 ; 
that  is,  twice  the  number  of  revolutions  of  a  high-speed  crank  shaft 
per  second  ought  to  exceed  the  natural  frequency  of  vibration  of  the 
shaft. 

It  is  not  necessary  to  give  other  cases  ;  the  mcst  complex  case  may 
be  worked  out  in  the  above  way. 


674  APPLIED    MECHANICS. 

Taking  the  very  simple  case  of  Art.  514,  we  assumed  y  to  be  known. 
If,  however,  F,  the  force  causing  motion,  is  given,  there  is  no  critical 
condition  unless  there  is  a  mass  at  the  place  A.  If,  however,  A  be  a 
fixed  point,  and  the  variable  force  F  acts  upon  the  mass  B,  we  have 
critical  conditions. 

In  the  above  I  have  neglected  the  masses  of  springs  and  of  shafts, 
because  these  are  usually  small. 

Examples  may  be  given  to  students  to  find  the  critical  speeds  of 
valve  rods  with  valves  and  other  masses  upon  them  ;  of  piston  rods 
which  have  crosshead  and  piston  masses  on  them.  A  number  of 
examples  will  be  found  worked  out  in  a  paper  published  by  Sankey, 
Chree  and  Millington  in  the  Proo.  I.  C.  E.,  Vol.  162,  in  which  the  mass 
of  the  shaft  is  taken  in'o  account.  The  paper  by  Messrs.  Frith  and 
Lamb,  published  in  the  lust.  El.  Engineers'  Journal,  Vol.  31,  ought  also 
to  be  consulted. 

In  the  above  fly-wheel  case  I  have  considered  all  the  masses  to  be 
rotating.  Unfortunately  this  is  not  true  in  a  steam  or  gas  engine,  and 
the  real  problem  is  more  complex.  The  departure  from  reality  is  so 
great  that  it  is  not  wise  to  trouble  ourselves  over  such  details  as  the 
masses  of  the  shafts  themselves. 

Exercise. — If  two  wheels  of  moments  of  inertia  m0  and  m^  are  con- 
nected by  a  shaft,  show  that  the  critical  frequency  q  is  such  that : — 
If  m0  is  held  fast  and  m^  allowed  to  vibrate,  let  the  frequency  be 

/i  =  —    k  /  —  ;  if  Wj  is  held  fast  and  w0  allowed  to  vibrate  let  the 

frequency  be  f0  =  ^-~./  -        where  c  is  the  stiffness  of  the  shaft; 

then  the  real  frequency  /  is  /=  ^  f^  +  fi2' 

Page  529. — Students  should  consult  the  Appendix  to  my  book  on 
"  Steam  "  to  see  how  the  rules  here  given  are  at  once  applicable  to  all 
forms  of  steam  turbines. 

Page  531. — In  designing  turbines  and  centrifugal  pumps,  &c.,  atten- 
tion ought  to  be  paid  to  the  fact  (see  page  81)  that  where  a  choice  is 
possible  we  ought  to  let  fluid  flow  in  the  direction  in  which  the 
bounding  walls  of  streams  converge  rather  than  diverge. 

Page  532. — The  proportions  given  above  for  the  Thomson  turbine 
are  not  quite  the  bc»st  under  all  circumstances.  In  any  turbine, 
whether  the  flow  is  axial,  or  radial,  or  combined,  if  R  is  the  average 

radius  of  entrance  to  wheel,  where  V  =/  fj  g  H  —  2  TT  R  w/60,  let  the 

component  velocity  normal  to  opening  be  v,  and  let  v  =  ^/  2^H  X  S/8 
where  I  took  s  =  1  and/=  1  in  the  Thomson.  Let  area  of  entrance 
openings  be  2?r  R  c  R  ;  that  is,  let  c  R  be  radial  or  axial  breadth  of 
opening  (in  the  Thomson  c  was  1/4),  then  Q=  2irc  n2t>  the  cubic  feet 
per  second.  If  p  is  the  total  horse-power  of  the  fall,  and  n  the  revolu- 
tions per  minute,  p  =  (H  133  Q  H.  It  follows  that 


APPENDIX.  675 

5/4      _  i 

I.  In  the  Thomson  and  Jonval,  n  —  23  H    '     P     ?  and 

B  =  2  36  P  !/2  H  ~3/4   because  s  =  1,  c  =  £,/  = 

are  usually  taken.  For  large  powers  and  low  falls  this  gives  n  too 
Hmall  for  many  ordii  ary  purges'. s.  Hence  in  such  cases  it  is  well  10 
take  greater  values  of  s  and  c. 

II.  In  the  advertisement  of  608  turbines  for  falls  from  H  =  3  ftet  to 
H  =  10  feet,  and  K  from  0'375  feet  to  2-25  feet,  and  powers  from  3/4 
1o  1202  actual  horse-power,  I  have  found  that  with    wonderful  con- 
sistency in  all  cases 

»  =  5CHH5/4P    -1/3,    K  =  0-963    p'/2   H-3/4 
Rim  speed  0'9  ^ g  H,    so  =  T51. 

III.  For  very  high  falls  and  small  powers,   to   diminish  n   it  is 
advieable  to  depart  from  the  ordinary  rule  in  the  opposite  way  from  II, 

Take  s  =  1,  c  =  ^. 

Exercise  1.— A   fall   of  10  feet,    100    total  horse-power.      Find  the 
speed  and  size  of  wheel— 1st,  if  cs  =  0'25;  2nd,  if  cs  =  2. 

Answers— 1st  case,  n  =  41,  R  =  4  23  feet. 
2nd  case,  n=  116,  R—  1-5     „ 

Exercise  2. — A  fall  of  500  feet,  5,000  total  horse-power.  If  we 
require  a  speed  of  450  revolutions  per  minute  find  cs  and  the  size  of 
wheel. 

Answer— c  S  =  0  0865,  R  =  2-68  feet. 

Exercue  3.  —A  fall  of  3  feet,  10  total  horse-power.  Find  c  s  and  a 
if  n  muit  be  80  revs,  jer  minute. 

Answer — cs  =  1-95,  K  =  1-17  feet. 

P*ge  538. — When  this  was  written  it  was  sutpos:d  that  in  air  or 
steam  tiowing  from  an  orifice  the  velocity  can  never  exceed  that  of  the 
velocity  of  sound.  I  showed  in  Nature,  of  October  29,  1903,  that  in 
an  ( xpanding  orifice  like  the  Laval  nozzle  there  may  be  much  greater 
velocities. 

It  is  quite  easy  for  the  student  to  work  it  out  himself.  Us'ng  the 
formula  in  line  7,  page  538,  take,  say,  w  =  1,  p^  =  14,400,  or  100  Ibs. 
per  sq.  in.,  and  calculate  a  table  of  values  for  A  for  the  following  values 
of  p.  Now  calculate  v  fiom  (5)  of  page  537.  Evidently  I  take  A  as 
the  cross-section  of  a  steam  tube  at  a  place  where  the  pressure  is  p ; 
A  gets  smaller  to  a  minimum,  its  value  in  the  throat,  and  then  gets 
larger  in  the  expanding  mouthpiece.  There  is  no  great  error  in  the 
formulas,  as  there  is  very  little  friction  before  A  reaches  its  minimum 
value,  but  in  the  mouthpiece  there  is,  of  course,  excessive  friction,  and 
the  tabulated  v  must  be  greater  than  in  reality. 


G76 


APPLIED    MECHANICS. 


1 

p  Ibs.  per 
sq.  in. 

A 

sq.  feet. 

V 

feet  per 
second. 

p  Ibs.  per 
sq.  in. 

A 

sq.  feet. 

V 

feet  per 
second. 

100 

oc 

0 

40 

•00524 

1963 

90 

•00732 

G58 

30 

•00a09 

2252 

80 

•00541 

994 

20 

•00743 

2654 

70 

•00489 

1245 

15 

•00889 

2910 

60 

•00483 

1456 

10 

•01170 

3220 

57-85 

•00481 

1512 

5 

•01430 

3506 

55 

•00484 

1573 

2-5 

•03306 

42H 

50 

•00488 

1708 

— 

— 

— 

If  all  the  pressures  are  doubled  the  values  of  v  are  the  same.  As 
was  to  be  expected,  very  curious  vibrations  occur  in  an  expanding 
nozzle  when  the  angle  of  divergence  is  too  large.  As  in  many  other 
phenomena  in  which  fluid  friction  plays  a  part,  the  student  must  rely 
upon  actual  trial  to  get  good  results. 


1KDEX. 


The  References  are  to  Pages. 


Absorption  Dynamometers,  233 
Acceleration,  13,  262 

Angular,  25 

Linear,  13 

in  Mechanism,  583 

inS.H.M.,  548 

Accumulator,  187 

Advantage,  Mechanical,  59,  88 

Air-vessel,  506 

Algebra,  1 

Alloys  of  Copper,  288 

of  Iron  and  Manganese,  287 

of  Iron  and  Nickel,  287 

Aluminium,  285,  289 

Bronze,  289 

Amperes,  92 
Amplitude,  20 

of  Vibration,  565 

Analogies,  Linear  and  Angular  Motion,  593 
Angle,  Measurement  in  Degrees,  Radians, 
23 

of  Repose,  110, 112, 125 

Sine,  Cosine,  Tangent  of,  23 

of  Twist,  225,  349 

Angular  Acceleration,  25 

Motion,  586 

Ratio,  571 

Velocity,  25 

Aqueduct,  208 
Arch,  113,  391,  457 

Least  Thrust  at  Keystone  of,  165 

• Line  of  Resistance  of,  163 

Load  Carried  by,  163 

Masonry,  166 

Metal,  478 

Middle  Third  of  Section,  163 

Rib  of  Iron,  166 

Thrust  at  Crown  of,  163 

Arched  Rib,  162 
Area,  Centre  of,  136 

of  Curve  by  Integration,  17 

of  Curve  by  Plauimeter,  7 

of  Irregular  Figure,  10 

Moment  of  Inertia  of,  136 

Projection  of,  27 

Areas  of  Figures,  6 

Armstrong's  Hydraulic  Crane,  194 

Ash,  278 

Attwood's  Machine,  64,  245 

Average  Curvature,  22 

Space,  Time,  272,  274 

Velocity,  12 


The  References  are  to  PageSi. 

Axial  Flow  Turbine,  531 
Axis,  Neutral,  381 

of  Suspension,  560 

Axles,  Railway,  83 

Ayrton-Perry  Coupling,  239 

Ayrton  and  Perry's  Dynamometer,  226 


Babbit's  Metal,  289 

Bailey's  Testing  Machine,  294 

Balance,  Chemical,  118 

Compensation  of,  559 

of  Watch,  558 

Spring,  40 

Balancing  of  Hoists,  201,  203,  204,  206 

of  Locomotive,  609 

of  Machines,  606 

of  Reciprocating  Body,  611 

Bali-Bearings,  69,  86 

Ballistic  Pendulum,  499 

Barge,  208 

Barker's  Mill,  486 

Basin,  Water  in,  539 

Bauschinger's  Experiments,  310 

Beam,  127 

Bending  Moment  Diagrams,  Sis 

Cases,  415,  428 

Cast  Iron,  Weight  of,  9 

Curvature  of  a,  386 

Fixed  at  the  Ends,  413,  443 

Not  Rectangular  in  Section,  417 

Shear  Stress  in,  459 

Shearing  Force  in,  414,  427 

Stiffness  of,  431 

of  Uniform  Depth,  446 

of  Uniform  Strength,  419 

Well-known  Rules  about,  410 

Without  Compression,  471 

Bear,  Punching,  179 

Bearing,  Diameter  of,  68 

Friction  at,  68 

Bearings,  Ball,  69,  86 

of  Fans,  Centrifugal  Pumps,  Dy- 
namos, 70 

Beech,  278 

Bell-mouthed  Pipe,  522 

Belt,  Centrifugal  Force  on.  387 

Length  of  Crossed,  26 

Length  of  Open,  26 

Slip  of,  86 

Belting,  Creep  in,  86 

Strength  of,  231 


678 


A.PPLIED    MECHANICS. 


Tlie  References  are  to  Pages. 

Belts,  Difference  of  Pulls  in,  226 
Bending,  380 

and  Crushing,  457 

Moment,  128,  383 

Moment    in    Beams    (more  difficult 

portion),  441 
Moment  Diagrams,  Six  Cases,  414, 

427 

of  Prism  369 

of  Struts,  464 

Theory  of,  318 

Unsymmetrical,  387 

Bends  in  Pipes,  519" 
Bevel  Gearing,  36,  95,  102 
Bicycle  Problems,  49 

Resistance  to  Motion  of,  50 

Bicyclist,  33 
Bifilar  Suspension,  564 
Blue  Prints,  2 
Block,  Pulley,  98 

Snatch,  102 

Blocks  and  Tackle,  98 

Board  of  Trade  Rules  for  Railway  Girders. 

305,  402 

Body  Turning  about  an  Axis,  114 
Boiler,  Bursting  of,  181 
—  Cylindrical,  320 

Spherical,  320 

Strength  of,  319 

Testing  of  Water-tube,  93 

Bolts,  100 

Tightening  of,  311 

Boundary  Condition,  372 

Boussinesq,  M.,  379 

Boy's,  Prof.,  Quartz  Fibres,  299 

Braced  Triangular  Pier,  148 

Bracket,  459 

Brake  Block,  240 

Horse-power,  91,  95 

Knot,  236 

Brass,  288 

Weight  of,  9 

Bricks,  276 
Bridge,  Forth,  299 

Suspension,  161 

Bronze,  288 

Brotherhood's  Water-pressure  Engine,  191 

Bulk,  Diminution  in,  316 

Modulus  of  Elasticity  of,  316 

Bull  Engine,  important  example,  257, 

269 

Bullet,  14,  79,  500 
Bursting  of  Boiler,  181 
Buttress  Thread,  101 
Buttresses,  166,  458 


Calculus,  Differential  and  Integral,  15 
Calorific  Power  of  Coal-Gas,  91 
Cam,  37 
Canal,  170,  206 
Candle,  496 


The  References  are  to  Pages. 

Cannon,  Momentum  of,  486 
Carbon,  280,  281,  285 
Case-Hardening  of  Wrought  Iron.  L'.S.". 
Castings,  Chilled,  183,  284 

Cooling  of,  282 

Internal  Strains  in,  282 

Malleable,  284 

Softening  of  Hard,  281 

Strain  in,  183 

Cast  Iron,  280 

Density  of,  281 

Shell,  173 

Toughening  of,  284 

Weight  of,  9 

Cast  Steel,  183 
Catenary,  169 
Cauchy's  Theorem,  363 
Cedar,  278 
Cement,  276 
Centre  of  Area,  136 

of  Fluid  Pressure,  215 

of  Gravity,  6,  136 

of  Inertia,  136 

Instantaneous,  575 

of  Mass,  135 

of  Oscillation,  560 

of  Percussion,  499,  560 

of  Pyramid  or  Cone,  9 

Centrifugal  Force,  597,  606 

Force  in  Belts  or  Ropes,  327,  605 

Pump,  33,  264,  540 

Centripetal  Force,  603 

Acceleration,  603 

C.G.S.  System  of  Units,  267 
Chain  Gearing,  232,  572 

Hanging,  167 

Horizontal  Pull  of,  152 

Loaded,  161 

Chains,  Admiralty  Rule  for,  315 
Cliange  Wheels,  101 
Charcoal  Iron,  284 
Chemical  Balance,  118 

Energy,  42 

Chemistry,  290 
Chilled  Castings,  183 
Chipping,  488 

and  Filing,  56 

Hammer,  264 

Chisel,  Tempering  of,  291 

Chree,  Dr.,  371,  380 

Chromium,  287 

Chronometer,  649 

Chuck,  Elliptic,  577 

Circle,  Circumference  of,  6 

Circuital,  30 

Circular  Stream  Lines,  541 

Cliuure,  29,  122 

Clock,  38 

Closed  Polygon,  120 

Coal,  Energy  of  1  Ib.  of,  42,  48 

Co-efficient  of  Friction,  05 

of  Friction,  Alteration  of,  75,  230 

of  Viscosity,  76 

Collars,  Leather,  170,  176 
Colours  in  Tempering,  291,  292 


INDEX. 


679 


The  References  are  to  I 'ages. 

Columns,  473 

Combination  of  S.H.M.'s.  555 

Combined  Bending  and  Twisting,  353 

Commercial  Tests,  306 

Compound  Interest  Law,  20,  99,  230 

Pendulum,  559 

Wheel  and  Axle,  102 

Concrete,  183,  277 

Conduit,  516 

Cone,  Area  of  Surface  of,  8 

Rolling  of  a,  577 

Surface  of,  8 

Coned  Pulley,  37 

Cones,  Stepped,  26 

Conical  Pendulum,  546 

Confluent,  30 

Connecting  Rod  and  Crank,  222 

Construction  of  Gun,  830 

Materials  Used  in,  275 

Contracted  Vein,  533 
Contraction  of  Section,  306 
Cooling  of  Castings,  282 
Copper,  287 

Wire,  289 

Cores,  10,  282 
Cork,  Floating,  547 
Corrugated  Flue,  326 
Cosine  of  an  Angle,  23 
Cottars,  101 
Cotton  Press,  184 
Couple,  127 
Coupling,  225 

Ayrton-Perry,  239 

Dynamometer  96 

Flange,  227 

Rod,  471 

Crab,  102 
Crane,  88 

Double-Powered,  195 

Hook,  458 

Hydraulic,  193 

Crank  and  Connecting-Rod,  222 

Overhung,  355 

Creep  in  Belting,  86 
Creeping,  299 
Cricket  Bat,  500 
Critical  Velocity,  83 
Crosshead  of  Engine,  113,  547 
Crushing  and  Bending,  457 
Cupola,  282 
Curvature  of  a  Beam,  386,  405 

Definition  of,  22 

Radius  of;  22 

of  Steel  Spring,  388 

Curve  of  Cosines,  555 

Elastic,  390 

Integral  of,  18 

Logarithmic,  508 

of  Sines,  78,  555,  565,  568 

Curved  Rollers,  522 
Curves,  Damping,  78 
—  Troohoidal,  573 
Cutting  of  Metals,  45 
Cylinder,  Rotating,  369 

Thipk,  183,  323 

W* 


The  References  ure  to  Pages. 
D 

Damping  Curves,  78 

-  of  Vibration,  566 
D'Arcy's  Experiments,  85 
Dash-pots,  239 
Deadloads,  305 
Definition  of  a  Radian,  23 
Definitions  of  Trigonometrical  Ratios,  23 
Deflection  due  to  Shearing,  334,  460 

-  in  Beams,  407,  429,  431,  460 
Delta  Metal,  288 
Descriptive  Geometry,  124 
Diagonal  Pieces,  162 

-  of  Velocities,  583 
Diagram  of  Accelerations,  583 

-  of  Bending  Moment,  six  cases,  413- 

Diagrams  of  Shearing  Force,  427 
Differential  Co-efficient,  16 

-  Pulley  Block,  Efficiency  of,  104 

-  Pulley  Block,  103 
Diminution  of  Swing,  563? 
Displacement  of  a  Ship,  55 

-  Relative  and  Absolute.  85 
Distance  of  Rubbing,  68 
Draining  of  Fields,  516 
Drawing  Office,  1 

Drill,  Watchmaker's,  291 

Dunkerley,  Mr.,  Stability  of  Shafts,  477 


Dynamo  Machine,  43,  92 
Dynamometer  Coupling,  96 

-  Froude's,  Thorncycroft,  234 

-  Hefner-  Alteneck,  235,  240 

-  Raffard's,  86 

-  ;  Transmission,  9S 

-  Transmission  and  Absorption  ,  233 


Earth,  Pressure  of,  167 

Rankine's  Rule  for,  349 

Earthquake,  497-617 
Eccentric,  222 
Economical  Efficiency,  89 
Economy,  General,  33P 
Eddies,  518 
Effect  of  Friction,  60 
Efficiency,  95 

Economical,  89 

Hydraulic,  529 

of  Screw  Jack,  110 

of  Turbine,  522 

Elastic  Curve,  390 

Elasticity,  More  Difficult  Theory  of,  361 

of  Bulk,  Modulus  of,  317 

Shearing,  333 

Young's  Modulus  of,  299 


680 


APPLIED    MECHANICS. 


The  References  are  to  Pages. 

Electric  Energy,  92 

Lamps,  43 

Motor,  42 

Supply  Station,  92,  94 

Electrical  Horse-power,  92,  95 
Electricity,  290 
Ellipse,  Area  of,  8 
Elliptic  Trammel,  577 

Chuck,  577 

Elm,  278 

Energy,  Chemical,  42 

Indestructibility  of,  59 

Kinetic,  Potential,  40,  242,  505,  511 

Loss  of,  due  to  Friction,  67 

Loss  of,  in  Change  of  Strain,  303 

Lost,  Fluid  in  Pipe,  53,  84 

of  1  Ib.  of  Coal,  42-43 

Pressure,  505-511 

in  a  Rotating  Body,  247,  254 

Storage  of,  in  Fluids,  326 

Store  of,  188 

Stored  up  in  any  Machine,  253 

Strain,  85 

of  Strained  Spring,  41 

Transmission  of,  Hydraulically,  170 

Waste  of,  in  Conductors,  43 

of  Waterfall,  59 

Engine,  Bull,  257 

Hero's,  486 

Steam,  Gas,  Oil,  43 

Water  Pressure,  189-191 

Willans',  611 


English,  Colonel,  55 
Epicyclic  Train,  35 


Epicycloids,  573 
Equilibrant,  30 

of  Forces  in  One  Plane,  122 

Equilibrium,  Conditions  of,  121 

in  One  Position,  104 

Equi-pollent  Loads,  366 
Equi-potential  Surfaces,  218 
Errors  of  Observation,  62 
Escapements,  554 
E wing's,  Prof.,  Extensometer,  294 
Expansion,  Cubical,  Co-efficients  of,  5 

Linear,  Co-efficients  of,  4 

of  Girder,  133 

Experiments  by  Bauschinger,  310 

by  D'Arcy,  85 

by  Fairbairn,  309 

by  M.  Tresca,  301 

by  Reynolds,  Prof.  O.,  81 

by  Tower,  M.  B.,  73,  81 

by  Wohler,  305,  309 

on  Attwood's  Machine,  245 

on  Balancing,  608 

on  Crane,  88 

on  Deflection  of  Beams,  431 

on  Fluid  Friction,  77,  79 

on  Forces  at  a  Pin  Joint,  151,  152 

on  Friction,  60 

on  Friction  of  Cords,  Belts,  228 

on  Friction  of  a  Machine,  63 

on  Friction  and  Speed,  71,  72 

— -  on  Guns,  14 


The  References  are  to  Pages, 

Experiments  on  Hydraulic  Jack,  176 

on  Inclined  Plane,  104 

on  Jack,  63 

on  Loaded  Links,  160 

on  Moments,  115 

on  Shearing,  332 

on  Strain,  293 

on  Triangle  of  Forces,  56 

on  Twisting  of  Wires,  349 

to  Find  Co-efficient  of  Friction,  67 

to  Find  Energy  in  Rotating  Body, 

247 

to  Find  Modulus  of  Rigidity,  563 

with  Discs  in  Fluids,  78 

with  Gas-engine,  91 

with  Oil-engine,  91 

with  Pulley  Block,  103 

with  Small  Carriage,  262 

Extensometer,  Prof.  Ewing's,  294 


Factor,  Load,  94 

of  Safety,  304 

Fairbairn's  Experiments,  309 

Falling  Weight,  Work  Doue  by,  29 

Fatigue,  559 

Ferguson's  Paradox,  86 

Fir,  White,  277 

Firwoods,  277 

Flat  Plate,  330 

Flexible  Pipe,  197 

Flexm«l  Rigidity,  634 

Flexure,  Non-uniform,  378 

Flow  of  Gas  from  Orifice,  537 

of  Metals,  335 

of  Water,  Sudden  Stoppage  of,  192 

of  Water  through  Notch,  514,  536 

Flue,  Corrugated,  331 
Fluid  Friction,  76 

Friction,  Cause  of,  80 

Fluids  in  Motion,  505 

Steady  Motion  in,  532 

Storage  of  Energy  in,  326 

Whirling,  216 

Force  Acting    on  a    Body,  important 

example,  265 

Centrifugal,  597,  606 

Constant,  267 

Due  to  Pressure  of  Fluids,  214 

Lines  of,  217 

Moment  of,  115 

of  a  Blow,  490 

of  a  Blow,  Time  Average  of,  273 

of  Friction,  64 

Polygon,  124 

Shearing,  383,  385 

Space  Average  of  a,  45 

Time  Average  of  a,  45 

Forces  Acting  at  a  Point,  123 

Graphical  Representation  of,  30 

in  One  Plane,  Analytical  Rule,  121 

in  Space,  Examples  on,  143,  144,  145 

— —  in  Structures,  Determination  of,  150 


INDEX. 


681 


The  References  are  to  Pages. 

Forces  Not  All  in  One  Plane,  124 

Parallelogram  of,  30 

Resultant  of,  29 

Triangle  of,  30,  56 

Forging,  Hydraulic,  170 
Forth  Bridge,  299 
Four-Bar  Kinematic  Chain,  577 
Fracture  Surface,  306 
Frame  of  a  Machine,  35 
Framed  Structures,  143 
Francis,  Mr.,  536 
Frequency,  20 
Friction  and  Speed,  71 

at  Bearing,  68 

at  Different  Speeds,  73 

at  Joints,  112 

at  Leather  Collar,  171,  176 

Between  Cord  and  Pulley,  228 

Circles,  113 

Co-efficient  of,  65 

Cumulative  Effect  of,  99 

Effect  of,  60 

Experiments  on,  60 

Fluid,  76 

Fluid,  Experiments  of,  77,  78 

Force  of,  64 

Internal,  495 

Laws  of  Fluid  and  Solid,  80 

Never  Negligible,  248 

of  a  Pivot,  68,  71 

of  Pulley,  58,  248 

of  Railway  Brake,  75 

of  Ships,  56 

on  Inclined  Plane,  107,  108 

Quasi-Solid,  177 

Skin,  54 

Statical,  75 

Wheels,  69 

Frictional  Loss,  Reduction  of,  68 

Frictionless  Hinges,  151 

Froude,  54 

Froude's  Dynamometer,  234,  238 

Fuel  Consumption,  93 

Fuller,  Prof.,  164 

Functions,  Graphing  of,  21 


Gulileo,  244 
Gas-Engine,  43 

—  Experiment  with  a,  91 

Gas  Flowing  from  Orifice,  537 
Gauche  Polygon,  121,  143 
Gauge  Pressure,  505 
Gearing  Chain,  232,  572 

Spur,  Bevel,  95,  102 

General  Plane  Motion,  575 
Geology,  275 
Geometry,  Practical,  1 
German  Silver,  289 
Girder,  393 

Bending  Moment  in,  393,  896 

Boa»-d  of  Trade  Rules  for,  305 

Calculation  of  Forces  in,  151 


The  References  are  to  Pages. 

Girder,  Diagonal  Braces  of,  393,  396 

Flanges  of,  393,  396 

Railway,  401 

Rolled,  419,  421 

Shearing  Force  in,  393,  896 

Web  of,  393,  397 

Glass,  279 

Sudden  Cooling  of,  279 

Toughened,  280 

Graphical  Statics,  1,  149, 151 

Graphing  of  Functions,  21 

Greenhill,  Prof.,  on  Stability  of  Shafts,  47o 

Grid,  Hydraulic,  212 

Guide  Blades,  529 

Gun  Experiments,  14 

Lifting  of,  199 

Making  of,  325 

Metal,  288 

Tensions  in,  183 

Wire,  325 

Gyration,  Radius  of,  138,  560 
Gyrostat,  498 


Hammer,  Chipping,  264 

Tilt,  500 

Hammering  of  Iron,  285 
Hanging  Chain,  167 
Hardening,  Case,  285 

of  Steel,  290 

Harmonic  Motion,  20 

Law,  20 

Heat,  290 

Energy,  Unit  of,  42 

Heavy  Disc,  Vibration  of,  568 
Hemp-Packing,  177 
Hero's  Engine,  486 
Hinge,  Quasi,  166 
Hinged  Arch,  114 
Hoists,  Balanced,  178 

Balancing  of,  201,  203,  204, 

Hotel,  Mill,  Warehouse,  20 

Horse-power,  39,  89 

Brake,  91 

Electrical,  92 

Indicated,  92,  95 

Hydraulic  Crane,  193 

Efficiency  of  Turbine,  528 

Forging,  170 

Grid,  212 

Intensifier,  186 

Jack,  63,  178 

Jack,  Efficiency  of,  175 

Limestones,  27t» 

Mean  Depth,  85 

Power  Company,  94, 187 

Power,  Loss  of,  53 

Press,  170 

Ram,  501 

Transmission,  52 

Hydraulics,  170 
Hypocycloids,  573 


682 


APPLIED    MECHANICS. 


The,  References  are  to  Pages. 

I 

I,  for  Various  Sections,  251 
Idea  of  Law  of  Dependence  of  two  Vari- 
able Things,  63 

of  a  Rate,  14 

of  Slope,  15 

of  Velocity,  12 

Impact,  487,  505 
Impulse,  264 
Inclined  Plane,  99,  104 

Friction  on,  107 

Indiarubber,  305 

Cup,  177 

Shaft,  318 

Spring,  41 

Indicated  Horse-power,  95 
Indicator  Diagram,  40,  274 

Centre  of,  149 

Inertia,  40 

Moment  of,  135 

Principal  Moments  or,  137 

Rule  for,  7 

Instantaneous  Centre,  575 
Integral  of  a  Curve,  18 
Integrals,  Table  of,  16 
Intensifier,  186,  188 
Involute  of  Circle,  573 
Inward  Radial  Flow  Turbine,  531 
Iron,  Alloys  of,  287 

Cast,  280 

Charcoal,  284 

Galvanised,  287 

Pig,  284 

Rolling  and  Hammering,  285 

Wrought,  284 

Irregular  Figure,  Approximate   Area  of, 


Jack,  Hydraulic  or  Screw,  63,  178 
Jet  Pump,  516 
Joint,  Masonry,  458 
Joints,  Riveted,  336 

with  Friction,  112 

Joule,  42 

Journal,  Friction  at,  68 

Journals  and  Footsteps,  78 


Rater's  Pendulum,  560 
Kelvin,  Lord,  302 
Kelvin's  Analogy,  358 

Tide-Predicting  Machine,  555 

Kerosene,  Energy  of,  42 

Keys,  101 

Keystone  of  Arch,  165 

Kinematics,  1 

Kinetic  Energy,  242,  505-511 

of  Any  System,  588 

Kinetics  of  Mechanism,  220 


The  References  are  to  Pages. 


Kirchhoff,  366 
Knife  Edges,  118 
Knot  Brake,  236 
Kohlrausch,  303 


Laboratory,  Mechanical,  1 

Ladder,  Example  on,  134 

Lake  of  Water,  512 

Landing  Stage,  206 

Larch,  278 

Lathe,  Screw-Cutting,  101 

Law  Connecting  Variable  Things,  63 

of  Crane,  89 

of  Friction  for  a  Machine,  63 

of  Moments,  115 

of  Worth,  58 

Laws  of  Fluid  and  Solid  Friction,  80 
Leather  Collars,  170,  176 

Friction  at,  171 

Level  Surface,  22 
Lever,  116 
Limestones,  275 

Natural  Hydraulic,  276 

Line  of  Resistance,  113,  126,  161,  163 

Projection  of,  27 

Vertical,  22 

Lines  of  Force,  217 
Link  Motions,  584 
Link  Polygon,  129 
Links,  Loaded,  160 
Live  Load,  305 
Load  Factor,  94 

Rolling,  402       '     . 

Loaded  Chain,  161 

Links,  160 

Loads,  Equi-pollent,  366 

Suddenly  Applied,  306 

Loam,  282 

Local  Strengthening,  306 

Lock  Nuts,  311 

Locomotive,  Pull  of,  76 

Logarithmic  Curve,  568 

Logarithms,  1 

Loss  of  Energy,  by  Impact,  488 

of  Energy  due  to  Friction,  67 

of  Energy  due  to  Fluid  in  Pipe,  84 

of  Energy  in  Change  of  Strain,  303 

of  Energy  per  Ib.  of  Water,  518 

of  Power  at  Different  Parts  of  Engine 

90 

Love's  Treatise  on  Elasticity,  361 
Lowell  Formula  for  Rectangular  Notch, 

515,  536 
Lubricant,  66 
Body  of,  75 


M 

M,  for  Various  Sections,  251 

of  a  Wheel,  2-17,  248,  254 

Machine,  Bailey's  Testing,  294 


INDEX. 


683 


The  ntfrrences  are  to  Pages. 

Machine,  Cotton  Baling,  186 
Design,  TO 

Kinetic  Energy  Stored  up  in,  253 

Riveting,  170 

Machinery,  Quick  Speed,  530 

Water  Pressure,  83 

Machines,  Balancing  of,  606  • 
Forging,  Welding,  Punching,  Stamp- 
ing, Shearing,  196,  198 

Shearing,  335 

Steadiness  of,  252 

Magnet,  564 

Mahogany,  278 

Mainspring  of  Timekeeper,  554 

Malleable  Castings,  284 

Manganese,  284 

Bronze,  289 

Marbles,  275 
Masonry  Joint,  459 
Mass,  40 

Centre  of,  135 

Materials,  Behaviour  of,  290 
Used  in  Construction,  275 

Weights  of,  9 

Mathematical  Tables,  24 
Mean  Depth,  Hydraulic,  85 
Mechanical  Advantage,  59,  88 
Hypothetical,  102 

of  Hydraulic  Jack,  174 

Mechanism,  220,  570 

Four-Link,  35 

Kinetics  of,  220 

Quick  Return,  578 

Mechanisms,  Acceleration  of,  583 

Velocity  of,  583 

Memel,  277 
Mensuration,  1,  6 
Metal,  288 

Arches,  478 

Babbit's,  289 

Delta,  289 

Flow  of,  298,  301,  336 

Gun,  289 

Muntz,  289 

Sterro,  289 

White,  289 

Mills,  Driving  of,  515 

Milne,  Prof.,  401 

Mitis  Castings,  287 

Modulus  of  Elasticity  of  Bulk,  317 

of  Rigidity,  333,  563 

of  Section,  398 

Young's,  299,  305 

Moment,  Bending,  128,  384 

Law  of,  115 

of  a  Force,  115 

of  Inertia,  135,  136 

of  Inertia  of  a  Circle,  141 

of  Inertia  of  Area,  149 

of  Inertia  of  Rectangle,  141 

of  Inertia  of  Sections,  398 

of  Momentum,  500,  528,  587 

Work  done  by,  115 

Momental  Ellipse,  142 
Momentum,  263 


The  References  are  to  Pages. 

Momentum,  Important  Example,  274 

of  Cannon,  4S6 

of  Shot,  486 

Tangential,  524 

Mortar,  276 

Motion,  Fluids  in,  505 

of  Rotation,  127,  592 

of  Translation,  121,  592 

Periodic,  546 

Produced  by  Blow,  498 

Motions,  Link  Value,  584 
Moulder,  9 
Moulding,  282 
Muntz  Metal,  88 


N 


Neutral  Axis,  381 

Surface,  381 

Newton's  Second  Law,  602,  604 

Nickel,  287,  289 

Non- Redundant  Frame,  Criterion  for,  148 

Norway  Spruce,  277 

Notch,  514 

Gauge,  515,  536 

Rectangular,  536 

Triangular,  514,  536 

Nozzles,  513 
Nuts,  311 


Oak,  English,  278 

Oil  Engines,  Experiments  with,  91 

Engines,  43 

Sperm,  66 

Tester,  Thurston,  81 

Orifice,  512 

Gas  Flowing  from,  537 

Rectangular,  535 

Sharp-Edged,  513,  534 

Triangular,  535 

Water  from,  512- 

Oscillating  Cylinder  Engine,  578 
Outward  Radial  flow  Turbine,  531 
Overhauling,  96,  104.  109 
Overhung  Craak,  355 


Packing  Hay,  185 
Parabola,  162,  168 
Parabolic  Rib,  162 
Paraboloids  of  Revolution,  218 
Paradox,  Ferguson's,  36 
Parallel  Motion,  Watt's,  C9 
Parallelogram  of  Forces,  30 
Patterns,  10,  282 
Peaucellier  Cell,  578 
Pendulum,  Ballistic,  499   500 

Compound,  559 

Conical,  546 


684 


APPLIED    MECHANICS. 


The  References  are  to  Pages. 

Pendulum,  Equivalent  Simple,  559 

Impulse  given  to,  554 

Simple,  242 

• Simple,  Time  of  Swing,  551 

Percussion,  Centre  of,  499 
Periodic  Motion,  546,  592 

• Not  S.H.M.,  554 

Time,  20,  546 

Time  of  Balance,  558 

Permanent  Axes,  607 

Set  in  Wire,  300 

Perry's,  Mr.  James,  Syphon,  517 

Phosphor  Bronze,  288 

Phosphorus,  284,  287 

Physics,  290 

Pianoforte  Wire,  289 

Piezometer,  173 

Pig-Iron,  Puddling  and  Refining  of,  284 

Pile  Driver,  273,  489 

Pin,  Resultant  Force  at,  112 

Pine,  Red,  277 

Pins,  101 

Pipe,  Bell-Mouthed,  522 

Flexible,  197 

Resistance  to  Motion  of  Fluid  in  a, 

76 

Strength  of,  182,  319 

Suddenly  Enlarged,  5 

Wooden,  183 

Pipes,  Bends  in,  519 

Flow  of  Water  in,  53 

Piston-Rod,  309 

Pitch  of  Screw,  100 

Plane  Motion,  576 

Planimeter,  2,  7 

Plastic  Elongations,  301 

Plasticity,  301 

Plate,  Flat,  330 

Platform,  Weight  of,  402 

Plotting  on  Squared  Paper,  62,  80,  89,  91, 

92,  229 

Poisson's  Ratio,  343 
Polygon,  Closed,  120 

Gauche,  121 

Link, 129 

of  Forces,  124 

Unclosed,  124 

Poitlaml  Cement,  277 

Potential  Energy,  40,  242,  505,  511 

Poundage  of  Steam,  95 

Power,   Loss  of,  at  Different   Parts    of 

Engine,  90 

Misuse  of  this  Expression,  89 

of  a  Stream,  515 

Transmission  by  Shafts,  224 

Precession  of  Top,  596 
Press,  Cotton,  184 

for  Packing  Hay,  1S5 

for  Warehouses,  185 

Hand,  185 

Hydraulic,  170 

Pressure  Energy,  505,  511 

Gauge,  505 

of  a  Fluid,  215 

of  Earth,  167,  348 


The  References  are  to  PagtA. 

Pressure  of  Water,  167 

on  Immersed  Surface,  215 

Principal  Moment  of  Inertia,  139 

Stress,  354 

Stresses,  364 

Principle  of  Work,  120 

Prints,  10,  282 

Prism,  Centre  of  Gravity  of,  9 

of  Elliptic  Section,  376 

Twisting  or  Bending  of,  309 

Volume  of,  9 

Prismatic  Body,  Volume  of,  9 
Projectile,  245 
Projection  of  Area,  27 

of  Line,  27 

Propeller  Shaft,  82,  100 
Propulsion  of  Ships,  52,  525 
Protractor,  23 
Puddling  of  Pig-iron,  284 
Pull,  How  it  is  Exerted,  292 
Pulley  Block,  98 

Block,  Efficiency  of,  98 

Coned,  37 

Friction  of,  58 

Rim  of,  283 

Pump,  505 

Centrifugal,  33,  264,  508 

Double- Acting  Force,  506 

Efficiency  of,  43 

Force,  506 

Jet,  516 

Lifting,  505 

Mechanism,  578 

Punching,  335 
Bear,  179 


Quantity,  Scalar,  29 

Vector,  29 

Quartz  Fibres,  299 
Quasi-Hinge,  166 

Rigidity,  497 

Solid  Friction,  177 

Quicklime,  276 

Quick  Return  Mechanism,  578 


Rack,  573 
Radial  Gears,  584 
Radian,  Definition  of,  23 
Radius  of  Curvature,  22,  388 

of  Gyration,  136,  251,  560 

Rafford's  Dynamometer,  86 
Railway  Axles,  83 

Brake,  Friction  of,  75 

Girder,  133,  402 

Ram,  Hydraulic,  503 

Rankiue,  319 

Rankine's  Rule  for  Earth,  349 

Rate,  Algebraical  Representation  of,  16 

Ratio,  Velocity,  5J» 


INDEX 


685 


The  References  are  to  Pages. 

Rectangular  Notch,  536 

Orifice,  534 

Red  Pine,  277 
Redundant  Bars,  395 
Refining  of  Pig-iron,  284 
Regulation  of  Turbine,  529 
Relative  Velocity,  32,  34,  35 

Viscosities,  569 

Resilience,  41,  308 

Compressive,  316 

Shear,  316 

Tensile,  316 

Resistance,  Line  of,  113, 114, 161,  430 

of  a  Moving  Train,  44 

to  Motion  of  Fluid  in  a  Pipe,  76 

to  Rolling,  85 

Wave,  54 

Resolved  Part  of  a  Vector,  29 
Resultant  Force,  29,  129 

on  Forces  in  One  Plane,  122 

Reynolds,  Prof.  O.,  73,  78,  81-4,  541 
1Mb  Arched,  162 

Parabolic,  162 

Rigidity,  Flexural,  634 

Modulus  of,  333,  563 

Quasi,  497 

Torsional,  634 

Rim  of  Wheel,  Volume  of,  9 
River  Weaver,  206 
Riveted  Joints,  336 

Work,  297 

Riveting  Machine,  170 
Rocks,  Slaty,  Stratified,  275 
Rolled  Girder  Section,  419,  421 
Rollers,  Curved,  572 

for  Girder,  133 

Rolling,  106 

and  Hammering  of  Iron,  285 

of  a  Cone,  577 

Load,  403 

Resistance  to,  85 

True,  571 

Roof,  Calculation  of  Forces  in,  151,  153 

Principal,  133 

Rope,  Centrifugal  Force  in,  322 

Weight  of,  315 

Ropes,  Wire,  232 

Rotating  Body,  Energy  in,  247 

Cylinder,  368 

Rotation,  Motion  of,  127 
Rubbing,  Distance  of,  68 
Rules  for  Beams,  410 

for  Deflection  of  Beams,  431 

for  Flow  of  Water  through  Notch, 

515,  536 

for  Periodic  Time  in  S.H.M.,  549 

for  Simple  Pendulum,  551 

for  Struts,  466 

for  Strength  of  Pipe,  182 

Rupert's  Drop,  183,  280 


S 
Safety,  Factor  of,  SO 


The  References  are  to  Pages. 

Safety-valve,  116 
Sandstones,  275 

Green,  Dry,  282 

Scalar  Quantity,  29 
Scraped  Surfaces,  73 
Screw,  The,  99 
Cutting  Lathe,  101 

Jack, 63 

Jack,  Efficiency  of,  110 

Piles,  101 

Pitch  of,  100 

Propeller,  101 

Propeller  Blades,  289 

Square-Threaded,  109 

Thread, Whitvvorth,  Sellers  Buttress 

101 

Seasoning  of  Timber,  101 
Section,  Change  of  Shape  of,  374 
Sections  of  Structures,  159 
Seizing,  73 
Sellers'  Thread,  101 
Sense  of  Vector,  29 
Shaft,  Hollow  Round,  354 

Indiarubber,  318 

Overtwisted,  290 

Section,  an  Equilateral  Triangle,  377 

Shafts  of  Various  Sections,  357 

Stability  of,  475 

Strength  of,  351,  352 

Subjected  to  Twisting  and  Bending. 

353 

Whirling  of  Loaded,  476 

Shape  of  Stream  from  Orifice,  534 

of  Teeth,  572 

of  Worm  Thread,  574 

Sharp-edged  Notch,  536 

Orifices,  534 

Shear  and  Twist,  332 

Strain,  332,  341,  350 

Stress,  332 

Stress  in  Beams,  460 

Shearing  Force,  128,  383,  385,  631 

Force  in  Beams,  414,  427 

Machines,  335 

Ship,  496 

Displacement  of  a,  55 

Heeling  Angle  of,  260 

Righting  Moment  of,  260 

Ships,  Friction  of,  56 

Horse-power  of,  54 

Models  of,  55 

Propulsion  of,  52,  525 

Resistance  to  Motion  of,  54 

Speed  of,  53 

Shot,  Momentum  of,  486 
Silicon,  284 

Bronze,  289 

Silver,  Alloys  of,  289 
Similar  Figures,  Area  of,  8 

Structures,  Similarly  Loaded,  424 

Simple  Harmonic  Motion,  20,  222.  546 

Harmonic  Motion,  Amplitude  of,  20 

Harmonic  Motion,  Frequency  of,  20 

Harmonic  Motion,  Periodic  Time  of, 

20 


686 


APPLIED    MECHANICS. 


The  References  are  to  Pages. 

Simple  Harmonic  Motions,  Combination 

of,  555 

Simpson's  Rule,  7 
Sine  of  an  Angle,  23 
Sines,  Curve  of,  78 
Skeleton  Drawings,  221,  570 
Skew  Bevel  Wheels,  572 
Slider  Crank  Chain,  577 
Sliding,  106 

Contact,  571 

Slip  of  Belt,  86,  232 
Slope  of  a  Curve,  15 
Smith,  Prof.  R.  H.,  45,  582 
Snatch  Block,  102 
Soapy  Water,  Use  of,  237 
Space  Average  of  a  Force,  45 

Average,  Time  Average,  272,  274 

Rate,  Time  Rate,  209 

Specific  Gravity,  10 
Speed  of  Bullet,  14 

of  Commercial  Shies.  53 

of  Train,  12 

Sperm  Oil,  66 
Sphei  e,  Surface  of,  8 

, Volume  of,  8 

Spherical  Shell,  Thick,  325 
Spinning,  106 

Tops,  498 

Spiral  Flow  of  Water,  539,  540 

Spring,  Vibration  of,  550 

Spring  Balance,  40 

of  Indiarubber,  41 

of  Steel,  41 

Spiral,  Vibration  of,  550 

Springs,  613 

Buffer  Stop,  613 

C,  643 

Carriage,  618,  646 

Clock,  613 

Cylindrical  Spiral,  624,  633 

Different  Forms  of,  CIS 

Elongation  of,  630 

Flat  Spiral,  624,  027,  628,  6?9 

Hardening  and  Tempering  of,  C22, 649 

Materials  Used  in,  620 

Phosphor  Bronze,  622 

Resilience  of,  620,  641 

Spiral,  622 

Tubular  Spiral,  642,  643 

Uses  of,  620 

Vibration,  615 

which  Bend,  462 

Spruce,  Norway,  277 
Spur  Gearing,  95, 102 

Wheel,  36 

Squared  Paper,  1,  13,  15,  21,  44,  60,  89,  90, 

229,  293,  300,  565 

Square-threaded  Screw,  Efficiency  of,  109 
Statical  Friction,  75 
Steadiness  of  Machines,  252 
Steady  Motion  in  Fluids,  532 
Steam  Engine,  43 

Turbine,  531 

Steel,  285 
Basic,  286 


The  References  are  to  Pages. 

Steel,  Bessemer,  285 

Cast,  183 

Castings,  Annealing  of.  286 

Crucible,  286 

Hardening  of,  290 

Ingots,  285 

Martin,  289 

Rails,  286 

Shear,  285 

Siemens',  286 

Spring,  41 

Strength  of,  286 

Tempering  of,  28 

Weight  of,  9 

Step,  68 

Stepped  Cones,  26 
Stiffness  of  Beams,  432 
Stilling  of  Vibrations,  565 
Stone,  275 

Artificial,  276 

Stoppage  (Sudden)  of  Water  in  a    IMpe, 

192,  501 

Storage  of  Energy  in  Fluids,  321 
Stores  of  Energy,  42,  509 
Straight  Line  Law,  62 
Strain,  293,  296,  312 

Energy,  85,  307,  489 

in  Castings,  183 

Nature  of.  311 

Potential^  362 

Shear,  331,  341 

Strained  Spring,  Energy  of,  41 
Stream  Lines,  Circular,  511 
Strength  of  Boiler,  319 

of  Chains,  315 

Modulus  of  Sections,  399 

of  Pipe,  182,  319 

of  Ropes,  315 

Stress,  296 

Principal,  354 

Shear,  331 

Structures,  Determination  of  Forces  in, 

156 

Method  of  Sections,  159 

Struts,  151,  463 

Bending  of,  464 

Euler's  Formula,  465 

Rules  for,  467 

Shortening  of,  298 

with  Lateral  Loads,  470 

St.  Venant,  296,  318,  348,  306,  370,  379 
Suddenly  Applied  Load,  306 
Sulphur,  284 

Sun  and  Planet  Motion,  36 
Surface,  Level,  22 

of  Fracture,  306 

Scraped,  73 

Surfaces,  Equi-potential,  218 
Suspension,  Axis  of,  500,  561 

Bi  filar,  564 

Bridge,  161 

Switchback  Railway,  41 
Syphon,  79 

Lubricator,  83 

Mr.  James  Perry's,  517 


INDEX. 


687 


The  References  are  to  Pages. 
T 

Table  of  Constants,  654 

I.  of  Sections,  397 

of  Strength  Moduli,  397 

Tables,  Mathematical,  24 

Tangent  of  an  Angle,  23 

Tangential  and  Normal  Components,  106 

Teak,  278 

Teeth  of  Spur-wheel,  572 

of  Wheels,  423 

of  Worm-wheel,  572 

Telegraph  Wire,  Dip  of,  108 
Tempering  Colours,  291,  292 

of  Steel,  290 

Template,  218 
Testing  Machine,  I 

Machine,  Bailey's,  294 

of  Water-tube  Boiler,  93 

Theorem  of  Three  Moments.  454 
Theory  of  Bending,  318 

of  Fluid  Friction,  80 

Thick  Cylinder,  183,  321,  328,  367 

Spherical  Shell,  325 

Thin  Cylinder,  324 
Thomson's  Jet  Pump,  516 

Prof.  J.,  Dynamometer,  235 

Prof.  J  ,  Overtwisted  Shaft,  290 

Triangular  Notch,  514,  536 

Turbine,  526 

Whirlpool  Chamber,  522 

Thorneycroft's  Dynamometer,  234 
Thread,  Whitworth,  100 
Threads,  Shapes  of,  101 
Thurston  Oil  Tester,  81 
Tide-predicting  Machine,  Kelvin,  555 
Tide,  Rise  and  Fall  of,  554 
Tie  Rod,  151 
Tilt  Hammer,  500 
Timber,  Felling  of,  278 

Preserving  of,  279 

Seasoning  of,  277 

Strength  of,  279 

Time  Average  of  a  Force,  45 

Time  Average,  Space  Average,  272,  274 

Time,  Periodic,  546 

Time  Rr.te,  Space  Rate.  269 

Tin,  289 

Top,  Precession  of,  596 

Torque,  77,  95,  127,  260,  351 

Torsional  Rigidity,  634 

of  a  Shaft,  352 

Tortuosity  of  Path,  606 

Total  Energy  of  1  Ib.  of  Water,  515,  533 

Toughened  Cast  Iron,  284 

Glass,  280 

Tower,  Mr.  Beauchamp,  Experiments  of, 

73,81 
Train,  Moving,  44 

Epicyclic,  35 

Speed  of,  12 

Tran i car,  Pull  on,  44 

Work  Done  on,  38 

Trammels,  Elliptic,  577 
Translation,  Motion  of,  121 


The  References  are  to  Pages. 

Transmission  Dynamometers,  92,  ! 

of  Power  by  Chains,  233 

of  Power  by  Shafts,  224 

Travelling  Loads,  436 
Tresca,  M.,  Experiments  by,  301 
Triangle  of  Forces,  56 
Triangular  Orifice,  534 
Trochoidal  Curves,  573 
Tube,  Built  up,  323 

Strengthening  of,  324 

Tungsten,  287 
Turbines,  515,  526 

Arranging  of,  530 

Axial  Flow,  531 

Efficiency  of,  43 

Guide  Blades,  529 

Hydraulic  Efficiency  of,  529 

Inward  Radial  Flow,  531 

Outward  Radial  Flow,  531 

Regulation  of,  529 

Steam,  531 

Tweddell's  Hydraulic  Tools,  199 
Twist,  Angle  of,  349-352 

and  Shear,  332 

Twisting,  349 

and  Bending  in  Shafts,  353 

in  Wire,  349 

Moment,  350,  352,  353 

of  Prism,  369 


Unclosed  Polygon,  124,  129 
Unit  of  Heat  Energy,  142 
Useful  Work,  39 
U-Tube,  Motion  in,  553 


Valve  Motions,  584 
Vanes,  Radial,  521 

of  Centrifugal  Pump,  508 

Vector,  29,  123 

Clinure  of,  29 

Resolved  Part  of,  29 

Sense  of,  29 

Vectors,  Difference  of,  30 

Sum  of,  80 

Vehicle,  Pulling  Force  on,  108 
Velocity,  12,  14 

Angular,  25 

at  Any  Instant,  12 

Average,  12 

Critical,  83 

Ratio,  59,  103,  220 

Relative,  32,  34,  35 

of  Rubbing,  74 

Venant,  St.,  296,  318,  348,  366,  370,  379 

Vertical  Line,  22 

Vibrating  Adjustable  Masses,  561 

Weight,  41 

Vibration,  Amplitude  of,  565 
Damping  of,  566 


G88 


APPLIED    MECHANICS. 


The  References  are  to  Pages. 

Vibration  Indicator,  590 

of  Heavy  Disc,  568 

of  Spiral  Spring,  550 

of  Strip  of  Steel,  552 

Stilling  of,  5(55 

Viscosities,  Relativ"  569 
Viscosity,  41 

Co-efficient  cf.  76 

Volts,  92 

Volumes  of  Solids,  88 

Voussoirs,  163,  165 


w 

Waste  of  Hydraulic  Power,  53 
Watchmaker's  Drill,  291 
Water  Flowing  Spirally,  539 

Lake  of,  512 

Pressure  Engine,  189,  191 

Pressure  Machinery,  83 

Pressure  of,  167 

Waterfall,  Energy  of,  59 

"  Waterwitch  "  Steamship,  486 

Watt's  Parallel  Motion,  69 

Sun  and  Planet  Motion,  36 

Wave  Propagation,  367 
Weighbridge,  117 
— —  Graduation  of  Lever  of,  118 
Weights  of  Materials,  9 
Wertheim's  Experiments,  302 
Wheels  and  Axle,  101 

and  Axle,  Compound,  102 

Friction,  69 

Teeth  of,  423 

Teeth,  Shapes  of,  221 

Value  of  Train  of,  102 

Worm,  102 

Whirling  Fluids,  216 

Loaded  Shaft,  476 

Whirlpools,  518 

Chamber,  Thomson's,  522 

White  Fir,  277 

Metal,  75,  289 

Whitworth  Thread,  100 


The  References  are  to  Paget. 

Willans'  Engine,  611 
Wind  Pressure,  156 
Windmills,  101 
Wire,  Area  of  Section,  10 

Copper,  289 

Drawing  Through  Die,  293 

Experiment  on,  293,  300 

Gun,  325 

Pianoforte,  289 

Ropes,  232 

Telegraph,  168 

Telephone,  289 

Wohler's  Experiments,  305,  30V 
Wooden  Pipe,  183 
Work,  38 

by  Expanding  Fluid,  19 

Cutting  of  Metals,  46 

Done  by  a  Moment,  115 

Done  by  Turbine,  521 

Law  of,  58 

Lost  in  Friction,  67 

Principle  of,  120 

Useful,  39 

.Workshop,  1 

Worm  and  Worm  Wheel,  102,  221 

Thread,  574 

Wrought  Iron,  284 

Case-hardening  of,  285 

Charcoal,  284 

Forging  of,  284 

Lowmoor,  284 

Red-short,  284 

Staffordshire,  284 

Weight  of,  9 


field  Point,  300 

Young's  Modulus,  299,  305 


Zinc,  287,  238 


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